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Writing about the mistakes on the web and getting "stuck into" faults in text books, makes you wonder what should be covered in a 3 year course on electronics.
If I had my way, the course would start 10 years before, with the encouragement to build at least one project a week and get it to work.
I have proof from thousands of readers who have built the kits from Talking Electronics Magazine, that electronics can be learnt by CONSTRUCTION.
First of all, it is a much-faster way to learn and if you account the costs of a course, it is CHEAPER.
I have talked to students after a 3 year course who "Don't do soldering."
I have University Professors who say you can achieve 3v on the base-emitter junction of a transistor or who cannot draw the simplest circuit from memory. Or cannot properly explain how a transistor amplifies.
I have seen graduates of an electronics course who have hardly constructed a single electronics project.
If you look at most of the courses, they consist of stale, uninteresting, irrelevant material such as making an XOR gate out of other gates or half-adders.
Show me one project or product using this and I will give you $10,000. Any product needing timing or logic gates will use a microcontroller. A flashing LED uses a microcontroller. A whistle-keyfinder uses a micro and these cost 10cents in a $2.00 product. They use a microcontroller because it is cheaper than using discrete components.

You can buy a fully pre-programmed PIC chip "die" for 6 cents. Obviously this is a price (within China) for a special MCV08A (which is a much-better version of the PIC12F629) that has been developed especially for China. China has already ordered over 30 BILLION PIC chips - mainly in the form of a COB for all sorts of devices, including toys and this technology/pricing has not been released to the external market. We are still paying exorbitant prices for a PIC chip (33 cents to 80 cents),

These courses need to update their content and concentrate on modern materials, chips and circuitry that will be needed for present-day designs.
It should revolve around small, medium and large microcontrollers.
Then cover building blocks to interface to these devices.
Then cover chips that perform specific functions.
With this information, the student can emerge with the knowledge to design much of the equipment for auto, health, domestic and recreational use.
In these days of high expenses for each week of study, it is pointless covering topics that "bit the dust" 10 years ago.
But this is all the lecturer knows. That's why he keeps re-gurgitating it to the class.  One of my friends is teaching SOLAR INSTALLATION and none of the other lecturers know anything about the subject.   

YES.  I am saying the complete opposite of a University course is the ANSWER.
It's a pity mere practical skills are not recognised. The paper degree is the almighty power.
That's why you have to bow to the powers of the recruiting officer and follow both paths, Knowing your ability comes from experimentation; and access to the "job Market" comes from your "CV."
You have to feel competent and capable within yourself and this comes from actual construction and testing.
This way you won't make a fool of yourself and undertake a project way beyond your capability.    

There's an old saying:
If you can:  DO 
If you can't: TEACH

The whole concept of Talking Electronics website is to explain electronics to electronics enthusiasts who want to learn electronics but do not want the mathematics.
It would be wonderful if all electronics enthusiasts understood complex mathematical notation, but this isn't the case.
The two understandings are diametrically opposite to being successful and it's a bit like expecting all stunt-car drivers to be able to do tricks on a unicycle.
Now you can see what I mean.
The wonderful part of electronics allows you to build a circuit and take "real-time" quantities and CRO observations.
These far-outweigh any hypothetical, calculated or simulated results.
The biggest problem with any new design is getting the circuit to work.
No simulation-software or text-book is going to help you.
That's why all the year's spent on the mathematical approach is going to get you down.
The only way to learn electronics is from the bottom-up.  Not from the top-down.
I know this is a radical approach, but you should learn the mathematics AFTER you have built and studied hundreds of circuits. The mathematics only makes sense afterwards.
Do you know why this is not done?
Because if you build a circuit and it does not work, 99% of the instructors will not be able to diagnose the fault.
Talk is cheap. Anyone can stand in front of a blackboard and churn out notes. The real skill comes from diagnosing a circuit and pinpointing the fault.
I am personally against trying to work out the operating point of a circuit and current values because all the values used in the equations are "GUESSED."
You have absolutely no idea of the current-gain of the transistor you are using as any batch has a range that will either be double or half the guessed-values.
Why spend time working things out mathematically when the circuit will have to be constructed and final values determined after seeing the results on a CRO.
But the biggest absurdity of most courses is the lack of basic content. The lessons cover circuit analysis without covering any of the details of circuit-design.
If you look at our eBook: The transistor Amplifier you will see over 100 circuits on designing a simple transistor amplifier.
You cannot possibly go into analysing a circuit before learning how to design a circuit and build at least a dozen or more circuits to see how they perform.
That's why University graduates emerge from a 3-year course without ever touching a soldering iron.
One of the most important topics is INTERFACING.  Connecting a microcontroller to the outside world. Or interfacing ANYTHING.
Normally this involves BUFFERING the output to drive a high-current load but it can also involve an input stage to amplify a low voltage transducer.
There are also a number of other BUILDING BLOCKS that use transistors: such as oscillators, constant-current circuits, filters, etc and covering these is much more important than analysing an amplifier with lots of mathematics.
These courses still have a long way to go and there are hundreds of "tricks" to designing that I have covered in the eBooks on this website.  None of these have ever been mentioned in any class notes.
Anyone finishing a course will be be making a complete fool of themselves by falling into simple traps that apprentices have already discovered.
That's why this website gets 7,000 visitors each day. They keep coming back for more.
That's why you are reading this topic.
You will learn MORE in this topic than in anything anywhere else.
You learn more from other peoples mistakes - and also from your own mistakes.
And that's the factor that is missing from all the text books. They never tell you what will happen if you use higher or lower values or what will happen if a component is missing or goes faulty.
That's why teachers have no experience in fault-finding a students work.
That's why they don't promote the idea of building things !!
Here's an example of what I am talking about:

I've bought an industrial pedestal fan but even its lowest speed is far too windy and noisy for my use.
The voltage is 240, at 50Hz. The fan is 200W. How can I slow it down?

For a 200 watt fan, full load current = 200/240 = 0.83 Amps.
To drop 100 volts at 0.83 amps, you need a capacitive reactance of 100/0.83 = 120R.
The reactance = 1/(2 x Pi X freq X C) so
C = 1/ (2 X 3.14 X 50 X 120)
C = 26.54 uF

The answer above has a major mistake. The current will not be 0.83 amps when the capacitor is included. It will be 0.42 amps.
A 100n capacitor will pass 7mA at 240v and only 3.5mA at 120v.
To pass 0.42 amps will require 120uF.

Using the correct figures for the equation above:   

For a 200 Watt fan, full load current = 200/240 = 0.83 Amps.
To drop 120 volts at 0.42 amps, you need a capacitive reactance of 120/0.42 = 285R.
The reactance = 1/(2 x Pi X freq X C) so
C = 1/ (2 X 3.14 X 50 X 285)
C = 112 uF

If the respondent to the question had built a capacitor-fed supply, he would not make that simple mistake.
That's why construction is so important.
You will notice I have provided an alternative approach to the MATHEMATICAL APPROACH. It's very important to approach a problem in two different ways so you can back-up your answer.
This is the point I am trying to get across.
I am providing a second-approach to everything you are doing, so you don't make a fool of yourself.

Here's another incorrect response to a question in an electronics forum:

Is there any device or component that can clamp 200V to 25V for 200ms without failure?

Loading the output WILL generate heat and 1/4W resistors just wont cut it.. you will need to get 20W or higher resistors..
Quick calculations..
1K resistor = 40W power dissipation = 35W in the resistor & 5W in the Zener( Max according to its Datasheet)
2K resistor = 20W power dissipation = 17.5W in the resistor & 2.5W in the Zener (bit better, with room to breath) However at 1mA measured load (25V supply) the Vdrop in the Resistor is 2V giving you 23V output.. (and at 10mA it's 20V, giving you 5V output) ...

The above is absolute RUBBISH.

You just need a 24v 400mW zener diode and a current-limiting resistor.  The zener will breakdown when a minimum current flows and as the voltage rises from 0v, a low voltage will appear across the diode. As the voltage rises, this voltage will appear at the junction of the diode (zener junction) and it will not rise above 24v. There is a certain amount of leakage across the junction and the voltage will not rise to 24v until a few milliamps flows.
The value of the current-limiting resistor is worked out by knowing the minimum current required by the zener and
the data sheet specifies this as 5mA.
This means the current-limiting resistor will have 200 - 25 = 175v across it.
We do not know the the source of the 200v and some assumptions have been made when making the following calculations:
If we want a 1 watt current-limiting resistor, it will need to be: 1 = (175 x 175)/R = 30,625   use 33k resistor
We don't know if the voltage is AC or DC but about 1 watt or 2 watt will be sufficient.
Certainly NOT 20 watt to 40 watt !!

The original poster has now stated the source of the 200v spike is coming from the coil in a coin comparator and we can state the current and waveform will be so small and short that almost no energy will be delivered. This means a 0.25 watt resistor will be suitable.
Furthermore, it is not recommended to put a zener directly across the coil as this will inhibit the operation of the coil and prevent the received signal providing the correct information.
Also we have not been told if the 200v signal is produced by the coil when it is collapsing or if it is a fault in the energising circuit.
You cannot simply put a zener across the coil to limit the flyback (in the same way as a diode is placed across a coil) because a zener is just like an ordinary diode but with a PIV of a very low value.
If you use a zener, it will have voltage of 0.6v in one direction and 24v in the other.
This is simply not going to work.

Here is a 240v LED Light:
The problem is only half the LEDs are illuminated at a time. This means you are using twice as many LEDs to get the same brightness  !!!!


The current through the 220R will be:  7v/220 = 31mA
There are 9 LEDs.  Each LED will get 3mA  !!     Not very bright  !!

You cannot connect a regulator to a capacitor-fed power supply. (it is pointless - "waste of time")

Here's the reason:
The output of the 78L12 will be 12v.  The current through the 470R will be 12v - 4v = 8v.   = 8/470 =  17mA
The 78L12 takes about 3mA  so the input current to the regulator will be 20mA.
The output of the power supply will be 7mA for each 100n of C1 =  33mA   The voltage on the input of the regulator will rise until it takes 33mA.
The maximum input voltage for the regulator is 35v.
The voltage will keep rising until the regulator takes 33mA and when the voltage reaches 35v, the regulator component will explode.
You have to understand how a capacitor-fed power supply works. It is not like a normal power supply.
We have covered this in the 30 LED Projects section.

Here's another junk circuit from ELECTRONICSHUB.ORG

The LUMILED's drop about 3.6v each.  3.6v x 5 = 18v.  The supply is only 12v  !!!!!!
The circuit has never been tested  !!!!

Here's another junk circuit from ELECTRONICSHUB.ORG

The circuit consumes 10mA when sitting around doing nothing.
The switch should have been on pin 8 and 4.

Here's a project from Makers Shed.
Makers Shed is supposed to assist you in building things. But this project, along with the other projects they sell, has no circuit diagram and a very high price-tag. The kit costs $17.00 and the postage is $21.00.
Makers Shed states: New Fun Kits from Technology Will Save Us!  I don't think $38.00 will save us !!

Here is the circuit diagram, generated from the pictures of the circuit board:

This is a really STUPID circuit:

More current flows when the LED is not illuminated because the transistor is diverting the current and WASTING It.
When light falls on the LDR, the transistor is turned ON and current flows through the collector-emitter of the transistor and turns OFF the LED.
The following circuit takes less than 1mA when not illuminated and the 10k resistor can be increased so the circuit takes less current. 

Iím trying to get real time data off of a rectifier by collecting it on a microcontroller and passing it to a computer. I am using a voltage divider to get voltage (A0 PIN) and the drop across the shunt resistor for current (A1 PIN). The problem is that when I apply rectifier I only get signal of the A1 PIN and no signal on the A0 Pin.

The poster on an electronics website is trying to produce a circuit that will monitor the voltage across a load and also the current.

This is the circuit required:

Here is answer from STEVE LAWSON:
Come to think of it, Colinís circuit doesnít make sense unless the voltage divider is also the load ó otherwise what current are you measuring? Or to put it differently, why are we interested in the current flowing through the voltage divideró itís merely for measurement purposes.
Steve Lawson doesn't understand the concept AT ALL.
The LOAD is placed on the output terminals and the shunt resistor monitors the current while the voltage divider resistors reduces the supply to a maximum of 5v so it can be fed into a microcontroller.

His further reply produced more lack of understanding:

This is a terrible circuit.
There is no reference point. The reference point is normally the 0v rail.
The voltage divider resistors are around the wrong way. You normally pick off a small percentage of the supply voltage.

I clicked on the banner above on an electronics website and found it went to a CYBER SQUATTING WEBSITE that wanted to sell the name for $2,000 !!!!
You can buy  website for $15.00 from Godaddy for 2-years.
There is nothing I hate more than cyber squatters who demand outrageous prices for websites.
Fortunately a website name is not important as most visitors come to a site via a Google search.
It has put most of the cyber squatters out of business as they were charging up to $30,000 for names they thought would be essential for running a business.
The most USELESS, SILLIEST invention on the web turned out to be the most important factor in bringing it together: GOOGLE.
Without Google, the web would have been a total flop and ripe for fraudsters.

Here's a promotion from a PCB manufacturer:

Up to 64 sq. inches total (1625mm) PCB for $120.00

64 square inches is an area.  1625mm is a LINE with NO DIMENSIONS.

64 square inches is equal to  64 x 25.4 x 25.4 = 41,290 square mm  (mm2). or if it is say 8 inches x 8 inches or 20.32cm x 20.32cm or 412 square cm (cm2).

If you can keep your PCB to within 100mm x 100mm, the cost of producing 10 boards is $20.00 including postage from:    This includes double sided boards, 1.6mm green, HASL and only a single routing around the board. If you have two PCBs on the 100mm x 100mm, the item is called a PANEL and you have to allow 70 thou between boards so you can cut them apart yourself with a hax-saw and linish the edges with sandpaper. 
Both David and Martin from Elecrow provide assembly for your boards at amazingly low prices. 3 cents for each pad of a surface mount device
plus $10.00 for set-up and $20.00 for a glue-stencil for surface-mounting. They will supply components for a minimum of 100 boards and this is quicker, neater and easier than trying to assemble the boards yourself.
Obviously you get a sample of 10 boards yourself and prove the circuit works. But after that, the production is left to Elecrow for an absolutely professional result.
Send a photo of your final project to us for inclusion in this article.
We are getting about 3 boards a week from Elecrow, as one idea generates another and you never run out of ideas.

Here we have the Indian magazine AGAIN, producing the most dangerous circuit on the market.
The neutral lead is connected directly to the microphone.
there will be very little insulation inside the microphone and if the mains leads are connected around the other way, the microphone will be 340v LIVE. Yes, the shock will be 340v. And if you are holding the microphone firmly, the shock will kill you.
This type of circuit has been banned from western magazines for over 10 years. It's about time the Indians stopped producing any type of capacitor-fed supplies as we have already covered their dangers. 
All the circuit does is turn on a LED. It only needs a few parts to do this. NOT a circuit connected to the mains.

This is one point that has never been mentioned before.
You cannot put 100mA through a mini trim-pot. When the pot is turned fully clockwise, the resistance is very low and about 100mA will flow to the two IR LEDs. The contact of the wiper on the track will not reliably pass 100mA.
It will burn a spot on the track and go open-circuit.
This is a point worth remembering.
The solution is to just use the 56R or decide on what level of current you need and use a fixed resistor.
Most IR controllers work to 10 metres.  How far from the TV are you?
This circuit just needs 2 transistors and few components - not an op-amp, but that's another story.

You only need 1 x 470R as only one output is high at a time.
A very poor design. No pin numbers, no component values and no understanding of how to lay-out a circuit.
A circuit is NOT a diagram to correspond with pin numbering.
It is a diagram with the least number of "jumpers" - to make the circuit as easy as possible to understand.
People without this understanding should not be presenting circuits on the web.
You should be able to see the output of the 555 connects to the clock-input of the 4017 and each output of the 4017 goes incrementally to the set of LEDs. There should be almost NO lines that cross. That's the skill of presenting a simple circuit. The layout is just a JUMBLE.

This circuit could be very clever or a complete waste of time. But looking at the design, the 4017 outputs Q0 when turned ON and thus the 1k current-limit resistor on the output LEDs has no voltage across it and the voltage divider into the clock of the chip is 1k:100k and thus the clock-line will not go HIGH and the 4017 will not clock.
This circuit is just a MESS and you should NEVER draw chips a block but rather as gates and the 4017 should just have all the outputs on one side to make it easy to see what is happening.
Also, the circuit is too large to fit on the screen and this makes it very difficult to see the whole circuit. Here is the circuit. Click on the circuit for FULL SIZE.

Click for FULL SIZE

How much volume will you get out of a BUZZER with 33k in series ????
It only needs one or two mistakes and a circuit will NOT work. The author is not locatable to solve these problems so I suggest the circuit should be avoided.
It is debatable if 4 transistors are needed on the front-end. The fourth transistor is just an emitter-follower and this type of stage is needed when the output needs to drive into a low-impedance load. The input of a chip is very high impedance this stage can be removed.
There may be a lot of other unnecessary components but until you draw the chip as a set of gates, it is impossible to work out what is happening.
That's why I say the author of this circuit has absolutely no comprehension of electronics.
The whole idea of electronics is to make a complex circuit "look simple."
That's how you get others to understands it.
That's the skill of presenting a project.
That's why I don't use any complex terminology. It only steers reader away.
So many text books turn a simple design into a Boolean Expression  - it's no-wonder they lose their readers with frustration.
The same with the circuit above. If it is simplified and drawn to fit the page, the reader can see what is happening. At the moment, even
I cannot see what the output will produce. A circuit has to tell you all these things without having to refer to the text.

Here's another over-designed circuit from ELECTRONICS FOR YOU July 2014:

The output of the microcontroller goes to a row of LEDs and 100R resistors. The current will be over 32mA for a chip that has a maximum of 25mA per drive-line.
What is the point of including the ULN2003 chip when the current for each segment is about 10mA?  The chip is capable of supplying up to 25mA per segment.
Why use a crystal?   Who needs accurate timing? 
The 7812 regulator is not needed.
The transformer should be 0-9v and the power LED with 680R will be very dull.
Just a badly designed circuit form Dr D.K. Kaushik and Ashok Sharma who know very little about circuit-designing. Dr D.K. Kaushik is principal and Ashok Sharma is technical assistant at Manohar Memorial (P.G.)  College, Fatehabad (Haryana).

Here is another FAILURE from ELECTRONICS FOR YOU July 2014:

Only the MCP73831 has totem-pole FETs on the
STATUS pin to drive LED1 and LED2.

Here is the trackwork around the BATTERY MANAGEMENT CHIP

When the board is turned over, the chip
is soldered as shown above.

There is no such MOSFET as IRFZ820A. It is IRF820A.  But when you look at the specifications, it is the wrong MOSFET to use.
The minimum ON resistance between Drain and Source is
one ohm. If the current through the LED is 300mA, the voltage drop will be 300mV. The battery is 3,700mV (3.7v) and a white LED has a characteristic voltage drop across it of 3.6v.
The LED will not take its full rated current and the MOSFET is not required.
The MOSFET should be something like IRF20 with an on-resistance of less than 0.1 ohm.
It's just another UNTRIED project from someone who does not know sufficient about electronics to put a project into an electronics magazine. 
You can buy the chip for 80 cents at an electronics supplier (plus $10.00 shipping) or up to 20 chips on eBay for $4.00 including shipping !!  That's why you cannot beat eBay !!

Here is another circuit from ELECTRONICS FOR YOU July 2014, with mistakes:

This is a copy of my circuit, which I designed over 20 years ago.
I can see some of the features I designed that are different to any other circuit.
But the circuit has two major faults. The 22p should be 22n and the inductor L2 has no effect as the main amplitude of the signal appears at the collector of the output transistor and the inductor is designed to keep the output tank circuit (L1 and C6) from the power rail to produce a very high output. In the circuit above, the inductor is simply reducing the output to the line.
Obviously the author has absolutely no idea what he is designing and it's just a "Copy and Show" project.
The other major fault is connecting the project across (parallel to) the telephone line.
Although the line is about 8v to 15v when the handset is lifted, this voltage will be pulled down even further when the circuit is connected in parallel.
The "bug" should be connected in SERIES as the phone line has enough voltage (50v supply) to allow the phone and bug to be connected in series.
The voltage and current of a phone line is complex, because the 50v supply is classified as a HIGH IMPEDANCE SOURCE.
In other words it is a 50v supply with a low-current capability.
The 50v supply has a relay at the exchange with a coil resistance of 1k. It is a special type of relay called a slugged relay. It is activated quickly but does not release quickly. This allows the line to be "held" when the decadic pulses are sent down the line. This is the "old type" of dialing with the "turn-the-dial" or "rotary dialler" that opened the line ten times per second to dial the phone number.
This relay does two things. It limits the current to 50mA and activates when the current is more than 10mA.
Another relay (called a step-by-step) detects these pulses and produces a result of the first two numbers. In other words it selects a line after a choice of one hundred possibilities.
This is all old technology but the exchange still delivers the same impedance to your phone and the short-circuit current is up to 50mA and the operating current is about 20mA - 30mA.
Putting the "bug" in parallel is an absolute disaster.
It " hogs the line" and does not allow the phone to "hang-up."
Unless you have studied the phone line and studied FM circuits, you are making a FOOL or yourself, when presenting a project.
The author has absolutely no idea about series and parallel connection on a complex set-up such as a phone line and no idea what he is doing with FM transmitters. 

Here is another circuit from ELECTRONICS FOR YOU July 2014:

The circuit drives a 5v relay from 2.5v. 
Where can you get a 2.5v supply ?????
Instead of messing around with a complex circuit, you can get a 1.2v solid state relay for $4.00.
The relay requires about 20mA to "turn ON" so you will need a resistor in series with the input pin to limit the current to 20mA. The two input pins are marked positive and 0v because they connect DIRECTLY to a LED. The input current illuminates the LED and activates a light-detecting TRIAC. The characteristic voltage-drop across the LED is between 1.2v and when the current reaches 50mA, the voltage-drop is 1.4v. This means you MUST include a current-limiting resistor to limit the current to between 20mA and 50mA. The relay will turn on at 8mA, but you should test it before using this low value. This will allow you to use a supply voltage form 1.2v to 5v or more.
The MAX4624 is a 16-pin surface mount chip.
The max4625 is 8-pin surface mount and costs $4.60 plus postage.
Just another disastrous circuit from EFY. 

To Sani Theo,
Technical Editor,
Electronics For You

Yes, of course I am annoyed.

All the Australian and English magazines publish any corrections and mistakes to circuits BUT you fail to inform your readers of the DISASTROUS mistakes.

You don't even have the intelligence to pass the mistakes to the authors of these articles and get a response.

Thank goodness you have stopped Professor Mohan Kumar supplying projects.

His projects were riddled with mistakes and I have finally got him banned from supplying articles to electronics forums.

You have got to improve your editing of projects and get an EXPERT to analyse them before publication.
I have monitored the building of the projects in EFY for the past 12 months and found the construction to be virtually ZERO.
Everyone I contacted found the projects DO NOT WORK.
And it is NO WONDER.

Who do you think you are trying to FOOL ??

No-one in the Western world takes any of the projects seriously and this is borne out by the fact that so few subscribe to the magazine.

Unless you improve the quality of the simple projects, you will NEVER get any respect.

I have not even bothered to look at the more-complex projects because they mostly involve a microcontroller I do not purchase.
But if the simple projects are not constructed, I suspect the more advanced projects are avoided too.

Your Kit-N-Spares section is also a disaster with projects lacking an overlay and hand-written scribble on the PC board.
Just take a look at the boards and see what an insult they represent to the electronics world.
Lastly, look at the postage. You want $50 USD to post a $5.00 kit ?? China posts things for FREE !!
You (India)  say(s) you want to present as an advancing technical (technological) market and have all the technical expertise to take-on world-wide challenges and yet the technical contributors to the magazine show little aptitude to quality designs and no-one has challenged the faulty circuits.
You are fooling yourself with the technical expertise in the country and failing to deliver a quality product to the electronics experimenter.  

I am trying to protect the uninitiated beginner who looks at your faulty design s and thinks they are the correct way to design a circuit.
I get hundreds of emails from Indian readers who visit my site and ask for guidance. They are stagnated by the terrible lack of technical material emanating from Indian publishing houses.
It is fortunate the internet is correcting this.     

I have been sending you emails for the past 12 months on these subjects and you have not replied. You have not updated the circuit boards, ANY of the projects (nothing has been added for the past 12 months) and you have not changed any of the images. Neither have you looked into postage costs for the kits.

What more can I say ???

Colin Mitchell

This project has been on the web for 3 years and no-one has realised it DOES NOT WORK:

The LEDs are exactly like a 3.6v zener across the battery. They will take a lot of current if a current-limiting resistor is not included.

Problem number two:
The 6v solar panel needs to develop 5.5v to overcome the back-voltage developed by the battery when charging. The LM317T needs 1.5v across it and if the charging current is 100mA, the 8R2 develops 820mV across it. This is a total of 7.8v. The 6v panel will not develop 7.8v.  

The LM317 is in constant-current mode and will limit the current to 140mA but at 100mA the solar panel cannot produce enough voltage to provide any charging current. The whole circuit is a MESS.

This is another complete MESS from D Mohankumar.


The solar panel is 24v.
The 12v zener will limit the output voltage to 12v + 1.2v =13.2v  How can you charge a 12.6v battery with 13.2v ??
Another untried, untested non-working circuit from PROFESSOR Mohan Kumar.

Battery Charger
This is one of the worst battery charger circuits I have seen.
To start with, the PNP transistor, 5v6 and string of diodes puts a zener voltage of 0.6 + 5v6 + 3v on the supply rail and if the battery is removed and the supply rises above 9.2v the transistor will be DESTROYED  !!
The whole circuit is equal to a current-limiting resistor of unknown value and the circuit will effectively deliver all the current from the solar panel to the battery.
If the solar panel can deliver 20 amps, this current will flow into the battery and BLOW IT UP.
Let's see what the circuit does:
As soon as the supply reaches 9.2v, the PNP transistor turns ON and this turns ON the emitter-follower transistor Q2 and raises the gate voltage on the MOSFET.
The supply only has to rise a few more millivolts and the PNP transistor is fully turned ON and the MOSFET will be fully turned ON. So the difference between the MOSFET being not-turned-ON and fully turned ON is only a few millivolts change in the voltage on the supply rail.
Now let's look at the MOSFET.
When (if) it turns on fully, the battery is connected directly to the solar panel and the supply rail is 4.8v plus the "floating voltage" produced by the battery when it is charging.
This means the battery voltage can be as high at 5.5v.
This is not high enough to get the circuit to turn on, so the MOSFET allows the difference between 5.5v and 9.2v to be dropped across it.
This is purely a VOLTAGE DROP and you can consider the voltage drop to be identical to a zener and the MOSFET will not limit the current.
We do not know the voltage of the solar panel or its current capability and this will be the limiting factor of the circuit.
But as soon as you remove the battery, the transistor will blow up.
The first transistor only starts to come into operation when the supply reaches 5v6 + 5v5 + 0.2 + 0.6v = 12v and it is clear that the supply can NEVER go above 9.2v without blowing up the circuit.
Thus the whole circuit can be replaced with a current limiting resistor and the battery will not be destroyed.


Thanks for these articles on: "So-called experts with all the fancy degrees!"
Iím 72 years of age. Iíve met so many of them in my working lifetime, they were the bane of my life!


The problem with this circuit is the 10M resistor.
It is only allowing 1 microamp to flow into the base of the transistor and this is causing two problems.
1 microamp is not enough to maintain the polarisation of the electrolytic and it will lose its capacity over a period of time.
Because the electrolytic is not maintaining it polarisation, it will become leaky and the leakage current will cause 1 microamp to flow AT ALL TIMES and the circuit will never turn off.
In fact the transistor is making the circuit between 100 and 300 times WORSE by requiring the 47u to be charged via a 10M resistor.
Without the transistor, the electrolytic could be charged via a 470k to 1M and the timing would be 10 times more reliable.
The circuit has a SCHMITT TRIGGER action.
The uncharged capacitor turns on the transistor and the gate voltage is LOW. The MOSFET is not turned on.
As the 47u charges, the base current falls and the transistor begins to turn OFF. The gate voltage rises and and the MOSFET begins to turn ON. This reduces the effective "turn-ON" voltage on the gate and the transistor continues to all the gate voltage to rise. The MOSFET now starts to draw current and the voltage on the top of the MOSFET starts to drop and this reduces the current through the 10M resistor. This turns the transistor off slightly and the voltage on the gate rises at a faster rate.
This action circulates around the two components and very soon we have a condition where the transistor is fully turned OFF and the MOSFET is fully turned ON.
This action has not been due to the electrolytic gradually charging but the voltage on the top of the 10M reducing from 12v to about 4v. 
This action may over-ride the poor design with the 10M and by using a tantalum or low-leakage electrolytic, the circuit may function.
However, the point still exists, not to use a 10M resistor to charge an electrolytic.  

Here we have the case of a 3v3 microcontroller turning on a LED that is connected to a 5v rail. 

Many responders on a forum said this cannot be done and should not be done. Others said it will damage the microcontroller and other absurdities.
The fact is this: It is quite acceptable to control the LED as shown. The LED will have a characteristic voltage of about 1.7v to 3.6v across it before any current flows and this means the voltage across it when the 5v rail is delivering 5v and the output of the microcontroller is about 3.2v, will be 1.8v. A red red will begin to turn ON and so a green or orange LED can be used. A white LED will be ideal.
The LED will indicate two things at the same time. When it is illuminated, it will indicate 5v rail is alive and the 3v3 microcontroller is operating.


Here is a simple circuit to test if a capacitor has "dried out" or lost some of its ability to deliver a high current.
ESR means Equivalent Series Resistance and most capacitors appear with a Equivalent Series Resistance of a
a fraction of an ohm or a few milliohm. 
In general, this value is not important and you simply compare the old electro with a new electro and see if the reading is the same.
The circuit is a very good design and works well.
It was placed on an electronics forum and two moderators/responders, Chuckey and Audioguru (with thousands of postings), gave this reply:
The 555 oscillator is feeding positive and negative AC to the capacitor being tested. Reverse polarity is bad for a polarized capacitor.

I replied:
Who said the capacitor is seeing a "reverse voltage" across its terminals?

The 555 has a 22n series output capacitor that has an output to the capacitor under test that swings positive THEN NEGATIVE then positive THEN NEGATIVE again over and over.
So the capacitor under test gets reverse polarity half the time unless its value is MUCH higher than 22n.

This is RUBBISH. It is amazing how a person who has been in electronics for over 30 years and replied to more than 34,000 enquiries can make such a false statement.
The problem is he cannot "see" how the circuit works.
The energy supplied to the electrolytic during the charging cycle will be removed during the discharge cycle and thus the positive lead of the electrolytic will simply increase to a small voltage and decrease to ZERO.
Here is an animation of what is happening:

Can you see any "negative" voltage entering the electro????

If you cannot see a circuit working, you have no possibility of testing, designing and repairing it.
Simply connect the circuit your are investigating to a CRO and reduce the timing to a low frequency and see what is happening - BEFORE making a stupid statement on a website.
One of the amazing features of a capacitor is this: It will convert a low-current high-voltage into a high-current low-voltage. That is exactly what is happening here.
The result of a low-current high-voltage is called ENERGY and it requires a lot of current to increase the voltage on the electro for each volt produced on the leads.
That's why the small capacitor produces a high voltage across it during the charging cycle.
The two responders are getting mixed up with a circuit consisting of a capacitor in series with a resistor. Or a capacitor in series with a diode. When the output of the 555 goes LOW with these two components connected, the voltage on the lead of the capacitor DOES fall below the 0v rail.
But that's why you have to LEARN ELECTRONICS.
You cannot afford to make a mistake like this and broadcast your ignorance to everyone else.
This is only one of many mistakes made by the posters and that's why I have decided to monitor the forums and correct their glaring mistakes.
The most disturbing part is their inability to realise their mistake after reading my corrections. Hopefully others will learn.

Here's another mistake from  Audioguru

"A 12V lead-acid battery that measures 12.5V with no load is almost dead."

This is totally FALSE.
A 12v lead acid battery should read 12.6v after it has been charged.
During the charging process it will generate a "floating charge" or "floating voltage" up to 15v due to the resistance of the bubbles of gas and also the type of material used in the plates. This voltage is the reason why they call some batteries "maintenance free" or "sealed" because they do not produce gas bubbles until a high voltage is developed across each cell. If you keep the charging voltage below this voltage, the battery will not produce gas and can be sealed.
You cannot measure the battery voltage immediately after charging due to the "floating voltage" it shows on a voltmeter. You need to wait 10 minutes and then apply a load for a short time to settle the activity within the cells.


Here's an example of how NOT to draw a circuit:

The circuit may work but it is impossible to see how it is working. Do do not know if the blinking is coming from the inputs or if it is generated by the components. The gates are SCHMITT TRIGGERS and this symbol is missing from the diagram.

The basic oscillator is shown in the diagram below.


The whole diagram should show the 6 oscillator circuits so you can see if it is gates with a low signal or a HIGH signal and see what is happening with ALL the gates. I have absolutely NO IDEA how some of the gates are connected and a properly laid out circuit diagram will provide these details.
This is important if you want to modify the circuit or locate a fault when the circuit does not work.
The diagram above is absolutely USELESS. It is just a JUMBLE. It is partially a layout diagram showing where the components are placed or how they are connected to the chip.
That's NOT the purpose of a circuit diagram.
It is entirely designed to show HOW THE CIRCUIT WORKS.

556   How do deal with a 556 circuit
It's hard enough to work out the operation of a 555 circuit. It's IMPOSSIBLE to work out a 556 circuit.

Rather than wasting time learning about the 556 chip, it is much easier to explode the 556 into two 555 circuits:

The circuit is now easy to see. All the connections are in the same places as when using a 555. Do not be confused with pins 2 and 6 on a 556.  The lower pin (on the circuit) detects the LOW voltage and the upper pin detects the HIGH voltage. The numbers of these pins are swapped when using a 555 / 556 and this is very confusing.
What is the purpose of complicating things with a dual 555 when a simple 555 will provide the required mark-space ratio.  There is NO SKILL in making a complex circuit, where a simple design has already been produced.



Here is a typical circuit posted on a forum. It simply DOES NOT WORK.
The circuit is supposed to detect the sound from a speaker to trigger a 555.
The output from a speaker is typically 20mV and may rise to 100mV when detecting a loud sound. The BC547 needs at least 650mV
Resistor R1 is not needed.
The pot will be damaged with turned to zero ohms and pin 7 goes LOW.
It's circuits like these that give the web a BAD NAME.
They are untried, untested and show the designer has absolutely no idea of electronics.


The circuit drives a cooling fan (12v 1.2w) and the negative temperature resistors are placed on hot items. These reduce in resistance when the items get hot and the fan increases in speed.
The fan takes 100mA when connected to the 12v supply.
The whole circuit is overdesigned and quite unnecessary, but lets look at the absurd design.
Firstly, what is the function of R8? The text says the voltage across this resistor will be up to 1.35v but when 100mA flows, the maximum voltage will be 100mV. !!!    Why specify 1watt resistor ???
What is the function of the 10 ohm resistors?  
The resistance (or impedance) of the fan is about 120 ohms. 
The NTC resistors will drop to 42 ohms when all are detecting 100įC. This makes the whole resistor network equal to about 10 ohms.
If you remove the 10 ohm resistors, the NTC resistors only have to detect an additional 10įC to achieve the same overall resistance. Or you can use 330R NTC resistors.
What is the function of diode D1 ??  What is it protecting ?
What is the function of LED2??  The fan will always be operating.

When a door opens, the transmitter sends a 40kHz signal to a receiver.
The circuit is over-designed and takes about 10mA when sitting around doing nothing. This is a waste of current the circuit needs a power supply.

The circuit can be simplified to:

There is NO SKILL in designing a complex circuit when a simple design has already been provided on the web.
The transistor circuit finds the natural frequency of the transducer because it contains a crystal that resonates at 40kHz. This saves providing any timing components and produces the highest amplitude due to resonance and the frequency is exactly 40kHz. This is the simplest and best design.

This circuit above is poorly designed and has a number of fundamental faults.
It will be very difficult to turn ON the second transistor to the point where the collector voltage is below 1.8v, but higher than about 1v, by adjusting the current into the base.
It needs to sit in this range for the circuit to have the best sensitivity.
If you turn the transistor ON too hard, the circuit will require a lot of energy from the 4n7 to remove this turn-on current and turn the transistor OFF.
Don't forget, the next two transistors rely on current form the 4k7 to produce an output. The transistor does not drive the output, the 4k7 delivers the energy to the diode pump. It is not really a diode pump because it is really pulsed DC and not AC driving the diode pump section.
The capacitor charges when the second transistor is turned OFF or nearly turned OFF.
The slightest change in temperature or supply voltage will alter gain of the second transistor and either turn it ON more to decrease the sensitivity of the circuit or turn it OFF more and cause the circuit to false-trigger.
The next point to note is the sensitivity of the diode pump has been reduced by the inclusion of the third transistor. This transistor is an emitter-follower and does NOT assist in the operation of the circuit. In fact it reduces the sensitivity of the circuit.
The diode pump is perfectly capable of turning ON the output transistor as it only requires only a very small current into the base (less than 0.5mA) and the 4k7 will deliver this current.
The 1k on the base of the output transistor is far too low for a transistor delivering 50mA collector current and should be 10k to 47k.
The 1k on the base of the first transistor has no effect because the 40kHz transducer is a high impedance device so it will not increase the base to ground impedance and the signal out of the transducer is so small that 1k will have no effect.
The 2M2 pot should be 2M as pots are only available in 1M and 2M.
The  whole circuit shows a lack of understanding of the principles of sensitivity.
The circuit needs to be very sensitive to detect the very weak 40kHz signal and the stages should be AC coupled to allow for variations in temperature and supply voltage.
The second transistor should be self-biased and AC coupled to the diode pump so no adjustment is needed.
The third transistor can be eliminated as the circuit only needs voltage gain because the output current is only about 50mA.
The third transistor is only providing current gain, which the circuit does not need and if the base resistor of the output transistor is increased to 10k to 47k, we will ncrease the gain of the circuit by 10 times to 47 times.
If the second transistor is AC coupled, ALL of the signal from the 40kHz transducer will be passed to the diode pump.
If the signal on this transistor is 1,500mV, the output will respond if the third transistor is removed.
This requires 30mV into the base.
To get 30mV on the collector of the first transistor the transducer needs to produce less than 1mV. I don't know how this relates to the required sensitivity of the circuit but it represents the maximum sensitivity for the circuit and it is AUTOMATICALLY self-adjusting and AUTOMATICALLY delivering the maximum overall gain without any need for adjustment.
This is how to work out the requirements of the circuit.
It is a VOLTAGE REQUIRING circuit not a CURRENT REQUIRING circuit.
And secondly, it is very difficult to DC-control 2 stages. This is because each stage has a gain of about 100 in DC conditions and 100 x 100 = 10,000. The change in current on the base of the second transistor is being multiplied 10,000 times by the time it reaches the LOAD and picking the correct level of DC BIAS is very difficult.  
The operation of this circuit is much more-complex than you think and you need to read our comprehensive article on designing transistor stages to see how to design stages correctly.
You NEVER try to control the bias on a stage via the base resistor.

This circuit shows 4 areas of bad design and although the "circuit will work" it is obviously not designed by a professional and not a design to be presented to beginners in electronics.


Here is another over-designed project.
Basically it will supply 3v3, 5v and 9v2 from a 3.7v Li-Ion battery.
But let's look at the design in REALISTIC TERMS.
The whole project is just an expensive way to get different voltages from a 3.7v battery.
It uses a PTN04050C boost converter module that takes the 3.7v and converts it to any voltage to 15v.  This module costs $16.00 !!
We now take the output of this converter (9v2) to a UA78M33  - a 3.3v linear regulator to deliver 300mA. This process wastes 70% of the energy !!!! We are converting 3.7v to 9.2 and then reducing it to 3.3v   ????    The data sheet for this IC does not tell you know the wattage dissipation for the device and this is very deceptive as it will probably only dissipate 1 watt. Using it on a voltage as high as 9.2v  @ 300mA will dissipate 1.77 watts.  
The 7805 also wastes nearly 50% in delivering its voltage and current. And this waste ends up as heat.
For both regulators, the designer of the circuit has not provided any mention of a heatsink and obviously does not know anything about heatsinking a device.
Lets look at the problem.
To work out the maximum current for a 7805 with 9.2v input, the thermal resistance is measured in degree C rise per watt. For a TO220 (7805 and uA78M33) case the max Tj (junction temp) =125 C, and TRjc (thermal
resistance-junction to case with no heatsink) = 50 deg C/watt.
Now we can do some calculations:
The author claims the 7805 will deliver 800mA
9.2 volts input voltage, 0.8 amp load, 25 deg C ambient;
9.2 - 5 = 4.2 volts across regulator. 0.8 amps = 3.36 watts dissipation.
Temp rise, junction to ambient = 3.36 x 50  = 168 deg C.
Add ambient temp, Tjunction = 168 + 25 = 193 deg C!
The specification is 125įC max so you need a heat sink.

For the uA78M33 the wattage dissipated is 9.2 - 3.3 = 5.9 x .3 = 1.77 watts and it will get to 115įC.
This is very hot and simply a waste of energy.

This project will cost more than $30.00 (plus shipping charges) for a something that could be achieved with a number of rechargeable cells and two regulators. You can buy a 18650 rechargeable battery on eBay for $1.50 each (post free) and a charger for $2.50 (post free) that will charge 2 cells at a time. Buy 6 cells and you will have 3 cells for the project and 3 cells being charged. These cells are 4,000mAHr and are much larger than the 1200mAHr cell used in the project and will last 4 times longer. By using 3 cells you have a project that will last 12 times longer at less than one quarter the cost.
It's simple ECONOMICS.

You wonder why electronics magazines don't show photos of the projects. Simple.  They do not build the prototype. Here's a typical example.  Where is C1 ????

Why go to the effort of producing a PCB layout for the magazine and not provide the trackwork for the reader?
Why not show a photo of the complete project?   Because they NEVER get the board made and never build the project.   It all called "CLOUD" designing.  It's all illusionary.

I have supplied over 20 major corrections and faults to the circuits in Electronics For You and have not received a single reply or comment from the technical editor or any of the designers.
The magazine offers $20.00 for each correction and this entitles me to $400 for my efforts. No payment has been provided to date.
But the important point is this.
The magazine should be including corrections and improvements in the following issues, just like the Australian magazines.
This shows responsible editorship and displays an understanding that EVERYONE MAKES MISTAKES.
It is fortunate that some of the "Professors" that I took to task, do not present their rubbish in the magazines any more and have desisted from presenting more of their designs on the web.
Their circuits simply DID NOT WORK and one professor said the voltage between base and emitter could be as high as 3v.
Obviously he has NEVER built a circuit in his life and when he saw my pages of corrections to his circuits, he pulled the plug on his website.
My main aim is to cover the basics of electronics as many designers lack these concepts and when confronted, are amazed there are design-rules and tolerances that make every circuit reliable.
It's funny that I don't have any disagreements with circuits designed by those with technical expertise.
I am not trying to find fault with circuits that don't have a fault.
I am highlighting circuits that are TECHNICALLY INCORRECT and showing how NOT to design a circuit and how NOT to make a fool of yourself.



Electronics for You has failed to reply to my emails.  The CEO has failed to reply, the Technical "team" has failed to reply and the "Technical Editor" has failed to reply.
Electronics for You is a LOST CAUSE.
I didn't expect a reply. They have failed to reply to any of my emails for the past 4 years. I only expected them to fix up Kits N Spares websites where they have the same photo for 4 different kits, images of PC boards with "prototype" scribbled on the edge and boards with no overlay but hand-drawn numbering.
This is an absolute disgrace and would not be accepted for a prototype shown to electronics enthusiast. And they still want $50.00 postage for a $3.00 kit. It is obvious they don't get any overseas enquiries. Who is going to pay $55.00 for junk kit.
They have removed their electronics forum and sale of components. The forum was so messy and so under-used that you could not follow any of the requests.  
If you look-up webtraffic: you will find their penetration outside India is 3.9% !!! It speaks for itself. They have been on the web for 17 years and only have a visitor tally that is twice Talking Electronics.
They claim a readership of 100,000 but they are selling their subscription for half-price and including a soldering kit, so they are effectively making no money on the subscriptions.
They can only do this to try and maintain readership to produce phony readership figures for the advertisers. 
It's sad they don't have any respect for their hobbyist market. But they simply do not have any technical personnel to improve the website or provide any technical information. They can only come up with stupid replies such as: "we can post your corrections on our website after we publish our circuits."  I offered to review the articles BEFORE publication so the terrible designs would not be published. They obviously did not understand my offer. 
You can tell you are talking to an absolute IDIOT when he says he has found your email in his SPAM folder.
This is the sort of intelligence you are talking-to at EFY.  Why have a SPAM folder where you can miss important emails ???
There are some people you can help and there are some people you CANNOT help.
After 4 years of sending them corrections to their circuits, they have still failed to "pull themselves out of the mess."
I contacted their advertisers and distributors and did not get one single satisfied customer. They are all paying enormous amounts each month for advertising, with little or no return.  But they are all living in the hope that next month will improve. WISHFUL THINKING.

Here is a statement from Electronics for You website:
Nearly 75% of the engineers passing out from various institutes are unemployable as they are severely deficient in practical hands-on training.
This confirms what I have been saying for years. It is too late to enter a course to get training.
Training starts at home when you are 8 years old.
Training starts with making projects.  Lots of projects. It is the ONLY way to learn.
Electronics for You provides a Basic Electronics courses at a cost of 8 days salary for a 2-day course and a PCB design course showing how to make a single-sided board for a power supply.
A comment from a student: "
I would also like to take up an advanced PCB design course with some training on 2-layer and 4-layer PCBs, which is currently the need in the semiconductor industryĒ
What is the purpose of the PCB training course???? Who designs single-sided boards???
I use 40 year-old CAD package because it is simple to use and does not "fall-over."
I can also combine boards from 20 years ago when making a panel.
You must use a package that is not frustrating.
On top of this you must use a PCB maker that is CHEAP.
I pay $2.00 per board for 10cm x 10cm and get 10 boards. Some manufacturers charge $20.00 per board. This is a RIP OFF. Every board is double-sided, PTH and silk screened. Every board is perfect. Cheap boards are absolutely perfect and anyone who says differently does not know what they are talking about.
I produced the first magazine in Australia to present projects using tinned boards with solder mask and legend.
All the other magazines sold bare copper boards with no overlay!!!!!  That's how far things have come.

Here's the clincher.   I sent Electronics for You 4 articles and the reply was:  "most of them are written for those who are beginners w.r.t. microcontrollers, and the applications discussed have already been published though through a separate design."
I don't know where he gets that STUPID comment from. None of my projects have even been presented before and no-one has explained how the circuits work to the extent of my description.
When you get IDIOTS like that being in charge of areas that are beyond their capability, you get a magazine that fails to deliver anything but JUNK.
The 3.9% acceptance by the rest of the world is an indication of how it is viewed. Even after 17 years of publishing, it cannot get a following that represents a percentage.

This is the sort of email I get every day from visitors to the site:

Dear Dr Mitchell,
I am teaching myself electronics and have been using your materials (pdf) on your website. I am writing to thank you for all that you have done in preparing these excellent teaching materials as an investment that benefits others.
When I was a kid I used to use the 500 in 1 electronics project kit and found it useful. I also found that while there are books out there (on Amazon) that teach electronics they tend to be very focused on the behaviour of each component and discuss them in depth. Though this is valuable I don't think it helps to teach how to construct circuits as a whole based on a problem or a requirement. I find you teaching materials such as 'transistor circuits' useful and like the step by step approach.

This message is also to encourage you and to show my appreciation.


Marc Edwards

Marc R Edwards
Research Officer
Advanced Medical Imaging and Visualization Unit
Bangor University

I am not a Doctor, nor a Professor nor do I profess to know more than 1% of the electronics scene. But what I know is being delivered to you in a way that EVERYBODY can understand and showing how to "see" a circuit operating so you can analyse, test, design and FIX it. 

I offered the same explanations to the management at Electronics for You and they refused to partake in any way.
Imagine the annoyance of the subscribers to the magazine if they knew they were being denied decent explanations and only being offered RUBBISH circuits that have been designed by incompetent muddlers and thrown into the magazine by "technical" staff that don't know a thing about electronics. 
That's the only thing that disappoints me.
But it's the same throughout the world with doctors, politicians, banks and the courts.
They are ALL corrupt and everything they do and say is incorrect.
If you only knew the truth you would be horrified.
I have added a lot of my comment on these subjects in SPEED READING.

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