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I am not against anyone building any of the circuits on my website  . . . and not buying a kit. In fact, if you build the project yourself, and use junk-box components, it will cost you less and you will learn 10 TIMES MORE than buying a kit.
How do you think I learnt so much?
Kits were not available in my day.
And printed circuit boards had hardly been invented.
Everything was hand-wired on tag strips and then punched board came on the market.
Then we got that terrible strip-board.
And finally expensive Printed Circuit Boards.
But if you make your own Printed Circuit Board and use salvaged components (and get the circuit to work), you will have learnt things from the BASIC LEVEL. Not the UNIVERSITY LEVEL where you pretend to make a project and come out of a 6 year course without having to touch a soldering iron . . and have the attitude that every circuit you design will work first first-go.

All the successful people who have gone though my training have build dozens  . . if not hundreds  . . of circuits and it is only by construction that you gain an understanding of being able to look at a circuit and see what is happening. 

Here is the metal detector NAIL FINDER project from:
Hashem Pakdel

Here is a Power Supply project from October 2019 issue of DIYODE magazine:
Let's see how many faults and bad-designs you can get in a single project.

From first impressions, the project looks ok. But when you look deeper, it is riddled with faults.
The most glaring mistakes are these:
The bridge rectifier diodes are 1N4004 for a 1.5 amp power supply.  These are 1 amp diodes.
The second mistake is the rotary switch. It is designed for currents up to 100mA and the circuit requires 1.5amp to flow though the tiny contacts. They will be destroyed in 10 minutes. The rotary switch needs to be replaced with a flying lead and machine pins. At least they will not burn out.
The 2,200u electrolytic used in the project is 50v.  You can see it is very small and will have a very small ripple current capability.
When you use a high voltage electro, the ripple current capability is reduced. The manufacturers cannot get high voltage and high ripple current into the same package.
The current through the 0.82R and 1R will produce about 1.5watts of heat and this heat will flow through the leads and if the pads on the circuit board are not large, the solder will melt and the resistor will drop off.
The 4 diodes in the bridge are Schottky 3 amp diodes and normally they have a low voltage drop across them. But I have used these diodes and found the voltage-drop at high current to be nearly 0.9v. This means nearly 1.4watts will be dissipated in two of the diodes when DC input is applied and the pads are so small and isolated that the diodes will drop off the board.
When 19v DC is applied (as suggested in the article) and the output is 5v  @ 1.5amp, the losses in the project will be 18 watts.
I have not worked out where the losses will occur but it looks quite impossible for the heatsinks in the photo above to dissipate this wattage.
The small heat fin is 2 watts, the medium fin is 5 watts max and the large fin is 5watts because it is in an inefficient place.  The heat is generated UNDER the transistor not on the top and gluing a fin to the top will be very ineffective. The bridge is 3 watts.
The 5k, 1k pots and 270R resistor on the second regulator will have the capability of producing an output from 1.25v to more than 27v and thus the project is using less than half the rotation of the pot to get to 12v. 
The 10R resistor is 5watt. But the regulator will have to deliver only a small current as the MJ2955 is doing all the work.
But here is the major fault with the circuit.
There is no electrolytic on the input of the second regulator. The regulator needs something to "push against." Something rigid and stable so it can produce a constant output voltage as the current-requirement of the load, varies. This circuit has no electrolytic and the output variations will be enormous.  The circuit is worthless. 
Nothing is marked on the top of the board. You would have absolutely no idea how and where to place the components.
Normally, a LED has the voltage dropping resistor on the anode, so the cathode is connected to the 0v rail. This just makes it easier to service.
I explained to the author of this project that it is a "dog-breakfast" with the components all over the board and not following the layout of the circuit.
The project shows a complete lack of understanding of electronics and current. The 10R resistor will pass about 60mA and 0.25watt will be suitable. The 0.82 and 1R can be much smaller. The rotary pot will burn out. The 5k pot will use only a portion of its rotation, the PCB has no component markings . . .  just to mention a few.
If you look at Talking Electronics projects, you will see the layout of the components follows the circuit and that makes it easy to service.  This project is just a jumble  -  a jumble of mistakes.

Here are a few points on what to expect from this power supply.
A LM317 is capable of delivering 1.5amp but the input voltage must not be higher than 12v for an output voltage of 5v so the voltage across the regulator will be 7v and the current will be 1.5A, making a total loss of 10.5watt. If the losses are higher than this, the regulator will get too hot, even though you may have a very large heatsink, and it will close down and reduce the output voltage. 
When operating correctly, like this, the output voltage will drop (fluctuate) about 5mV to 25mV when the current fluctuates, from say zero current to 1 amp, due to the regulator driving say a 5watt amplifier or a resistive load that is being turned on and off.
The output will only have this low voltage-drop (fluctuation) if the input voltage is stable and does not drop below 10v as the regulator needs 3v or more across it AT ALL TIMES to provide the feature of a STABLE OUTPUT VOLTAGE.
If the voltage across the regulator falls below 3v, the fluctuations on the output become enormous, so it is the 2v (up our sleeve) on the input that we are relying on to maintain a low voltage-drop on the output.
Now the 2,200u on the input has the ability to store energy and if the input voltage is AC, it will be getting 100 pulses of energy each second.
You have to remember, the AC on the input is rising and falling and some of the time it is not present AT ALL. That's why the 2,200u is added to the circuit to deliver during these parts of the cycle.
But the 2,200u really has very little energy and if you are taking 1.5amps, the voltage across the electro will drop about 5v during part of the cycle. That's why, to get a minimum of 10v for the regulator, you need to charge the electro to 15v.
All this applies to a 5v output.
You can see, the 2,200u is really quite useless but when combined with the 3-terminal regulator, the regulator improves the performance of the electro about 1,000 times. It's called an "electronic improvement," or "electronic improver." It reduces the fluctuations (called "dips") that will be present on the electro by not allowing the peaks on the electro to be passed to the output and thus the output "thinks" the electro has say "10v across it at all times." In other words the output thinks the electro has a constant, fixed, rigid voltage across it at all times.
For any voltage above 5v, the same amount must be added to the voltage across the electro. Thus for 12v out, the voltage across the electrolytic must be 22v. 
Finally, the output voltage from a transformer has a rating.  Say the rating is 14v @1.5amp.
This is an AC rating and the output will be 14v  when 1.5amp is flowing.
But the secondary winding of the transformer has resistance and when current is flowing, there will be a voltage drop in (across) the secondary winding.
This drop may be 5v, so the manufacturer adds extra turns so the voltage is 14v AC  when 1.5amp is flowing. This means the output voltage will be 19v on NO LOAD.
When this winding is connected to a bridge, the voltage gets converted to pulses of DC and stored in the electrolytic. The conversion from AC to DC increases the voltage 41%, minus about 2v drop across the diodes in the bridge.
These voltages must be taken into account when designing a power supply and some of them are only known when you test a prototype. 
The 220u on the output has almost no effect (no improvement) as we already mentioned a 2,200u on the input sees a 5v drop.
If the regulator is not delivering a quality current and a voltage-dip of less than 25mV, the 220u on the output will make no improvement. In fact it will not alter the 25mV.

Here's another way to look at the circuit.
Suppose we have an electrolytic charged to 25v and it dips to 20v when delivering a current (due to a bridge and AC input).
If we were able to "tap" it at say 18v, (or any voltage below 19v) the output current would be perfect and no ripple-current would be present and no voltage-dips would occur. But we cannot do this by simply connecting a wire to the electro, so we need a 3-terminal regulator. The regulator is, in effect, picking off the voltage of the electro at say 18v and delivering it to the output. 


The battery pack has 6 cells in parallel and 3 cells in parallel.  What do you think is going to happen when you use the battery???
The amp-hr of the battery will be limited by the 3 cells.  This means the battery will only deliver about half the available amp-hr. 
And the 4 pin plug will only handle 100mA.  The battery will deliver 30 amps !!!
Pity the artist did not understand what he was doing!!!! 


The circuit works perfectly but it contains a technical fault that needs to be covered.
The emitter-collector junction of the BC557 and the base-emitter junction of the 2N2222A are directly across the supply.
If, and when these junctions are fully saturated, they will try to have a voltage across them of 0.2v and 0.7v If this were to occur, a very high current would flow and both transistors would be destroyed.
The only thing preventing this is the bottom 2M2 resistor. It limits the current through the base-emitter junction of the BC557 to about 1.5 microamps. If the transistor has a gain of 300, the emitter-collector current will be 450uA (less than 0.5mA) and the transistor will not be destroyed.
This is the same current that will flow in the base-emitter of the 2N2222A and is sufficient to turn on the transistor and drive the circuit, but not get damaged.
The circuit does not have a timing capacitor and relies on the time taken to saturate the inductor and then release its energy to the LEDs.   

I posted a comment about the 555 Metal Detector article and it was immediately removed. That's what they do to all my comments.
The circuit is a copy of one on my website but it has 3 mistakes. The 2u2 is up-side down, the 100u to the speaker should be 10u and the 1k pot will burn out. The coil has almost no details. Who winds a coil by saying the wire is 57 metres long!! Why use 51k when a beginner has enough trouble reading a 47k resistor.


Here is a DELAY CIRCUIT from a design engineer in India:

It has a number of faults and will not work at all.
The circuit is far too complex but looking at the base of T1, it will be at 0v when T2 is turned ON via the delay on its base.
The emitter is connected to the base of T5 and the emitter of T5 we will assume to be at 12v for the moment.
This means the base of T5 will be 12v and the emitter of T1 will be 12v.
This places the base-emitter of T1 in a reverse voltage situation between the base and emitter and this voltage can on be about 5v before the transistor starts to leak. This means the reverse voltage on the base-emitter of T1 will zener at 5v and the emitter of T1 will not go lower than 5v. Pin2 of the 555 needs to go below 4v for the 555 to start timing and must go above 8v for the chip to stop timing.
The circuit is such a mess that I don't know what will happen.
The second problem is the 1,000u on pin7 of the 555. Getting pin7 to discharge this capacitor will put a very high current through the discharge transistor and damage it.
The third problem is the output of the opto-coupler being able to turn off the BC558. The load on the output of the coupler is 12mA and I don't know the voltage across the output with this current. It may not be able to go as low at 0.5v   The reason is this: As you load the output of the transistor device, the transistor cannot fully saturate and the voltage between the collector and emitter will rise from 0.3v to 0.4v  or 0.5v or even 0.6v   It depends of the quality of the transistor and the current is designed to handle. When you are relying on unusual features like this for a circuit to work, you have to test a number of devices to see if the whole batch will work as required,.
The fourth problem is the 7 minute timing. Delays above 3 minutes are very unreliable with high value electrolytics and high resistances.  You are charging the 1,000u at less than 3uA and the leakage current for this type of electro can be 3uA so it will never charge. This circuit was designed to go into an air conditioner for a life of 20 years and and you can see the problems that will start to show after 12 months and the company will go bankrupt. 
The second relay and opto coupler are not needed.
And improved design will eliminate most of the components. It can be designed much simpler.

Here is another circuit from D.Mohankumar.
None of his circuits work and he has 2 million Indians going to his website each month and seeing faulty circuits that don't work and never will work. It is no wonder the poor Indians are hopeless at understanding electronics when they see this RUBBISH.

He says: Here is a simple 3.7 volt Lithium-Ion battery level indicator. Green LED lights when the battery is full with 3.7 to 4.2 volts. When the battery voltage decreases below 3 volts, Green LED turns off and Red LED lights. Just connect it to the 3.7 volt battery charger.

There is no current-limiting resistor for the green LED. Imagine what would happen to the green LED when the battery reaches 4.2v.   A green LED drop 2.4v  The diode drops 0.7v and the base-emitter junction drops 0.7v  That is 3.8v    A fully charged cell is 4.2v and after testing the circuit for 1 minute the green LED burnt out. It got so hot on the high voltage that the crystal overheated and died. I connected another green LED and the battery voltage dropped enough so the LED did not burn out. The circuit is badly designed and I suggest you do not build it. However the red LED did come on at 3v - - so just the green LED section has to be fixed.

Another failure from D.Mohankumar.
Basically the circuit will not work because the 100R resistor will allow 60mA to flow and produce good brightness.
The way the transistor works is this: It effectively reduces the 100k by a factor of 100 due to the gain of the transistor and this means it adds a 1k resistance to the 100R and less than 6mA will flow. The brightness will be a lot less than expected and you will wonder why. You have to design a circuit so that it works 100% better than expected to take into account all the differences in the features of the components.

Another failure from D.Mohankumar.
The circuit will not work because the LED and resistor will not get any voltage. A photo diode has a very high resistance and putting 470R in series has no technical understanding. Turning on a mechanical buzzer slowly like this does not work. Just try it yourself and see how the circuit fails.  Another untried circuit from Professor Mohankumar who has NO understanding of electronics.

240V LED
Here is a simple circuit that not been understood by any of the readers or authors:

The zener is designed to limit the current when the switch is turned ON and the mains voltage is at a peak. In this condition the current will be very high and we know a LED can be destroyed INSTANTLY.
Let me say the circuit is badly designed because it is too close to  destroying the LED when the conditions are such that the capacitor is uncharged and the switch is turned ON when the supply is at a peak. The current through the LED at this peak will be 90mA when the characteristic voltage of the red LED is 1.8v and the zener is 2.7v    This leaves 0.9v for the voltage across the 10R resistor. This will allow 90mA to flow.
A much better arrangement is to use a 100R resistor and the 4mA will flow normally and when a peak occurs the current will be 10mA.
This is a much better design.

The secret is understanding the resistor can be increased considerably without affecting the current.


Here is a zener regulator circuit.
The first thing you have to accept is this:   The voltage across the zener will be constant when the current taken by the load changes. 
Of course the voltage will change, but if you don't accept the variation, don't use a zener regulator.
The voltage will change because the zener gets hotter when it is passing more current and you need to provide good heatsinking.
But if you don't accept CONSTANT ZENER VOLTAGE don't argue with the following description.
To start with, all zener diodes have a tolerance of +-5% so a 9v1 may be 500mV  less or  500mV more.  This means the voltage has a range of 1v !!
That is called the initial setting of the output of the zener regulator. But when the circuit is in operation, the output voltage will change by only a few millivolts or maybe 100mV
On top of this, every 9v1 zener will produce a different characteristic voltage across it when taking 20mA so this is a start to understanding the wide range of output voltages from a batch of 9v1 zeners.
Suppose you set up a circuit with 20mA flowing through the zener. As you put a load across the zener, the current flowing through the load will be taken from the zener-current.  The current through the supply resistor WILL NOT CHANGE.   It will simply be split into two paths.
You can keep "robbing" the current from the zener until only about 2mA flows through it and the output voltage will remain constant.
If you did the same thing with a VOLTAGE DIVIDER made up of two resistors, the output voltage will drop 2, 3 or 4 volts. The zener circuit may drop 100mV or 200mV. It has much better features.    A zener regulator will be about 50 times better than a voltage-divider. For instance a voltage divider may change 100mV and a zener circuit will only change 2mV.
In reality the zener will go colder when the load increases and the natural characteristic voltage across it will alter and drop slightly. That will be main contributing factor to the change in output voltage. Good heatsinking will reduce this.

Before we leave this topic, here is a diagram from DIYODE magazine:

How can one end of a supply rail be 12v and the other end 9v ????   It is blatant, GLARING MISTAKES, like this that are not corrected and the editor is saying I am: "nit-picking." 
If I was a beginner, I would think it is possible to achieve the result above AND HOW MISLED I WOULD BE.

Here is a circuit from April 2021 of DIYODE Magazine.
The author is trying to explain how to design a SERIES REGULATOR circuit.

But he is missing the hidden facts and disasters of using a transistor such as a BD139. He does not understand the BASICS of electronics and that's why he has made so many mistakes in this articles.
The basics of a POWER TRANSISTOR is this:  The gain of these devices changes according to the current flowing.  It may have a gain of 250 when 15mA flows, and a gain of 100 when 150mA flows but the gain will only be 40 when 500mA flows and less than 25 when 1 amp flows.
Let us say the transistor has a gain of 40.
This means the current into the base must be 13mA.
This means 13mA must flow through the base-collector resistor. And then you need another 10mA to keep the zener in regulation.
When the output is delivering no current, the current through the zener will be 23mA. As the output current increases, the transistor "robs" the zener of current and theoretically when the output current is 500mA, the transistor will use 13mA and 10mA will flow through the zener. The zener will stay in regulation until the current through it drops to about 1mA but when it drops to zero, the output of the supply will drop.
The voltage across the base-collector resistor is 6.4v  so the value of resistance must be: 6.4/0.023 =  280R
But at 1 amp the gain of the transistor is about 25 and the base current must be 1,000/25 = 40mA plus 10mA = 50mA
The base resistor must be:  6.4/0.05 =  130R
The wattage dropped in the zener will be 0.05 x 5.6 = 280mW  and it will get slightly hot.
The circuit above will "fall over"  when more than 200mA flows and the output voltage will dip and drop appreciably. AND YOU WILL WONDER WHY!!!!
You must test everything before you put it in a magazine.
I don't know what the 330u is doing as you need another 1,000 on the output for 1Amp  and the 100n does nothing. He took none of the essential calculations into account when he decided the resistor should be 820R.

It is obvious the author has no idea how the zener works in this circuit as his calculations were totally incorrect.
The zener works like this:
The value of the resistor is worked out by allowing 10mA for the zener and the maximum current required by the transistor. Say it is 40mA.  The total current through the "supply resistor" needs to be 50mA.  This current will flow ALL THE TIME. 
When the output of the SERIES REGULATOR is zero, all the current we are talking about will flow through the zener. As the output current increases, the transistor will take (rob) the required current from the zener. When the output current of the SERIES REGULATOR is a maximum it will take 40mA from the zener and the zener will be left with 10mA.

Here is a circuit from June 2021 of DIYODE Magazine.
When some LEDs receive very bright illumination they produce an output voltage accompanied by a very small current.
In the following circuit, the OPTO-COUPLER has a bank of diodes on the output that produces about 10v and a current of 10 microamps.  This is not obvious from the diagram and you have to reference the component date to see its features. The top of the stack of diodes is positive but it not obvious that the opto coupler is connected around the wrong way.


This is the part of the circuit we are discussing:

The circuit has been rearranged so you see the MOSFET with a positive voltage on D
and a lower voltage on S. In addition you can see the
negative output of the opto-coupler is connected to G
and this will not turn on the MOSFET.

Here is the correct connection to the OPTO-COUPLER.

This modification is supposed to speed up the signal from the opto-coupler.
I cannot see how it works, and I say it does not work, so I have sent the request to Scott Williams, the technical editor of DIYODE for a clarification.  scott.williams@xentron    He has not replied to me.

Here is how the circuit works. Let us suppose the 10v from the opto-coupler rises slowly as the circuit is supposed to turn on at some special point. 
When the 10v is say 3v, the base will be 3v and the emitter will be 2.5v   For the transistor to turn ON, the emitter has to be higher than the base. 
When the voltage rises to 10v, the base will be 10v and the emitter will be 9.5v  So the transistor NEVER TURNS ON.
However the 10v from the opto coupler goes through the diode and into the Gate of the MOSFET and turns it ON.
The transistor does NOTHING.

Here is a stupid circuit from
They are trying the explain High-Side connection of MOSFETS. But if the 5v supply is capable of delivering current to the load, what is the point of including the MOSFET !!!!!

Here is a circuit with a number of problems.   Apart from the fact that is overly complex, the question is: why would you want to produce 330v DC for a LED lamps when most lamps operate on 5v from a USB connection??
And 240v LED globes work on 240v AC !!
The other question is: why select an expensive transistor?  The only two sources on the web are $7.50 each or 10 for $30.00!!
The circuit is designed for 7watt output, so why use a transistor capable of handling 5 amps??  But if you design the circuit to pass 5 amps, the voltage across the transistor will be 1v and the losses will be 5 watts and yet the transistor is the size of a BC547 and the data sheet says the transistor will handle 1.5 watts.   I would like to see a BC547 dissipating 1.5 watts  !!!   Maybe it will dissipate 300mW. 
The whole idea of selecting a common transistor is to allow the hobbyist to build the circuit from readily-available components.
The circuit is only suitable for 4v operation
To build this circuit you will need to know a lot about ferrite core transformers.  The primary winding is 16 turns for 4v.  Making it 4 turns per volt,  The secondary is 1360 turns, making the output 340v AC. I don't know why you want 330v DC.    What LED lamp operates on 330v DC????
The feedback winding Nf appears as an air core on the circuit, whereas it is wound on top of the primary.
Removing the air gap is not a good idea.  Without an air gap the transformer can be saturated and the energising current will simply go into heating the winding.
Where are you going to buy the wire with all the different diameters? 
Finally the 10k pot is in a low-impedance part of the circuit and the value is only suitable for a low power inverter. The value is too high for a low-impedance section.
Here is a basic run-down on its performance.
The circuit takes 2 amps. The voltage across a transistor is 1v.  This gives the transformer 3v. The gain of the transistor is 100.  The base current will need to be 20mA. The output of the feedback winding will be 3v minus the diode drops of 1v.  You have 2v remaining. The pot resistance plus the 68R must be 100 ohms.  The pot must be 32 ohms !!!   How are you going to select 32 ohms from a 10k pot !!!!   The whole circuit is so impractical !!!
It can be simplified by cross-coupling the bases of the two transistors to the primary winding and using base resistors. 
The whole idea of putting a circuit on the web is to help hobbyists produce a worthwhile project as cheaply as possible with readily-available components. 
There are a lot of cheaper, simpler designs, that I would recommend.

I can "see" how the whole circuit works and since there are no timing capacitors, the frequency depends on how fast the magnetic flux increases to a maximum in the core of the transformer.
The only component that cannot be "visualised" is the inductor between the bases of the transistors. The note says it increases the frequency to 20kHz, but we don't know the frequency before the inductor is fitted or the impact on the efficiency. Normally one end of an inductor is fixed and the other end is raised or lowered so that the inductor stores energy. In this case, one end can only be raised 0.7v and the other end can be raised to the same value. You would need to test the circuit with an Oscilloscope to see the result.

We are constantly being bombarded to get projects done in India.   Here is a recent project from an Assistant Professor !!!!   It is just a piece of Junk !!!    No-one can follow the circuit.   I have no idea what it does. And neither does anyone else.
For a few extra components you can achieve constant volume as shown in the circuit below.  

Here is his bio:
RAKESH JAIN received a Masterís degree in VLSI, B.E. in electronics and communication, and DIPLOMA in electronics. He is currently working as an Assistant professor in ECE department at Geetanjali Institute of technical studies, Udaipur. His research area is SENSOR and Microcontrollers. He has 26 copyright and 3 Indian patents. He has also been honored with Mewar Scientist Award2023.

Here is the circuit, properly drawn, so you can see and understand how it works
and see the faults in the design.

The design of the circuit shows no understanding of electronics. It is not sufficient to bundle a lot of components together, and if the circuit works, publish it.
Someone, like me, who has had 50 years of designing circuits is going to see the mistakes and problems and pull the circuit apart.
No thought has been put into the current taken by the circuit and this is one of the most important factors in a good hearing aid design.  Hearing aid batteries are very expensive.
The 3rd and 4th transistors are directly coupled and 3rd transistor is going to allow 3mA to flow through the emitter-base of the 4th transistor. A current-limit resistor is needed to reduce the wasted current. The other stages have high quiescent currents and the whole circuit should be redesigned. Stages should be "self-biased" to get them into the centre of their operating range. Not turned-on by any value base resistor you have in your "junk box."
Putting a 4k7 pot on the earpieces is going to reduce the output enormously.



Here's an amplifier with an unexpected feature. It is the connection of the base resistor to Q2.  Instead of it being connected to the positive rail, it is connected to the speaker.
This produces a surprising effect.
We start with Q3 turned ON via Q1 and the speaker receives current as the 470u electrolytic is charging. Eventually it will have a voltage almost equal to rail voltage across it. During this time, Q2 will not be turned ON because the D1 is keeping the base LOW.
Q1 starts to turn OFF and this puts less "pulling effect on the base of Q3 and the transistor starts to turn off because basically the emitter lead will not rise because the 470u is changed.
However the base of Q3 can rise and D1 allows the base of Q2 to rise via the 470R resistor. This turns ON Q2 and the emitter follows the base and pushes the 470u higher. This keeps occurring and the speaker "flips over" and now the current flows in the other direction and this pushes the cone in the opposite direction. 
Normally, when the transistor modes to wards the supply rail, the voltage across the base resistor reduces and the transistor has less effect.  But by connecting the resistor to the speaker, it still keeps the transistor turned on until the electrolytic is discharged. This produces slightly better performance than if the base resistor is connected to the supply rail.


The electrolytic can be thought-of as a battery to make the circuit easier to understand.

Here is a PNP circuit from YouTube.  The instructor said to replace an NPN transistor with a PNP transistor.
But the circuit is wrong. It will work, but the transistor needs an emitter-base voltage of .7v for the device to turn ON and this means the emitter-collector voltage will a minimum of 0.7v.   Normally the emitter-collector voltage will be less than 0.2v and so the transistor will get hotter than normal if the voltage rises to 0.7v or more. The base cannot be a 0v as this will not allow any current to flow through the 1k resistor and turn the transistor ON. So the emitter-collector voltage increases a small amount and now the 1k has a small voltage across it and base current will flow. The emitter-base voltage was measured as .697v and the base voltage .026v and the emitter-collector voltage = 0.723v
If you put a more powerful LED in the circuit and decrease the 470R to say 100R, the emitter-collector current will increase and the transistor will demand more base current. It does this by increasing the emitter-collector voltage to about 1.1v so that the emitter-base voltage of 0.7v  leaves 0.4v for the voltage across the 1k.
But now we have a voltage of 1.1v across the transistor and it gets very hot.

Here is the correct way to connect the PNP transistor:

The emitter-collector voltage will close to 0.3v and the transistor is allowed to operate correctly, over a wide range of current.

Here is a circuit from Ekeeda (an on-lime learning institute in India). They are describing, discussing and explaining how it works.  But it does not work.

The output of the first inter-stage transformer will be only a few ohms and to deliver enough current through the winding to turn on the second transistor will require very low base-bias resistors. These will be so low that most of the signal will be lost in them.

This is how the transformer should have been connected. No-one understood the mistake and they all found the video very helpful. This is why no-one is being taught electronics on YouTube.  The instructors don't know and those watching the videos are just being fooled with complexities or faulty circuits. 


There is no fault with this circuit but the instructor said the circuit comes ON with the LED illuminated and it would require another transistor to change this state.
A low value electrolytic across the collector-emitter of the first transistor would delay the turn-ON and create this condition and it is an exercise for you to try different values and see if this will work.  

Here is an ON-OFF circuit from YouTube where the author has not learned much about electronics.
Firstly, I could not work out what the circuit does until I redrew it in the conventional way.

You must draw a circuit so you can instantly see what each component is doing. This requires the positive rail at the top so you can see the voltage-drop each component. 
The circuit must be at the beginning of an article as it lets you know what it is doing  and you can decide if the project is of interest to you.


Here is a JUNK circuit from ETron Circuit Labs,  who are trying to sell a JUNK electronics course for $52.95 usd to unsuspecting hobbyists who do not know what JUNK instructions looks like.
The Continuity Tester circuit above contains ALL the faults and mistakes that a beginner could make.
For a start, the 555 does not work below 6v and the output is capable of delivering 300mA so the transistor and components are not needed.
The website shows a 9v battery connected to the circuit.  The circuit symbol is not correct for 9v battery.
A small speaker (8 ohm) will overload the output and it must be connected via a 10u electrolytic to prevent excess current.
But the biggest mistake is the position of the probes.  All the current of the circuit will pass through the component you are testing and if the component is delicate, it will be destroyed.
This circuit is an absolute disaster.

Triangular Output
Here is a circuit that only worked when you increase the resistance of one of the resistors.
And the output is nothing like a triangular waveform.

This is not how you design a reliable circuit. Circuits must be quite independent of voltage, transistor type and resistor values. The more tolerant it is, the more reliable the circuit will be.
The circuit did not work but by changing the 10k to 33k, a very unusual output was obtained. There was no output point identified on the YouTube video but a probe across the 10n and on the join of the two 10k resistors showed some unusual results.


Here is another JUNK circuit from ETron Circuit Labs.

I am picking out these circuits because they want to charge $70.00  for a JUNK basic Electronics course from people who don't have the slightest idea how to design circuits.  You can find all their circuit on the web and there are hundreds of FREE electronics courses.
The main fault is the 10R in series with the speaker.  The circuit is powered from a 9v battery and if you use an alkaline battery, it will deliver over 1 amp when heavily loaded.
The 8R and 10R will take over 500mA if the output of the 555 was able to supply this current.  But it can only supply 300mA and when you ask for a higher current, the output transistor does not fully turn-on. This mean it has a voltage across the output and 0v rail and when 300mA flows, the heat generated in the output transistor is excessive and the whole IC will get quite hot.
It does not matter they used a cheap, junk, Indian battery that was old and useless and their 555 did not get hot.
They present themselves as a teaching program and as such they should have an understanding of circuit design.  This, they have NOT. 

Here is a Tripler circuit from the web.  You cannot trust the web. The circuit does not work.


To understand how a capacitor gets charged, you have to look at the following circuit:

The 22u on output pin 3 gets charged when the output is LOW via the top diode. When pin 3 goes HIGH the top diode does not allow current to flow from the capacitor to the power rail.
This means the only path is the lower diode.
However, the lower 22u is already charged to to nearly supply rail by the two diodes, allowing current to flow to the electrolytic and charge it to nearly supply voltage. This means it is already charged to 15v.
The fully-charged 22u on pin 3 gets "lifted up" by the action of pin 3 and the left lead has say 15v on it and the capacitor has a charged voltage of 15v between its two leads. This means the right-hand lead is nearly 30v above the 0v rail and thus the second 22u gets charged from its present condition of 15v to a final voltage of 30v.
For this action to occur, we need a capacitor that gets charged and then gets "lifted" so that it can pass its energy into another capacitor. This action is called a "CHARGE PUMP."
The diodes are called "gating diodes" as one diode allows the capacitor to get charged and the other diode allows the charge to be passed to another capacitor. Each is like a "one-way" gate.   
You must have these 3 features to be present for the CHARGE PUMP to occur.
In the following circuit, diodes "a" and "b" and electrolytic "a" create a Charge Pump to charge capacitor "b" to about 22v. 
Capacitor "b" is called a STORAGE CAPACITOR. It is just like the 12v rail, but it is 22v.
We now have the second Charge Pump section consisting of capacitor "c" and diodes "c" and "d."
Capacitor "c" moves up and down by an amount of 15v via pin 3. When pin 3 is LOW, it gets charged by capacitor "b" and then it is "lifted up" by an amount of 15v via pin 3 and this extra voltage is passed to capacitor "d" to create a final voltage of about 31v.
This voltage is not produced after one cycle or even 2 cycles. It takes many cycles. This is because capacitor "c" takes energy from capacitor "b" and it takes many cycles for the capacitors to gradually charge. That's why the Charge Pump circuit takes a few seconds to deliver the 31v.

Now you know how the Charge Pump circuit works, you can see the first Tripler circuit only doubles the voltage.
The first stage consist of capacitor C3, C4 and diodes D1, D2.
C3 is charged via diode D1 and when pin 3 is low, C4 is charged via D2 from C3. But C3 is never charged to more than 15v and the first Charge-Pump stage does not see voltage doubling. The second section creates voltage doubling and that is all you get on the output.

CLASS-A AMPLIFIER with transformer
This is another mistake from a discussion on YouTube.
The instructor said the polarity of the voltage on the primary winding of the transformer reverses during the second part of the cycle and he says the voltage on the collector will rise above 12v.
This is not true.
I know this is very complex to understand and that's s why so many mistakes are made.
When the transistor turns ON, the current through the primary winding produces flux in the core of the transformer and the top lead will be 12v and the lower lead will be about 4v.
This is the incorrect way to describe the circuit:
When the transistor turns OFF, it effectively disappears from the circuit and the flux in the core collapses and produces a voltage on the primary winding that is OPPOSITE to the original voltage. This voltage can be 10 times higher and even 100 times higher.
This action is due to the natural and fast collapsing of the magnetic flux (field) and we call this effect FLYBACK.
During this time the voltage on the secondary will firstly be in one direction and during the second half of the cycle it will appear in reverse at a much higher value.
BUT this is not the case in the circuit.
The circuit is not a FLYBACK arrangement.
It is an audio amplifier and the transistor does NOT turn OFF or disappear.
It is always connected and it always controls the increasing and decreasing current through the primary winding.
This action is completely different to FLYBACK.
When the transistor is turning ON, the flux in the core increases and produces a voltage in the secondary winding with the output voltage as a positive on the top wire.
When the transistor turns OFF, the flux is reducing and produces a negative voltage on the top wire of the output.
In other words, the output sees a reversal of voltage during this complete cycle (the secondary voltage) and it is not very high  - or may not be very high - as it determined by the turns ratio of the number of turns on the primary and the number of turns on the secondary.  
So, the secondary voltage may reverse but the primary voltage simply decreases and increases.
That's the amazing part, you can get a reversal of secondary voltage via two different ways and that's why so many people get confused.

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