SPOT 
THE MISTAKES!
Page 12
 

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We start another page of faults, hints and corrections to circuits found on the internet.
I am surprised at the number of faults and poor designs produced by authors who boast a university degree and experience in prestige places such as NASAís Jet Propulsion Laboratory, as stated by our first designer: Chris @ PyroElectro.com
His circuit for a White Line Following Tank has a number of points worth mentioning. Here is his circuit:

The layout is a "dogs breakfast."
The first thing that "threw me" was the diode on the TIP120 transistor. I could not see what is was doing.
Then I had to work out that the TIP was a Darlington transistor.
There were other mistakes but the layout was so poor I had to redraw the circuit to make it easy to understand. The result is shown below:

The circuit is over-designed.
This is easy to see.
The TIP120 is pulled HIGH via a 330R resistor.
A photo-transistor receiving light has a resistance of about 300R. This means only a single amplifying stage is needed between the photo-transistor and the Darlington output transistor.
In addition, the base of the transistor driving the second motor could be taken from the collector of the first TIP120 transistor.
This would save 2 transistors.
Two diodes and a resistor have removed from the original circuit as they are not needed.
The circuit does work, as shown in the following photo and a full description of its construction can be found at:  http://www.pyroelectro.com  and an eBook on building Robots:
http://www.pyroelectro.com/book/Building-Robotics-&-Electronics.pdf
The only point I am making is the fact that the circuit had no component values ( I added the values from the construction-notes), the layout was incomprehensible and is over-designed.
The author was informed of these details and has failed to update the circuit.
I realise many authors are extremely embarrassed when confronted with faults and most completely fail to respond.
That's why it is so dangerous to surf the net and conclude that everything you see is correct and functional.
Before accepting any circuit, collect at least 2 or 3 similar designs and see if they correspond with each other.



This circuit is a good design but the 10k resistor (R3) is not needed and the 5v1 zener diode serves no purpose. The output voltage is being determined by voltage across the collector-emitter terminals of Q2 (about 5v) and the zener voltage. Why not simply use the voltage divider consisting of R3 VR1 and R4 to create about 10v across the collector-emitter of Q2 and remove the zener. The circuit will operate exactly as the original design. With these changes we have eliminated two unnecessary components. 

Here is a dangerous circuit from Alternative Energy Circuits Blog by occulist  - J McPherson

If both pots are turned to low resistance, they will create a short-circuit and will be damaged. A damaged lower pot will cause the output voltage to rise and damage the circuit you are supplying.
The answer by J McPherson:
2 pots are fine in the circuit.  If the 2 pots are set to near short condition - one will blow out like a fuse.
What an absurd statement to make.
It's designers like
J McPherson that should not be displaying their stupidity on the web.
If the lower pot fails, the output will rise and damage the circuit you are protecting!!!!!!!!

Here is a another circuit from Alternative Energy Circuits Blog by occulist - J McPherson

The output voltage will not be 15.2v 
The peak voltage produced by 24v AC secondary will be 34v. The transistor will only pass the positive portion of the waveform to produce pulsed DC on the output. The 47R and 170R voltage dividers will deliver a peak of 34/217 x 170 = 26v 
Any battery connected to the output will see pulses of 26v and the charging will never cease.
I don't see any purpose in this design and it is certainly not a "Regulated Battery Charger" as stipulated on the website.
The only thing that may reduce the charging current is the limitation of the transformer. It will possibly be a low wattage design.
The 47R needs only 0.5 watt.
It's absurd to use a high voltage secondary transformer then dissipate a lot of wattage in the external components.
The circuit is a very poor design and shows how NOT to design a battery charger.

J McPherson makes the statement:
"Lead acid batteries are very tolerant for their voltage inputs."   I don't know what he means but this statement but if you want a charger to turn off when the battery is fully charged, the detection-point is very critical. The actual voltage depends on the type of battery and must be within 100mV of the specified voltage, otherwise the battery will eventually DRY OUT. Some batteries have special compounds mixed with the paste in (on) the plates to increase the point at which gassing occurs. This is one of the features of a MAINTENANCE FREE battery.
 

Here is a another circuit from Alternative Energy Circuits Blog by occulist

The output is NOT 1.2v higher than the voltage of the zener diode.
occulist was notified of this and could not see the reason. He has not fixed the problem on his website.
That's why many websites are very dangerous. They contain faults that the unsuspecting reader will interpret as fact.
When the circuit is drawn as shown below, you can clearly see the output voltage is always 0.7v below the voltage of the zener. That's why it is important to draw a circuit so it makes the operation of the circuit easy to understand. For more details on The Transistor Regulator, see The Transistor Amplifier eBook.



NITPICKING
The only reaction I have had from J McPherson from Alternative Energy Circuits Blog is to say I am "nitpicking."
He claims everyone is clever enough to realise the circuits he has presented on his site contain faults and no-one will fall into the trap of putting two pots across the power rails and create a short-circuit, or build a battery charger without knowing the characteristics of the transformer.
In addition, he claims it is quite normal to use a 1M pot to create a resistance between 1k and 100R then measure the pot and substitute a resistor.
He obviously has absolutely no idea of the capabilities of the average reader on the web.
90% of readers are beginners and have no ability to correct a faulty design.
Neither J McPherson nor Jose Pinto have made corrections to their site. They are obviously so embarrassed that so may faults have been picked up that they have nestled into a corner and closed their eyes.
These "Spot The Mistake" pages are not "nitpicking" but highlighting the fact that many mistakes can be found on websites and in technical books.
After all, why do you think a book is read by a technical proof-reader before it is published?
I have been a proof-reader in the past and before many books go to press, they contain myriads of mistakes.
In some cases, every page needed corrections and these books were intended for junior school electronics classes. Imagine the problems if they went to press without corrections.
J McPherson fails to realise one important point. A circuit will not work unless it is 100% correct.
It is fortunate that these sites attract very few readers but if you go to forums such as:
http://www.electro-tech-online.com/  you will see the result of readers trying to build a faulty circuit.
It's only through a network of dedicated electronics professions who willingly assist on these forums that the beginner can find a resolution to his problem.
Faulty websites, untested articles in magazines and poorly presented text-books are a danger to the newcomer.  
They pick up wrong terminology and when the circuit does not work, they are likely to get frustrated and give up the hobby.
This especially applies to circuits in the magazine: Electronics For You. It nearly always  contains poorly designed projects and this makes it doubly-difficult for a reader to get a circuit to function correctly.

That's why we will keep adding to these pages.

The best way to learn a subject is by observing the faults of others.



Here's another disastrous design by D.Mohan Kumar
The circuit has NOT been tested as it DOES NOT WORK.
He is supposed to be a University Lecturer in India. No wonder Indian students go overseas for a real education. Nearly all of his circuits have a mistake. He has never replied to any corrections and continues to deliver faulty circuit, not only on the web, but in Electronics For You magazine.
Go to our eBook The Transistor Amplifier for the operation of a PNP transistor. The voltage between emitter and base is never more than 0.7v. What is point of putting a 10v zener across this junction?
Here is his article. Can you see where he has made a major mistake?

A PNP transistor BC 557 functions as a voltage controlled switch. As you know, PNP transistors forward bias and conduct only when its base is grounded. A Zener diode is used to control the switching of T1. In the 12 volt battery monitor, 10 volt Zener is used. When the battery voltage is above 10 volts (The breakdown voltage of Zener), the  Zener conducts and the Base of T1 becomes positive and it remains off. Buzzer will be also in the off state. Variable resistor VR adjusts the exact biasing of T1 at 10 volts and C2 buffers the base of T1 positive as long as Zener is conducting.

Here is his mistake: D. Mohan Kumar thinks the zener diode will break down at 10v and only a very small voltage will appear across it. This is not so. At 10v, the zener will break down and 10v will appear across it.  However the zener is connected between the emitter and collector of the PNP transistors and the voltage between these two leads cannot be higher than 0.7v as this is the characteristic of a transistor. So, the zener never sees 10v.

Here is the way to analyse the circuit:
As the supply increases from 0v, the transistor is turned on via the 4k7, when the supply reaches about 0.8v. And as the supply increases, the current through the emitter-collector leads increases. The supply can increase to 10v, 11v, 12v, 15v and the zener never sees more than 0.7v across its terminals.

The following circuit will activate the piezo buzzer when a 12v supply drops to 10v:

The two transistors operate as a Schmitt Trigger. 




Here's another design by D.Mohan Kumar
The circuit does work but can be improved. You can learn a lot from other people's mistakes.
The first mistake is the rating of the supply. The 14 - 0 - 14 3-amp transformer is an AC rating.
This means the voltage is 14v AC and when it is rectified, the voltage becomes 14 x 1.4 = 20v.
The 3-amp rating of the transformer is also an AC rating and since we have gained extra voltage from the voltage component of the rating, we must derate the current so the wattage (the volt-amp) rating of the transformer remains the same.
This means the AC current rating becomes 3 x 0.7 = 2.1amp DC.
The second fault is the 5k pot. The "pick-off" from the 12v zener is taken directly from the zener and this means the 5k pot will have no effect on adjusting the voltage.
The third improvement is the removal of the second transistor and a few components. These are not required when the circuit is re-designed.
The 470u will have no effect on smoothing the supply when a battery is connected. 
The supply is not regulated so you cannot say the output is 14v.  It is 20v when unloaded.


The original 3-amp charger circuit


The improved circuit


Here's another faulty design by D.Mohan Kumar

Here is his original article. The corrections are in bold RED. The circuit is a terrible design but here is the best that can be done with circuit:

Transformer X1, diodes D1, D2 and capacitor C1 forms the charger section.   X1 is the centre tapped 9-0-9 step-down transformer with 500 mA current. Diodes D1 and D2 are rectifiers to convert 9 volt AC to 9 volt DC. It converts 9v AC to 12.6v DC.  High value capacitor C1 removes AC ripple from 9 volt DC to get clean DC. The 1,000u is not needed. The circuit is charging a battery and the battery accepts pulses.  LED indicates the power on status and resistor R1 limits LED current to 20 mA. Under normal condition, 9 Volt DC passes through D3 and R2 to the battery for charging. Resistor R2 limits the charging current to 40 mA (9 / 220 = 0.4 Amps x 1000 = 40 Milli Amps) for slow charging since the charging is process is continuous. The current will be (12.6 - 6v) divided by 220 = 30 mA.  At the same time, the load gets power through the PNP medium power transistor BD 140 and functions normally. NO! It does NOT! The transistor will deliver only a few milliamp. See The Transistor Amplifier eBook for details on how to design this section of the circuit.   9 volt DC It is 12.6v DC. is used to charge 6 volt battery since 2-3 volts excess is necessary for normal charging and the circuits itself consumes some power. Output voltage will be around 6-8 volts.  It is 12.6v DC.


6v BACK-UP SUPPLY WITH FAULTS

Diodes D3, D4 and the battery form the backup section. When the mains power is available, D3 forward bias to provide power to the battery and the load. When the mains power fails, D3 reverse biases and D4 forward biases so that current flows from the battery to the load. D3 prevents the back flow of battery current to the transformer. Diodes D1 and D2 prevent reverse current to the transformer. D3 simply prevents the LED illuminating when the power fails.

Resistors R3, R4, zener diode ZD and transistor T1 are the components of the cutoff section. When the  battery voltage is more than 5.3 volts (4.7+0.7=5.4) zener conducts, taking the base voltage of T1 to ground. No. The base of T1 sits at 4.7v  Since T1 is a PNP transistor, the negative base bias allows it to conduct so as to provide load current. This is false.  When the battery voltage drops below 5.3 volts, zener stops conducting and the base of T1 becomes positive through R3, R4 and it switches off. This is false. This cuts off power to the load. This condition remains as such till the mains power regains. This is ALL false

You can see how many mistakes the author has made.
He has the wrong concept of how a zener diode works and no idea how to design a transistor output stage.
When the main is alive, the output current will be determined by the current though R4.
The current though R4 is 12v. - 0.7 - 4.7 = 7.2 divided by 10,000 = 0.72mA   If the gain of the transistor is 100, the current will be 72mA.
The authors circuit simply does not work at all when the main fails.
The emitter sits at 6.3v - 0.7v = 5.6v  and the base sits at 4.7v. This give a turn-on voltage of 0.9v. The battery only has to drop 0.2v and the circuit stops working. The output current will be a maximum of a few milliamps.

You cannot provide all the features the author is requiring when using a 6v battery. The diode- drops and transistor voltage-drops leave almost no voltage for the devices you are powering.



Here's another faulty design by D.Mohan Kumar

Here is his original article. The corrections are in bold RED. The circuit is a terrible design as it uses a 12v lighter battery to flash a 3v LED. A lighter battery has very little energy (about 23mAhr - 30mAhr) and is expensive. The circuit will not last very very long.

Here is a Dummy LED flasher to confuse intruders. It simulates the LED blinking of a sophisticated alarm system. It can flash for years both day and night using a single 12 volt alkaline battery (lighter battery). The battery will last just a few days!  If the life of the battery permits, it can flash continuously for more than 10 years. Theoretically it is true since the LED consumes less than 5 microamps.  This is untrue. The LED consumes about 20mA.  The interesting feature of the Flasher is that, the LED lights by taking current not from the battery, but from the capacitor. True.


The Circuit is simple and can be made match box size. It can fixed to a door, window or car using double sided adhesive tape. The CD 4093 is a Quad NAND gate with 4 identical Schmitt Trigger gates. Each gate has two inputs and one output. When both inputs are HIGH, the output goes LOW.

Only one gate is needed for this project. The two inputs are connected together and the output is connected to R2 and the LED. Capacitor C1 charges through resistor R2. Since R2 is a high value, the charging current will be low (around 50 microamps). During this time, LED will be off. When C1 charges to 66% of the supply, the gate changes state and the output goes LOW. The stored energy in the capacitor lights the LED. The flash rate depends on the value of R2 and the brightness depends on the value of C1. If a high bright red transparent LED is used, the flashes are visible even in day light. Resistor R1 reduces the current to the circuit to prevent current loss from the battery. This is untrue. The unused inputs are tied HIGH and the chip gets its charging current via the inputs. You can remove the supply to pin 14 and the chip works exactly the same.  This is normally NEVER done.

It is important to note that all the inputs and outputs of the unused gates should be connected to the positive rail to prevent floating.  This is entirely UNTRUE.  You must NEVER connect the outputs to either rail  - only the inputs to the POSITIVE RAIL

The circuit above is very poorly drawn. You cannot work out what the chip is doing.
The following circuit has an improved layout:


Tying the inputs HIGH reduces the current by as much as 70%

 

The original circuit has been copied from the following circuit:

The circuit above can be simplified as the 1N4148 diode is not needed. It is designed to prevent discharging the electrolytic when the gate changes state, but the flash rate is very brief and the 330k will have no effect on discharging the capacitor.
The 470R is not needed as we want all the energy from the electrolytic to drive the LED.


Another faulty design by D.Mohan Kumar  The main fault is the circuit takes 10mA via the 555 when sitting around doing nothing.

A better design uses transistors and the circuit takes less than 20uA.

The LED will be illuminated for about 15 seconds via the clap of the hands.
The quiescent current is about 20uA, allowing 4 AA cells to last a long time.
The circuit takes about 20 seconds to reset after the LED goes out. The 100u discharges through the 27k, 100k and 10k resistors.

Another faulty design by D.Mohan Kumar  The main faults are the lack of a "Stop Resistor" on the Photo Darlington transistor and the wrong symbol for the transistor. 


IR Beam Alarm

The circuit can be set-up on one side of a doorway to detect persons entering or leaving a premises. The other side of the doorway should have a mirror to reflect the beam. The piezo buzzer is an active device with a transistor drive-circuit enclosed in the unit. It produces a tone when a voltage is applied.

Another design by D.Mohan Kumar  The main fault is the unusual layout. Every type of circuit has a prescribed way to lay it out. This allows you to instantly recognise the type of circuit.
He is his layout:

Here is his terrible description of how the circuit works. Remember, he is a University Professor in India with 20 year's teaching experience and has presented over 200 circuits to all sorts of magazines in India. His description is completely flawed and meaningless.  It is the worst description I have ever seen. It teaches nothing and leaves out the important functions of how the circuit changes state. It is no-wonder Indian students have no understanding of how circuits work. The teachers themselves are confused.

Here is a simple Dancing LED circuit. The LEDs turn on/off alternately giving a dancing appearance. It is a simple astable multivibrator using two NPN transistors. It works on a principle of charging and discharging capacitors C1 and C2. Current from the positive of the battery flows through the first set of LEDs D1 - D3 to the collector of T1 through resistor R1. Resistor R1 limits current through the LEDs to protect them. The current through R1 charges capacitor C2. It then discharges through the base of T2 and resistor R4. This gives base current to T2 and it conducts. As a result, the second set of LEDs D4 - D6 light as the current flows through T1. Capacitor C2 again charges and the cycle repeats.
The same thing happen on the other side. This gives alternate flashing of the LEDs.

This description is utter rubbish. No-where does he mention the charging of C1.   

It is very difficult to see what the circuit is doing, when it is drawn incorrectly.
All circuits need to be drawn correctly so it is easy to see what it is doing.
When it is drawn correctly, you can see it is an astable (non-stable - free running) multivibrator.

 

Here's how the circuit works:
When the power is applied, both transistors have 47k resistors connected to the bases and both will start to come on. But one transistor will come on slightly faster than the other. This is called "racing" and the transistor that comes on slightly faster will pull the positive lead of the uncharged 10u towards the negative rail and this will reduce the turn-on voltage to the other transistor.
We now have the case where the left transistor is tuned ON via the second 47k and the left three LEDs are illuminated.
How, here the part that no-one has mentioned before.
The first transistor is also turned on by the second 10u charging via the three LEDs and 100R. This add to the turn-on of the transistor and adds to its "saturation" In other words it is tuned on more than just the effect of the 47k resistor.
This causes the collector-emitter voltage to be very low.
During this time the first 10u electrolytic is also charging (in reverse) and the second transistor is starting to see base-emitter voltage.
The second 10u is getting nearly fully charged and the "charge-current" starts to reduce.
This turns off the first transistor very slightly and the collector voltage rises. The positive end of the 10u rises and takes the negative lead higher. This action starts to turn on the second transistor and the collector voltage falls. The positive lead of the second 10u falls and takes the negative lead lower. This starts to turn off the first transistor and the first 10u rises to turn on the second transistor. This is how the two transistors change state.
We mentioned the first 10u was initially charged in the reverse direction by about 0.5v as the positive end was 0.2v about the 0v rail and the negative lead had about 0.7v on it.
When the first transistor turns off, the positive lead rises due to the characteristic volt-drops of the three LEDs and it starts to charge with a high current. This current firstly discharges the electro and starts to charge it in the other direction. This fully saturates the second transistor and makes the collector-emitter voltage very low.
 As the charging current reduces the second transistor becomes less tuned-on and the collector-emitter voltage rises. As we said above, this is how the two transistors change state.

This description is a far-cry from Professor Mohan Kumar's understanding of the issue.
The operation of this type of circuit is much more complex than you think.



Here's Another stupid design by D.Mohan Kumar
Here is is his description. He has absolutely no concept of how to explain the operation of the circuit.
This simple circuit can protect your luggage. It gives a loud warning beep when somebody tries to take your luggage.
A thin wire loop is used to keep the alarm off in standby mode. Preset VR1 is used to adjust the voltage at pin 3 of IC1
(this is NOT true - how can you adjust a voltage when a wire loop has zero resistance). IC1 is used as a comparator. Its non inverting input (pin3) is connected at the junction of wire loop and VR1. Its inverting input (pin2) is connected to a potential divider comprising R1 and R2 which gives half supply voltage (4.5 V) to the inverting input. Output of IC1 is connected to the base of PNP transistor BC 557 through the current limiting resistor R3. Resistor R4 limits current to the LED and Buzzer.

 

Normally current passes through the wire loop to pin 3 of IC1. So pin3 of IC1 gets more current than pin2. (this is NOT TRUE) This makes the output of IC1 high. The high output from IC1 inhibits the transistor from conducting. So LED and buzzer remain off. When somebody takes the luggage, wire loop breaks and current to pin 3 drops (The inputs are voltage controlled - not current controlled).  As a result voltage at pin 3 becomes zero and the output of IC1 becomes low. This allows T1 to conduct and buzzer sounds and LED lights. Tie the wire loop around the luggage so it breaks when the luggage is pulled.

The circuit above is over-designed, and it has a major fault. If the 10k pot is turned fully, it will create a short-circuit with the wire loop. In addition, the circuit takes a high current when not doing anything. In addition, the buzzer will never work as it sees only 1.8v.
This circuit has never been tested and Professor D.Mohan Kumar should stop making a fool of himself by publishing such rubbish on his website. It's no wonder Indian students look to overseas to get an education.
Half of Talking Electronics 9GB of traffic each day comes from India. It is unbelievable how India is advancing in this technological age.
I get emails from Indian businessmen who was to make cheap LED lights for villagers who don't have electricity, so you can see the enormous range of social benefits in this country.
On the other hand I see Indian lecturers on videos trying to teach electronics. I can hardly understand what they are trying to present. It's no wonder students don't understand.
Now, back to the answer for this circuit:
All that is required is two transistors and three resistors, as shown in the circuit below:


LUGGAGE GUARD

Here's Another disaster by D.Mohan Kumar

He claims to test all his circuits. This is one of the poorest designs for an FM transmitter. See our article Spy Circuits  for a full description of how to design this type of circuit.
Remember, this is a University Professor from India, describing the circuit. It is the most inaccurate description I have come across and quite a useless design as far as FM transmitters are concerned. .

Here is his wording for the details of each component.
1. Condenser MIC
The condenser MIC is used to pick up the sound signals. The diaphragm inside the MIC vibrates according to the air pressure changes and generates AC signals.
(He does not mention anything about the FET transistor inside the case. He knows nothing about this type of microphone). Variable resistor VR1 adjusts the current through the MIC and thus determines the sensitivity of MIC. (If the 10k is turned fully it will completely damage the microphone. Very dangerous. A very poor design). The condenser MIC should be directly soldered on the PCB to get maximum sensitivity (rubbish) . Sleeving the MIC inside plastic tubing can increase its sensitivity enormously.

2. Decoupling Capacitors
C1 is the first decoupling capacitor
(It is a coupling capacitor not a decoupling capacitor) impedes the different frequencies of speech signals. (not an accurate description).  C1 modulates the current to the base of transistor. The 4.7 uF capacitor isolates the microphone from the base voltage of the transistor and only allows alternating current (AC) signals to pass.
A large value capacitor induces bass (low frequencies) while a low value one gives treble (high frequencies). Capacitor C2 (0.01) act as the decoupling capacitor
(not really. It actually couples the base to the rail). Capacitor C3 across the transistor T1 keeps the tank circuit vibrating. As long as the current exists across the inductor coil L1 and the trimmer capacitor, the tank circuit (coil-trimmer) will vibrate (oscillate) at the resonant frequency. When the tank circuit vibrates for long time, the frequency decays due to heating (the losses are due to magnetic losses and losses in charging and discharging the capacitor). Presence of the capacitor C3 prevents this decay. A capacitor between 4 and 10 pF is necessary. (Poorly presented and reasonably inaccurate - he has not told you anything).

3. Resistors
Variable resistor VR1 restricts the current through the MIC. The voltage divider R1 and R2 limits the base current of T1
(not really - it puts 3v on the base) and R3 forms the emitter current limiter. The given values are necessary for the 2N 2222A transistor. (He has not mentioned the transistor is a common-base stage).

4. Transistor
2N 2222A is the common NPN transmitter used in general purpose amplifications. It has maximum power rating of 0.5 Watts. Over-powering of 2N 2222A can generate heat and destroy the device. So maximum power output should be around 125 milli watt.
(the output of this circuit is about 30mW)


5. Inductor Coil
The inductor used in the circuit is a hand-made coil using 22 SWG (Standard Wire Gauge) enameled copper wire. The length, inner diameter, number of turns etc are the important values to be considered while making the inductor. Then only the inductor resonates in the 88-108 band FM frequency.
(badly worded). For this circuit, the coil radius was selected as 0.26 inches (outer diameter) and 0.13 inner diameter. Coil can be wound around a screw driver (with same diameter) to get a 5 turn coil of 0.2 inch long. Remove the coil from the screw driver and use the 5 turn air core coil. Remove the enamel from the ends and solder close to the transistor.

6. Trimmer capacitor
A small variable capacitor with a value of 5 -22 pF can be used to adjust the resonant frequency of the tank circuit. The variable capacitor and the inductor coil form the tank circuit (LC circuit) that resonates in the 88-108 MHz. In the tank circuit, the capacitor stores electrical energy between its plates while the inductor stores magnetic energy induced by the windings of the coil.
Tank Circuit
Every FM transmitter needs an oscillator to generate the radio Frequency (RF) carrier waves. The name 'tank' circuit comes from the ability of the LC circuit to store energy for oscillations. The purely reactive elements, the C and the L simply store energy to be returned to the system
(badly worded). In the tank (LC) circuit, the 2N 2222 A transistor and the feedback 4.7 pF capacitor are the oscillating components (badly worded).  The feedback signal makes the base-emitter current of the transistor vary at the resonant frequency (badly worded). This causes the emitter-collector current to vary at the same frequency. This signal is fed to the aerial and radiated as radio waves.

FAULTS
Apart from the 10k pot damaging the electret mic, the circuit needs a 22n across the battery. This improves the output considerably.
The 10k and 4k7 voltage-divider resistors put 3v on the base and the transistor operates in the range of about 3v.  Using a 9v supply is a complete waste of voltage. The circuit will work with a 3v battery and a single 47k base resistor.
The 4u7 can be 100n with no difference in quality.
The 100R emitter resistor is too low. It should be 470R.
Overall a terrible copy of a single transistor FM transmitter that can be found on the web with fewer mistakes than is design.
See our article Spy Circuits  for better FM transmitters.



Here's another disaster from Professor D.Mohan Kumar  who knows nothing about electronics' design:

20 kHz IR Transmitter
This continuous tone IR transmitter can be used in Broken Beam Detector system to activate an alarm. The circuit is a simple multivibrator and an IR LED driver output. This transmitter is useful in Photodiode and Phototransistor based IR receivers.


Faulty 20kHz transmitter - do not construct!

IC1 is designed as a simple oscillator using the components R1, R2 and C1. With these components, output pulses will be around 20kHz. The effective range of the IR beam depends on the peak current to the IR LED rather than the mean current. Around 100-200mA current is necessary to increase the transmitting range. Hence the output pulses from IC1 are fed to a LED driver circuit comprising T1 and T2. When output of IC1 is high, T1 is driven into saturation via R5 and red LED turns on. When T1 conducts, T2 turns on to drive the IR LEDs. When the red LED turns on about 1.8 volts develop across the LED and around 1.2 volts develop across R6 (10.2v develops across the 100R resistor - this is the BIG MISTAKE!). Thus T2 acts as a constant current generator (NO  . ... IT DOES NOT) and with R6, the peak current flowing through IR LEDs is around 170 - 200mA (1.2 / 6.8 =0 .176 A or 176 mA) (The calculations are WRONG!!!!  The current though the IR LEDs will be more than one AMP!!!!   They will be  DAMAGED!!! )  When the output of IC1 goes low, T1 and T2 turns off and IR LEDs switch off.
IC1 is the CMOS low power version of 555 IC. It works on 5 -15 volts DC.

More faults:
1. The circuit may not turn off as the 7555 goes low but not below 0.6v.
2. What is the function of the 22k.  It should be on the base of the BC547 transistor.
3. C2 is not needed.
4. 6R8 is too low. It will instantly damage the IR LEDs.
5. Both transistors will be instantly damaged when the circuit turns ON. The current will be about 1.5 amps!!     -  through the 6R8 and emitter-base of the BC557 and collector-emitter of the BC547.
I have contacted
Professor D.Mohan Kumar  more than 10 times with all his previous faulty circuits, and yet he continues to pump out this RUBBISH on his web pages.
He never tests anything and I don't know why he fails to respond to any of my corrections.
He is just an embarrassment to the electronics industry and it is fortunate we don't have many fools like this in the teaching profession.  He is possibly the WORST designer I have come across.  
At least he provides a platform for the rest of us to study and learn.
Remember -  don't make a fool of yourself. Test everything before you release it.
I have things sitting on the bench at the moment - turning ON and turning OFF to the sounds of the TV - just to prove a circuit works over a long period of time and remains working when the battery voltage reduces.



Here's another circuit from: http://www.circuitstoday.com/

Never connect two transistors so that current can flow from the positive rail to the 0v rail  via the junctions of the transistors. If you supply sufficient current into the base of the first transistor, a larger current will flow via the collector emitter leads of Q1 and this current will flow from the emitter to base junction of Q2. 
This current can overheat and damage the transistors.

The answer is to add a resistor called  a SAFETY RESISTOR as shown in the following diagram. The resistor will not alter the performance of the circuit but it will limit the maximum current:

Here's a circuit from CIRCUIT DIAGRAM.ORG The website is owned by: Salman Feroz Ali
from Karachi   Pakistan

http://circuitdiagram.org/automatic-nimh-battery-charger-circuit.html

Here is his description of the circuit:

The diagram is a fully automatic NiMH battery charger circuit using a 7805 voltage regulator which is providing a constant current to charge 2 NiMH cells. The LED works as a charging indicator so when the cells become fully charged the LED will go off. The circuit can charge two NiMH cells at a time. The circuit has four different current sources: 50mA, 100mA, 150mA and 200mA which is selectable with the switch S, so you can choose the current source for your cells. For example if you want to charge 500mA NiMH rechargeable cells you have to select 50mA current, for 1000mAh NiMH cells select 100mA, for 1500mAh select 150mA and to charge 2000mAh cells select 200mA.

The only problem is the LED will never illuminate. The second transistor is always ON. The circuit cannot detect the exact voltage when the two cells are charged. Since the charging current is about 1/10th the rating of the cells, it will take about 14 hours to fully charge flat cells.

Here's another circuit from CIRCUIT DIAGRAM.ORG The website is owned by: Salman Feroz Ali  from Karachi   Pakistan

The base of the transistor is connected to a 220n capacitor. How does the transistor work?

I think the circuit is supposed to be:

Here's another circuit from CIRCUIT DIAGRAM.ORG The website is owned by: Salman Feroz Ali  from Karachi   Pakistan.  His whole website is filled with faulty circuits. They provide a good lesson on how NOT to design a circuit.
Here is an FM Transmitter.  I have listed the faults with the circuit. You can see a discussion on FM transmitters HERE.

1. The symbol for a microphone is a carbon microphone. These are no longer available. The symbol should be an electret microphone.
2.Turning the 10k pot fully will destroy the microphone. It should have a 1k "stop resistor."
3.The 1k connected to the 470n does nothing.
4. The coil should be 5 turns on 3mm former. The energy from the coil should match the energy from the capacitor.
5. The 62p does nothing.
6. The 3p on the antenna simply reduces the range.

Here's another circuit from CIRCUIT DIAGRAM.ORG

Putting a 1k in series with a 5v6 zener does absolutely nothing. It is just wasting current though the zener.

Here's another circuit from CIRCUIT DIAGRAM.ORG

The circuit looks ok, until you work out the voltage available. The supply is 9v DC.  Why have a bridge for 9v DC? The voltage into the regulator will be 8v. The voltage out will be less than 7v. The voltage after the 1N4007 will be less than 6.3v. The voltage drop across the 16R resistor will be 1.6v for 100mA charging current. This leaves less than 4.7v.  The battery needs at least 7v to start the charging process as the battery quickly generates a "floating charge" (or floating voltage) of 7v. The 9v DC transformer will not work.
A 9vAC transformer will only deliver 150mA as 2.4v will be dropped across the 16R resistor and this uses up all the available 13.5v.
The circuit really needs 12v AC.

Here's another circuit from CIRCUIT DIAGRAM.ORG

The Intercom circuit does not work. The main fault is the 220k bias resistor on the base of the second transistor. The value is too low. It should be 1M. 


INTERCOM (circuit does not work)


INTERCOM (this circuit works)

By changing the position of the base resistor on the second transistor to create self-bias, the circuit became very sensitive and you could hear a pin drop.
The 100n feedback capacitor must be reduced to 330p to produce the call-tone.
The other changes were only minor and the 470R on the output transistor reduced the quiescent current from 25mA to 15mA. 


Connecting two intercom circuits to produce two-way talk

Here's another circuit from CIRCUIT DIAGRAM.ORG

The circuit works but the layout needs to be changed so it conforms with "standard layout."


THIS CIRCUIT DOES NOT WORK!

We can now see the LED does not have a dropper resistor and will be instantly damaged:


THIS CIRCUIT DOES NOT WORK!

But the circuit above DOES NOT WORK.
The circuit must be redrawn as follows:


THIS CIRCUIT WORKS!

The circuit works on the basis of a high-gain amplifier being driven into saturation (fully turned-on), firstly by the very small amount of current delivered by the 2M and then from energy stored in an electrolytic.
When the energy from the electrolytic has been fully delivered, it cannot keep the amplifier fully turned on and it turns off slightly. This action removes the "turn-on" effect from the electrolytic and the amplifier begins to turn off. This action continues until the amplifier is fully turned off and is kept in the off state while the electrolytic begins to charge. The off-state is very long and the on-state is very short. This is how the LED produces a brief flash.

Here is the technical description of the operation of the circuit:
When the supply is connected, both transistors are off and the 2u2 electrolytic charges via the 2M2 resistor and 68R. When the voltage on the base of the first transistor rises to about .6v, it begins to turn on and the resistance between its collector-emitter terminals is reduced. This allows current to flow in the collector-emitter circuit of the second transistor via the 1k resistor. The second transistor conducts and the LED is illuminated. The current through the LED is limited by the 68R resistor and at this point in the cycle a voltage is developed across the 68R. The negative end of the electrolytic is "jacked up" by this voltage and the positive end pushes the charge on the electrolytic into the base of the first transistor to turn it on even harder. In a very short time all the energy in the electrolytic has been delivered and it cannot hold it ON any longer. The transistor turns off slightly and this has the effect of turning off the second transistor a small amount. The LED begins to turn off and the voltage across the 68R reduces. The negative lead of the electro drops a small amount and so does the positive lead. This action continues until the first transistor is fully turned off. This turns off the second transistor and the LED is extinguished. The cycle starts again by the 2u2 charging. The charge-time is considerably longer than the discharge time and this gives the LED a very brief flash.

Why doesn't the original circuit work?
The 50u is connected to the top of the LED. When the two transistors start to turn off, the negative end of the electrolytic has to be pulled to the 0v rail and this will pull the positive lead down too. The "pulling-down" effect has to be greater than the charging provided by the 2M2 so the two transistors get fully turned off.
In the original circuit, the LED can only pull the right lead down to 1.7v and this allows a turn-on voltage to exist on the base of the first transistor. The transistor remains ON and the LED remains illuminated. Two more points to note:
The 2M2 must be a high value otherwise the circuit will not turn off as it will start to charge the electrolytic during the time when it is being "pulled-down."
The 68R cannot be higher than 100R otherwise the circuit will not turn off as the current through this resistor this resistor must be a considerable amount so that the PNP transistor needs a fairly high base current . When the transistor does not get this current, the circuit is able to turn off.

Here's another faulty circuit from Professor D.Mohan Kumar  who knows nothing about electronics' design:

He says:
The battery charger section comprises a 0-9 volt 500mA step-down transformer, a full wave bridge rectifier (D1 through D4), a 470u (C1) smoothing capacitor and 7808 voltage regulator IC. IC1 provides regulated 8 volt DC for charging the battery. Resistor R1 restricts the charging current to 100-110mA
(wrong).
LED driver circuit uses a medium power PNP transistor BD 140 which acts as an automatic switch. When the mains power is available, battery is charged through D6 and R1. At the same time, voltage from the bridge rectifier passes through D5 and R2 to the base of T1 to inhibit its conduction. So both the LEDs remain off. When the power fails, T1 turns ON and the LEDs turn ON
(wrong).


High bright LED Emergency Lamp Circuit

Mistakes in the circuit:
When the power fails, the transistor will never turn ON. D5 will not pull the base of the transistor to 0v rail.  The transistor will just be left "floating."
Resistor R1 restricts the charging current to 10mA. It should be 10R.  The 100R resistors to the 1 watt LEDs will only allow allow 30mA, or 100mW to the 1w LEDs. R3 and R4 should be 10R  1 watt.
The value of resistor R2 is too high. It will never allow the transistor to pass 600mA.
This circuit has never been tested.
The following circuit is the simplest design needed.

The charging current is about 20-30mA. It will take about 7 days to charge the battery and this will allow illumination for 5 hours, once per week.
A charging current more than 50mA will gradually "dry-out" the battery and shorten its life.
If the project is used more than 5 hours per week, the charging current can be increased.
The 220R charging resistor can be reduced to 150R or 100R.

Here's another circuit from Professor D.Mohan Kumar:

He writes:
This circuit protects home appliances from low voltage. When the voltage in the AC lines drop below the normal voltage, the circuit cuts off power to the appliance and keeps it off till the normal voltage resumes. It is useful to protect appliances like fridges which may over heat if the line voltage drops below 180 volts. Low voltage increases the current consumption in a fridge due to over-stressing the compressor.
The circuit uses a zener controlled relay driver to switch the load. A 9v 500mA transformer provides power for the circuit. Any voltage drop in the primary of the transformer reflects as a corresponding voltage drop in the secondary. So when the line voltage is normal (around 230 volts), the zener conducts to bias the relay driver transistor T1.When T1 conducts (as indicated by the glowing LED), the relay is energized and switches the load ON. The load is connected through the Common and Normally Open (NO) contacts of the relay.
When the mains voltage drops below 200 volts, a corresponding voltage drop occurs in the secondary of the transformer and the zener turns off. T1 then turns off to de-energize the relay. The load remains off till the line voltage increases to normal.
Capacitor C1 gives a short lag before the zener switches to avoid false triggering of T1 if the line voltage fluctuates.
It is necessary to adjust the trigger point using the preset VR. Check the line voltage and if it is above 200 volts, slowly adjust the wiper until the LED turns on and the relay energizes.

Three faults/improvements are:
1. The zener diode is not needed.
2. R2 is not needed.
3. C3 is not needed.

Here's a  REGULATOR circuit that has not been tested:

At 12.8v input the voltage across the 1k base-bias resistor will be 0.1v. The current through the resistor will be 0.1mA. If the transistor has a gain of 100, the output current will be 10mA !!!
At 20v input, the output current will be 800mA.
The 5 amp fuse will not protect anything.

Here's a CLAP SWITCH with a few problems:


The electret mic should not be connected to the base without a load resistor.
Pin 2 of the 555 should have a 1M resistor to the positive rail to prevent false triggering. Pin 2 is high impedance and it is not capable of charging or discharging the 100n.

Another faulty circuit from Professor D.Mohan Kumar:

1. The 3v1 zener is connected across the battery  !!!
2.  Where can you get a 3v1 zener????
3. Where can you get AC187 ?   They have not been manufactured for 15 years!! 
4. They cost $5.00 each !!!!!!   Plus postage
5. What is the impedance of the speaker?

This is another untried circuit from Professor D.Mohan Kumar. He keeps pumping out worthless circuits and destroying the value of the internet.

At least you are being shown how NOT to design a circuit. And this is a most-valuable lesson.



Here's another faulty circuit from Professor D.Mohan Kumar:

The first stage is badly designed.
A piezo element is called a HIGH IMPEDANCE DEVICE. It can produce a voltage when it is hit or bumped but the current is very small.
The load resistance on the first stage is 1k. This is classified as a LOW VALUE.
A transistor merely amplifies the current it receives via the base and allows about 100 times more current to flow in the collector-emitter circuit.
To produce a voltage-drop across the 1k resistor of 1v, 1mA must flow. The base must receive 10uA.  And the voltage from the piezo element must be greater than 600mV.
A piezo element produces about 20mV to 50mV when it detects sound. A heavy strike may produce a few hundred mV. But this circuit is very unreliable.
The author has then amplified the signal produced by the first stage.
Instead of making the first stage very insensitive, he could make it more sensitive and this would require only one stage.
See "Knock Knock Doorbell" for the correct way to design a piezo-detection stage.
The biasing components for the second stage are all incorrect.
The 10k pot is in the wrong place.
Sensitivity adjustment is normally placed as the collector load. It should be 100k.
The 4M7 is too high. It should be 1M.
Diode D1 is not needed. It serves NO purpose.
In fact it should not be used. It puts a high-impedance on the input of the IC when at rest and this will allow noise to enter the chip.
It's just a badly-designed circuit.
This type of design shows a complete lack of understanding of electronics.

Here's an absolute DISASTER from Professor D.Mohan Kumar:

This is the most dangerous power supply I have ever seen.
The two capacitors on the front end produce a voltage-divider. To see what I mean, remove all the power supply components, including the bridge, and connect the two 220n capacitors together as shown in the following diagram:

The peak voltage of a 230v AC supply is about 320v. Two equal capacitors will produce a voltage equal to half the peak value, at their join, as shown in the following diagram:

This means the capacitor-fed power supply will always be 160v higher than the neutral - no matter which way the power supply is connected.
A normal capacitor-fed power supply has one side connected to the neutral and the output is classified as "ground-referenced." This means the capacitor power supply can be touched (although it is not recommended as the input connections may be around the wrong way and the supply is really 340v above "earth").
However the supply shown above is always "180v HOT" - no matter which way the supply is connected.  
Professor D.Mohan Kumar thought he was producing a safer supply.
This is what he says in his introduction:
It is totally isolated from mains supply using two capacitors in the phase and neutral lines. So the connected device is safe even if the phase and neutral lines changes.
We know how wrong he is!!!  - that's why a little knowledge is  DANGEROUS THING. 

Here's a circuit from RED CIRCUITS. It is a set of LEDs with a constant current section made up of Q1 and Q2.
When Sw2 is closed, Q3 increases the current to the LEDs

There are a number of faults with this circuit.
The 33k in the constant current section will allow only a very small current to flow. The resistor should be 33R for 20mA.  This gives 10mA for each string of LEDs.
The two diodes are not needed.
D2 should be replaced with a 10R resistor to increase the current though the LEDs when Sw2 is pressed. The voltage between emitter-collector of Q3 will be about 0.2v and the voltage across the 10R resistor will produce about 25mA for each string.
See The Transistor Amplifier for more details on Constant Current.

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