Diode D1 only needs to be 1N4004.
T1 does not need to be a power Darlington transistor. It can be BC547.
The output voltage will never reach 15v. The Darlington will turn on when the output voltage is 11.5v + 1.2v = 12.7v.  Yes a 12v zener starts to "leak" at 11.5v. No-one has ever told you that a zener starts to turn ON at a lower voltage. That's why they cannot be used in a circuit as shown above.  You need a resistor between the anode and 0v to remove the leakage at 11.5v and wait for the major breakdown at 12v. 
The output will NEVER rise above 12.7v
Suppose the circuit is supplying 100mA to the battery. The voltage drop across the 100R resistor will be = 0.1 x 100 = 10v !!
Can you see how absurd this circuit is?
It has never been tried or tested.
 

When the switch is changed from one voltage to another, the output instantly increases to more than 18v and will blow up many circuits.
The BC557 transistors are not needed as the resistors can be connected to the 0v rail via the switch.
However the 220R resistor (R13) is the resistor that should be changed and the circuit needs to be totally redesigned.
This circuit comes from Electronics For You. This Indian magazine is fooling the readers by saying they have electronic engineers and have tested the circuit.
I have asked for the names of the "electronic engineers."  They have NONE.
It's rubbish like this circuit that makes Electronics For You a total failure as a magazine. They don't have anyone to check the accuracy of a circuit and "just because it works" they publish it.
Both Professor Vidyasagar and Professor D.Mohankumar supply projects to Electronics For You and we have shown how incompetent they are. Professor Vidyasagar has now removed his website completely from the internet, after I sent him over a dozen emails with his faulty reasoning and circuits that don't work. At least he has woken up to the fact that you can't fool "ALL THE PEOPLE ALL THE TIME."
D.Mohankumar's website


Here's another impractical circuit from Electronics For You:

This circuit allows you enough time to reach your bed and lie down before the bedroom lamp switches off automatically.
Mains power supply is reduced by the capacitive dropper arrangement. A 1F capacitor (C2) handles the timing part. When switch S1 is pushed to ‘on’ position for a few minutes, bulk capacitor C2 charges and the stored energy drives the internal LED of IC MOC3041 through resistor R4. 

Who is going to push the switch for "a few minutes"  
1F @ 5.5v is $3.00 to $19.00 depending on the supplier. eBay is the cheapest - it always is.
Where can you get a 1,000,000u  (1F) capacitor @16v ??

The circuit only charges the supercap to 8.5v and thus half the possible energy is wasted. The only 1F @16v supercap on the web is very large and costs about $100. What an impractical circuit. Doesn't Electronics For You test anything?

1F capacitor can store one coulomb of energy @ 1v.  Thus 1F @ 5.5v supercap can store 5.5 coulombs of energy.
But it can only  be discharged from 5.5v to 2v (as this is the characteristic voltage across a LED).   Thus the cap can supply about 3 coulombs.
1 coulomb is 1 amp x 1 second  
If we allow 20mA for the LED, it will be illuminated for (3 x 1,000) / 20 = 150 seconds.
It will take as long to charge the supercap as discharge it. The charging current is about 20mA and the 9v zener does not come into operation until the cap sees 8.4v.



Here's another circuit that doesn't work, from Electronics For You:

The LED illuminates when T1 is not illuminated.
Two major faults:
1. Current will flow through the 10k and T1 when illuminated (about 3mA) and this will flatten the battery.
2. What is the purpose of the IC ????  It serves NO purpose. Simply connect the LED to the collector of the transistor via a low value resistor.
Another pointless circuit from Electronics For You.

Here's another fault from Electronics For You (April 2013 issue):

The reset-line of a CMOS 7555 IC is connected to the gate of a FET and it requires almost no current to provide the reset feature.
The reset line is active LOW and this means any voltage above about 1v DOES NOT reset the chip.
When the voltage falls below 1v, the chip is reset. 
Even though NO CURRENT is required by the line, you MUST pull the line low via a resistor so that the gate sees a LOW VOLTAGE.
In the circuit above, the transistor pulls the gate HIGH to keep the chip activated, but when the LDR sees illumination, the voltage between the base and emitter is less than 0.6v and the transistor "turns off."
This makes the collector lead a HIGH IMPEDANCE LEAD and it is just like removing the transistor from the circuit. This leaves the reset line "floating" and there is nothing pulling the line LOW. It requires 100k to 1M from pin 4 to 0v.
This is a major fault in the design of the circuit due to the author not knowing the internal features of the chip.
Electronics For You also failed to pick this up because they have no technical staff to diagnose and check a circuit.
This leaves the experimenter in the invidious position (likely to arouse or incur resentment or anger in others) of finding the circuit does not reset and thinking he has made a mistake.

Here's another fault from Electronics For You (April 2013 issue):

7-segment displays should have a current limiting resistor on each line (a,b,c,d,e,f,g) so that each number (0,1,2,3,4,5,6,7,8,9) will illuminate with the same brightness.
In the circuit above, the common line has a 680R resistor and when a single segment is illuminated, 4mA will flow through the segment. When two segments are illuminated, each segment will get 2mA. When all 7 segments are illuminated each will get 0.5mA. The display will get duller and duller.
Obviously the circuit has never been tested.
Each line should have a 220R resistor and the common line is connected directly to 5v.
This project has been designed by Sani Theo, the Technical Editor for the magazine.
You can now see why the magazine has so many faults.

Here's another untried, untested circuit from Electronics For You (April 2013 issue):

The Gas Sensor is supposed to be connected DIRECTLY to 5v, not via 100 ohm resistor.
The Gas Sensor has a heater that draws 150mA @ 5v (750mW) and if a 100 ohm resistor is added, the heater will take 37mA. (the Gas sensor heater has a resistance of 33 ohms).
This circuit has never been tested and may or may-not work with 37mA heater current, but it will certainly take longer to make a reading because the idea of the heater is to burn the air-gas mixture and take a resistance reading from a ceramic substrate.
This circuit has been supplied by Professor Mohan Kumar and we have already identified him as being completely inept at designing electronics circuits.

Here's another untried, untested circuit from Electronics For You (April 2013 issue):

We start the discussion with just the LM317 3-terminal regulator.
It is designed to produce a voltage of 1.25v above the voltage on the adjust pin (pin 1) and it monitors this voltage accurately.
This means the output voltage will be 1.25v when the adj pin is connected to 0v.
If we add a 180R resistor and 1k pot, we can "jack-up" the output voltage until the output is about 2v less than the input. (This 2v is needed by the circuit inside the IC.) If this voltage is delivered to a load, the chip will deliver up to about 1.5amp.
Now we add the 10 ohm resistor.
This resistor is called the CURRENT LIMITING resistor and normally it is added as shown but the 180R and 1k pot is removed and the adj pin is connected between the 10R and diode. 
For a 10R resistor, the maximum current is 125mA
When this is done, the output voltage will rise to any voltage (that is about 2v less than the input voltage) and as the voltage rises, the current through the load will increase and also through the 10R resistor, (because they are in series).
As the current through the 10R increases, the voltage across it increases and when it reaches 1.25v,  the circuit inside the IC detects this and the output voltage stops increasing.
When we add the 180R and 1k pot, we have an unusual situation.
We can increase (jack-up) the voltage on the output to about 8v but when we add a load, the 180R alters the current limiting to 35mA.
If the circuit requires 35mA charging current, a 33R resistor can be used and the 180R and 1k pot can be removed.
You cannot add both features at the same time. If you try to set the output voltage, it will decrease rapidly as current is drawn.
Another untried, untested circuit from  Electronics For You.

Here's another faulty circuit from Electronics For You website:

You have to work out the total impedance of the circuit.
This is 31k for the .047u + 0.47u (in parallel) and 39k for R2.
This means the current through the circuit will be 3.4mA
How bright will the LEDs be with 3.4mA  !!!!

Here's another faulty circuit from Electronics For You website:

The output of a 555 can only deliver 200mA to a load and this means the impedance (resistance) of the load must be 60 ohms or more for 12v supply.
However we have another factor that must be taken into account. The chip can only dissipate 600mW.
This means the output must be derated to 50mA when the supply is 12v.
Neither of these factors have been taken into account with the circuit above.
If we take a frequency of 500Hz, the capacitive reactance of 470u is LESS THAN ONE OHM !
This makes the output less than 10 ohms !
Another untried, untested circuit from  Electronics For You.

Here's another terrible circuit from Electronics For You website.
This circuit has been designed by Sani Theo - one of the Technical people at Electronics For You. He has absolutely no idea how to design a circuit:

This circuit has lots of mistakes. Apart from the fact that it does not work, here are the technical faults:

1. The output of the 741 sits at half-rail-voltage. This is a very bad voltage to deliver to a digital gate as it causes the gate to oscillator at high frequency. The output of the 741 should go to a very low voltage to provide a genuine LOW to the Inverter gate.
2. The output of pin 6 is low impedance. There is no point in placing a 10k on pin 6 to 0v rail.
3. C2 serves no function. Pin 6 is low impedance and the electrolytic has no effect.
4. Gates N2 and N3 have no effect on the circuit and can be removed.
5. R5    22R is far too low to protect the output of the Inverter gate.
6. IC3 needs a 1k on the output as shown in the specification sheet.
7. R6 is far too low for the position shown. It should be 4k7 to 10k.
8. T2 is a Darlington Transistor and should be shown as a Darlington.
9. There is no current limiting resistor between T1 and T2 and more than 100mA will flow though the base of T2 as wasted current. 
10. There is no delay to allow the music IC to produce the Ding Dong sound.
That's 10 poor designs in a simple circuit from someone who touts himself as a Technical Editor. What a disaster.

Here's faulty circuit from Electronics For You from the year 2000:

The transistors are directly connected to the power rail via collector-emitter and base-emitter junctions and when the transistors are turned ON, these junctions have a very low voltage across them. This means a very high current will flow.
The actual current will depend on the base current from a previous stage and the CD4047 can deliver 10mA. If the transistor has a gain of 100, this means a wasted current of 1AMP will flow through the transistors, and especially the base-emitter junction, which is not designed to pass this high current. 

Here's a faulty animation from:
http://electronicsgurukulam.blogspot.com.au/2012/07/cathode-ray-tube-animation.html

The animation is entirely wrong.
The beam is deflected in the neck and then it travels in a STRAIGHT LINE to the screen.

Here is over-designed circuit by D. Mohankumar:

It detects if a cable is carrying the "mains."

All you need is 3 transistors and you can add a piezo diaphragm to hear the "mains."

Here is a clear description of how the circuit works:
Circuit C is a DIODE PUMP (VOLTAGE DOUBLER) from an AC source (such as the "Mains").

The circuit takes a number of cycles to get up to full output voltage and this is how it works:
The best way to consider an AC waveform is this: The bottom rail is 0v (or neutral) and the voltage on the top rail rises from 0v to 15v (in this case). The voltage then decreases to 0v and then the top rail jumps to below the bottom rail and it increases in the negative direction until reaching -15v. It then decreases until both rails are 0v and then it goes above the negative rail and increases to +15v.
We start with the input voltage rising and because the 22u is uncharged, the 10u starts to charge as soon as the 0.6v across the top diode is reached.
The 10u charges to about 10v and puts about 5v on the 22u - this is how the 15v is divided-up on the first cycle.
When the AC reverses polarity, the top diode does not have any effect but the lower diode becomes forward biased and it allows the 10u to charge to about 15v.
When the AC reverses again (so that the top input becomes positive), the 10u has 15v on it and incoming the AC voltage adds another 15v.
This means the positive lead of the 10u is 30v above the lower rail and it charges the 22u to about 15v - this is how the voltage is divided between the two electros in the second cycle.
This happens a few more times and eventually the 22u gets charged to 30v (minus 2 x 0.6v diode drops).
After 5 or more cycles, the 22u has about 30v across it and the 10u keeps "topping up" the voltage as follows:
Say the 10u has 14v across it, when the top input of the AC becomes negative, the 10u immediately goes to a position below the 0v rail and the diode connected to the 10u allows it to be charged to 15v, (the top diode effectively comes "out-of-circuit" as shown in diagram D. The circuit only receives a "charging-pulse" during the time when the top rail of the AC voltage is NEGATIVE. It delivers this energy to the circuit when the top rail of the AC is POSITIVE.

This PRE-AMPLIFIER circuit will not turn ON because the voltage on the base is 50mV and it should be 650mV.
The voltage-divider resistors on the base are incorrect. The 4k7 should be at least 47k. 
Using the correct ratio of base resistors will turn the transistor ON but the actual voltage on the collector will depend on the gain of the transistor. Using 3 resistors in a semi-bridge arrangement is not a good idea. It should have an emitter resistor bypassed with a 100u electrolytic.
It is much better to use a self-biasing stage as shown in the second diagram to produce approx mid-rail voltage on the collector.  

I have never seen  a more-ridiculous circuit in all my life. Apart from being over-designed, it contains a number of technical faults.
A BC549 will not pass more than 100mA at the best of times and in this circuit the base has a 2k resistor so the current is limited to less than 100mA. It is not a "power transistor" and has no current-carrying capability. Maybe a BC338 with 1amp capability is a better choice.
The charging-current through the inductor is an indication of the energy that will be available when the transistor is turned off and in this circuit the charging-time can be adjusted.
You only need to allow current to flow through the inductor until it is saturated. Any further current-flow is wasted. It is impossible to determine this without accurately measuring the flux produced by a totally isolated secondary winding. And this feature is not available in this circuit.
See our Joule Thief article for correctly-designed driver circuits.
What is the point of transistors Q4 and Q5? They serve NO PURPOSE.  They simply turn ON and allow energy to flow to the 10u and LED. You cannot pass all the energy from the inductor by simply adjusting the mark-space ratio when you are using the same adjustment for the "charging operation." It just needs a high-speed diode to pass the energy.
The inductor is overly complex. If you turn off the transistor quickly, the back emf (high voltage) produced will be sufficient to drive a white LED. You don't need and extra winding.
The 510 ohm resistor on the collector of one of the multivibrator transistors means current is being wasted for part of the cycle. 
The driver transistor should be "driven" by a previous stage and not simply "pulled high." What the designer has done is analogous to turning off a car's headlight by shorting across the battery. 
The characteristic voltage-drop across a white LED is between 3.3v and 3.6v. He suggests 3v and it indicates  he has not driven the LED to its capability.   He has used a 1 watt LED and delivered 30mA.
The 10u across the LED has very little effect as it gets charged to 3.6v and when the voltage drops to 2.8v, the LED is not illuminated. This small amount of energy from a 10u will hardly be noticed.
The designer claims the output of the multivibrator is triangular, whereas a multivibrator is known as producing square waves. 
The circuit below turns ON the transistor HARD to produce the highest output. The circuit above turns on the transistor very feebly and it will have a high collector-emitter voltage that represents wasted energy.
There is no point in having a tapped inductor. By driving the transistor correctly, a single inductor will produce sufficient voltage to drive the LED. The inductor used in the circuit is expensive ($5.00) and a simple hand-wound coil could be used.
Overall a badly-designed circuit that is overly complex and very inefficient. A good study in "how NOT to design a circuit."

When you see a simple single transistor circuit that performs the same function, you realise how stupid to try and guess the timing to "charge" the inductor. The following circuit does it automatically and by simply changing the number of turns on the transformer (and its size), you can increase the output. See the Joule Thief article for more details and a circuit using a transformer with an AIR CORE !

WET NAPPY ALARM

The main reason we included this circuit is due to the layout.
It is impossible to work out what the circuit is doing. The layout is just a jumble. The circuit is presented in the May issue of Electronics For You Magazine and is supposed to cater for beginners to show how a circuit works. I have no idea what the circuit is doing. How do you expect a beginner to learn from this layout ????

Here is the correct way to layout a circuit so the operation of each "building block" can be understood:

Each NAND gate is a 2-input Schmitt Trigger with the first gate wired as an inverter. The second Schmitt inverter is a high-frequency oscillator and the third gate is a low-frequency oscillator. These two oscillators are gated to the fourth block to drive a piezo diaphragm as beep-beep-beep.  
A 1k on the output is not needed and will only reduce the loudness of the piezo.

LED LAMP

Here we have the ridiculous situation of providing isolation via a transformer but producing the dimming feature via a rotary switch on the MAINS SIDE of the transformer.
The same switch could be placed on the secondary side with smaller wattage resistors to produce a much-saver circuit.
The purpose of the two 100 ohm resistors, 100n capacitor and 150mA fuse is unknown. [150mA fuses are very unreliable and when they are a slo-blo type, they do not burn out until 500mA flows.]
These items are not needed. The power lost in R2 is less than 5mW - why use a 2 watt resistor ????
The 10 ohms resistors are far too small to have any effect. We have already discussed using a low-value dropper resistor in series with LEDs in our 30 LED Projects eBook.
The whole circuit is badly designed and gives the wrong information to newcomers to electronics.
Sani Theo, the technical editor, doesn't know what he is doing and is just making a fool of himself and Electronics For You Magazine.
This is another stupid design from Electronics For You Magazine May 2013 issue.

When you are designing a circuit, you need to "THINK OUTSIDE THE BOX." This involves looking at the circuit and seeing if it can be improved. The circuit above is a typical example. It would not be accepted in Australia, where the rules for transformerless power supplies, switching the mains via rotary switches and, in fact, anything to do with power supplies, dimmers, battery chargers etc is tightly regulated and prevented from publication in hobbyist magazines.  
The rotary switch should be placed on the secondary side of the transformer where it will be perfectly safe to handle and the circuitry will be simpler.

All that is needed is a float with a pin at the top and bottom that moves the lever of a toggle switch. When the float falls it turns the pump ON and when it rises, the pump is turned OFF. The float can be a plastic Coca Cola bottle partly filled with water. A plastic pipe in the top and two pins to touch the switch. The pipe will need guides to all it to rise and fall.

This is another over-design from Electronics For You Magazine where they didn't "THINK OUTSIDE THE BOX."

MINI UPS

The 12v battery will drop 0.6v across D4 and a further 2v across the emitter-collector of the Darlington transistor T2. The output at "A" will be less than 9.4v and not 12v.
The 9v regulator will also drop out of regulation as it needs at least 2.5v across the IC.
This circuit is a failure as a back-up power supply.
What is the purpose of D3 ??? It is not needed.
The FULL CHARGE LED will come on slowly and not give an indication of the exact point when the battery is charged.
What is the point of ZD1, the 22k?  (20k) pot and T2 when the output has a regulator and 12v zener.
The whole circuit does not make any sense.

More Junk circuits from Electronics For You website:

This circuit detects when the kettle turns off. But the 47k will only provide about 2-3mA for the piezo buzzer and they require 10mA to 20mA.
The reed switch will be vibrating at 50Hz during the 5 minutes when the water is heating and this will eventually break the reed. They do not last very long. Each heating will create 15,000 vibrations.
Why have 2,200u to collect 3mA !!!

Just another JUNK circuit that has not been tested. 

This is a STUPID CIRCUIT:

If you push the wrong button you are not penalised. Simply push each button (in turn) 7 times and the lock will open.

ALARM

Here's another circuit from Professor D. Mohankumar email: mohanwordpress@gmail.com  that does not work.
The output of the op-amp goes LOW to turn on the buzzer. But the LED is around the wrong way and the transistor is an emitter-follower so that the buzzer will get less than 6v due to the voltage-drop across the LED, the base-emitter junction of the transistor, the drop across the 1k and the output of the 741 being slightly above the 0v rail.
He does not test ANYTHING !!!!!

MULTIPLEXING

Here we have a typical multiplexing arrangement where a microcontroller drives a 4-digit display.
Each display is accessed turn and it only gets turned on for 25% of the time. However you eyes have a feature called Persistence Of Vision (POV) that retains a previous image for a fraction of a second after it has been removed, so that the whole display will appear to be illuminated.
However to get a good brightness from the 4 displays, you have to drive them at 10mA to 15mA. The chip is only capable of delivering 10mA per output and it is really not suited to directly driving a display like this.
But when dropper resistors are included the drive-current decreases to a point where the display is almost useless.
That's the case in this example.  The LEDs drop 1.7v, the sourcing transistor drops 0.6v across the base-emitter junction and the micro has 0.3v between output and 0v rail. This means the dropper resistors have 2.4v across them.
The 10k base resistor can be divided by 100 (the gain of the transistor) and included as a 100 ohm resistor. This means the dropper resistors equal 430 ohms.
The maximum current that 2.4v can deliver via 430 ohms is 5mA. The displays will be very dull.
The emitter-follower used to drive each display should not have a resistor on the base. This transistor should be turned ON-FULLY so the 330 ohm resistors can provide the current limiting.
A very poor design with a number of faults. 

When no light falls on the photo transistor, it resistance is very high, the voltage on the base of TR1 is less than 0.6v and the transistor is turned OFF.
Capacitor C1 is charged via R3 and D1. This puts a voltage on the base of TR2 and the transistor is turned ON.
Full rail voltage appears across R6 and thus the voltage on the base of TR3 is less than 0.6v and the transistor is not turned on.
All the rest of the circuit is not active.
But a small current flows through R3, D1, R4, VR1, R5 and R6. This is an absurd way to turn off a circuit.
It's a bit like putting a short-circuit across a battery to turn of a globe or motor.
The circuit is effectively putting a short-circuit across the input of the switching transistor TR3 to keep the circuit OFF.
Here is a simpler circuit that takes NO CURRENT when the LDR does not detect illumination:

Electronics For You is still publishing circuits that do not work:

An inverter is contained inside the chip, between pins 11 and 10 with the input on pin11. When an inductor is placed between these two pins, an oscillation is produced and the frequency is determined by the delaying action of the inductor. But the inductor has a lot of "strength" in delivering this signal, (because the inductor has a very small resistance) so that if a finger is placed on pin 11, the amplitude of the signal is not decreased.
For a TOUCH PLATE to work, the signal must be very weak and only just able to be detected by the input of a chip. When a finger is placed on the plate, the signal is absorbed slightly by your body and the input registers "no signal."
This circuit has been tested and DID NOT WORK. Another Electronics For You failure.
Try 10k to 100k in series with the 100uH inductor. (You need to make the oscillator-pulses very weak so your finger will reduce the amplitude.)

Another Electronics For You circuit that does not work:

Apart from the circuit being far too complex, the circuit contains a major fault.

Here is the response to your claps:

Two claps: Turn Device 1 'on'
Three claps: Turn Device 1 'off
Four claps: Turn Device 2 'on'
Five claps: Turn Device 2 'off

The designer of the circuit failed to mention the fact that you cannot keep device 1 ON and turn ON device 2, because the output sequence of the 4017 controls the flip flops in the 4027 IC and as the pulses pass the flip-flop that controls device 1, the flip-flop is RESET (as the signal passes to the flip-flop that controls the second device).
On top of this, you need a lot of claps to perform a simple task.
A really messy design that DOES NOT WORK !!!
You can see a simple design using a microcontroller on our new website: Electronics Maker

DOOR ALARM
Here's a Door Alarm circuit that does not work.
The circuit takes 1mA when sitting around due to the 10k resistor feeding the reed switch and the output of the circuit (pin 11) goes HIGH when the alarm is not producing a tone and this turns on the transistor and causes a high current to flow in the speaker. The battery will go flat in an hour. 
It is impossible to work out the state of the gates when a block diagram is presented and all circuits should be drawn with gates, so you can see what is happening. The next circuit is drawn with gates, so you see the mistakes.

By making a few changes, as shown in the circuit below, it consumes less than 10 microamps when "sitting around."
Two resistors have been removed as they are not needed. The 10k has been replaced by 1M and the BC547 has been replaced by BC557.

Another Electronics For You badly-designed circuit, July 2013:
When the 22k mini trim pot is turned fully clockwise, the 9v supply will be directly across the FET inside the microphone and will possibly blow it up.

Another Electronics For You badly-designed circuit, July 2013:

Apart from the fact that the layout of the 7805 and driver transistors for the 3 digits on the display are all up-side-down and hard to follow, the PNP transistors for the display are badly configured. The drive-lines to the display should contain 220R resistors (7 x 220R) and the BC558 transistors should have no resistor.
The problem with the circuit above is this: As more segments on a particular display are illuminated, a higher current will flow though the 220R resistor and create a larger voltage-drop. This means the segments will get duller as more segments are illuminated.
The second poor design is the connection of the base of each driver transistor to an output of the microcontroller.
This arrangement is absurd, as a low on the base of each transistor will fully turn it on and not allow the current to pass to the segments of the display. I don't know how they expect the circuit to work. It has obviously never been tested. 
The next problem is the value of R11. The circuit suggests 15R. And the parts list suggests a 5watt high-power LED.
Let's go through the mathematics. If you have a supply of 16v DC from the rectification of the  12v - 0v - 12v transformer, the voltage across a 5watt LED will be about 3.6v. This leaves 12.4v across the 15R resistor and thus 825mA will flow. This gives the LED .825 x 3.6 = 3 watts. The wattage lost in the 15R resistor will be 12 watts!!
If you put two LEDs in series, the result will be: 3.6v + 3.6v = 7.2v The voltage across the 15R will be 8.8v. The current will be 590mA and the wattage will be 8.8watts  The wattage of the LEDs will be 4.2watts. Obviously the values provided by the author do not work. It is not good enough in the text to say: "select your own resistor." You need to provide details on how to select the value, especially when a high current is involved.

Another Electronics For You badly-designed circuit, July 2013:
This circuit contains 4 separate paths that will create a high current. We have shown one path consisting of semi-conductor junctions (with NO current-limiting resistor) and these junctions have a very low resistance when conducting. This means a very high current will flow and even though two of the three transistors are fully turned ON to drive the motor, the current-flow will be higher than required because no CURRENT LIMITING resistor has been included in the circuit. The current can be high enough to damage the transistors. This is a MAJOR fault.

Another Electronics For You badly-designed circuit, July 2013:
A protection diode is needed across the 7812 regulator to protect it from reverse voltage when the 12v battery is connected and the power from the mains is not present.
The 1,000u capacitor will be uncharged and the voltage on the output of the 12v regulator will be 12v when the battery is connected and may cause damage to the regulator. This 12v-difference may damage the regulator but if a 1N4004 diode is added, the voltage-drop across the diode will be only 0.7v and the regulator will be protected.

Another Electronics For You badly-designed circuit, July 2013:

The circuit above detects an over-voltage and trips an SCR to blow the 500mA fuse. This is a drastic way to save an over-voltage condition and is called "CROWBAR PROTECTION."
In 40 years I have never heard of a voltage regulator failing and producing an over-voltage, even though I have sold thousands of power supply kits and thousands of computer kits using a 7805 regulator.
So I think the circuit is solving a problem that does not exist.
However, apart from this, lets look at the circuit and see the bad design features.
Firstly, a 7810 regulator is very unusual. If you re going to produce a "general circuit," the components should be commonly-used values.
Secondly, the 15v transformer will deliver 20v to the regulator and this will produce 10 extra volts across the regulator. At 500mA this amounts to 5watts. This could be the cause of the regulator failing.
The 1N4007 diodes should be specified as 1N4002 or 1N4004 as 1N4007 are 1,000 volt diodes and are only specified when voltage spikes can be of this magnitude.
Now we come to the BC548 transistor. What is the function of this transistor? It is not needed. The zener can be connected directly to the gate of the SCR, saving 4 components.
The following circuit switches off the output when the voltage from the regulator rises above a specified value and reconnects when the voltage falls.
The voltage will rise if the bottom terminal of R2 is removed from the 0v rail.

Another Electronics For You badly-designed circuit, July 2013:

This is a very dangerous circuit. The output voltage will jump to 12v when the switch is moved from 3v to 5v.
DO NOT BUILD THIS CIRCUIT.
The switching should be between the adjust pin and output.

Another Electronics For You badly-designed circuit, July 2013:

There is no current limiting resistor when charging the NiMH battery from the 6v regulator and the charging-current will be unknown.
NiMH batteries are very complex to charge and a simple charger like this will cause damage.
NiMH cells should not be charged above 1.78v per cell (5.25v total) and this charger will keep charging as the input voltage is 5.4v.  Just a very dangerous circuit. Do NOT build this circuit.

Another Electronics For You badly-designed circuit, August 2013:

This project has been submitted by ABHISHEK KUMAR a final-year student with absolutely no electronics experience AT ALL.
The rubbish he is presenting is passed to 50,000 readers of the magazine and many will experience problems with the power supply.
Let's look at the first problem. When a transformer states 20v AC secondary, the voltage will be 20v at full load. All transformers have a term called REGULATION. This means the output voltage will be slightly higher than 20v when the current is say 1amp, because the as the current increases, the resistance of the secondary winding will cause a slight voltage-drop. This means the no-load voltage will be about 23v to 25v.
The author states the output voltage will be 28.2848 volts after the rectifier. What an absurd thing to say when the AC voltage can be anything from 20v to 25v. You cannot state an accuracy with more figures than the given data.
If we take the 7805 and try to draw 1 amp, the voltage across the 7805 will be 25 x 1.414 - 1.4v = 34v - 5v = 29v.
Wattage dissipated as heat will be 29 watts.
The regulator is designed to allow a maximum of 15 watts to be generated as heat and this means the regulator will only allow 500mA load.
The same applies to the 7812, with about 700mA load. I am not going into the finer details because the circuit is so badly designed that I would not construct it.
For the LM338, the situation is just as bad. The metal-can version of the regulator can dissipate up to 50 watts, and this is just like the heat of a soldering iron. 50 watts is a LOT OF HEAT! The plastic version (as shown on the diagram with leads marked 1,2,3) dissipates a maximum of 25 watts.
Suppose you adjust the LM338 to output 10v at 5 amps. This will produce 20 x 1.414 - 1.4 = 26.8v -10v = 16 x 5 = 80 watts.   The chip cannot deliver more than 3 amps due to the badly-designed power supply.
The output of a regulator should have a 100n and 10u electrolytic to prevent parasitic oscillations (1MHz oscillations). These capacitors need to be close to the regulator to have any effect.
None of this has been considered by the author or the Technical Director of the magazine Sani Theo.
I have repeatedly told Sani Theo to contact me before publishing rubbish like this, but he doesn't have the intelligence to get these circuits analysed before publication and continues to deliver this rubbish to the readers of the magazine. 

He has no electronics knowledge AT ALL.
You cannot connect a transformer to a battery and expect it to charge the battery without overcharging.
But most-important, you cannot connect a transformer to a battery as the transformer will BURN OUT.
The circuit above is one of the worst circuits I have seen. It does not matter if you are a Doctor, Professor, or student, test and check your circuit before sending it to a magazine, where lots of intelligent readers will see the badly-designed circuit.
Here's the point that no-one has mentioned before. A battery is just like a zener diode. It sits in the circuit just like a 15v zener diode when it is fully charged.
If you have a transformer capable of delivering more than 15v, current will flow into the battery. Just because the transformer is rated at 500mA, DOES NOT mean that 500mA will flow. A lot more than 500mA will flow and the actual current-flow is unknown, but will depend on the resistance of the secondary winding. It could be as high as 1 amp or 2 amp and the transformer will definitely BURN OUT.
No only will the transformer get very hot and eventually burn-out, but the current into the battery will not reduce when the battery is fully charged and it will start to "gas" and dry-out very quickly.
You MUST have a voltage regulator to supply the exact voltage when charging a battery or deliver a trickle charge so the battery will not "dry-out." None of this was mentioned in the article.
The remainder of the project uses 7-segment displays that are not carried by any parts-supplier. Obviously the author has dug up some "junk bits" without making any effort to see where they can be purchased.
I'm glad Dr D.G. Vyas  is not my lecturer.
 
Another Electronics For You circuit with mistakes, August 2013:

D4 should be reversed and D1, D2 are not needed:

When designing a circuit, ask yourself this simple question:  Is this component needed? remove the part and see if the circuit works. The combination of D1 and D2 will pass current, no matter if the polarity of the supply is positive or negative. Thus they are not needed. Another untested circuit from Electronics For You magazine.

Edge FX wanted me to recommend their site and provide a link.
Look at just one of their kits. A voltage multiplier:

Look at the bare connections at the end of the 240v power lead !!!!
Look at the 1,000v diodes soldered on top of each other !!!!
The kit contains 100u electrolytics. Can you imagine the current that can be supplied by the multiplier circuit !!!
It looks like the electrolytics are connected in series.  Why ???  You can get 100u  @ 415v
The kit costs $289 !!!!
Now have a look at where they are reading the output voltage !!!  They have the voltmeter connected to the 2,000v end of the generator !!!!
The voltage divider consists of 10 resistors to divide the 2,000v to produce 200v across a single resistor but they are reading the HIGH END and not the "EARTHY END" !!   The voltmeter will have 2,000v on it !!
Edge FX has NO IDEA how to design a kit and this kit is one of the worst, MOST DANGEROUS kits on the market.
You can imagine why I didn't recommend their site.

Here is another absurdly over-deigned, over-priced ($219) design from Edge FX:

The circuit is designed to increase and decrease the brightness of the LEDs. The LEDs are supposed to be a lamp, but they are next to all the extra circuitry and the project is not very practical.
The 8051 microcontroller is HUGE. All the project needs is an 8-pin PIC micro.
The LEDs have only a single current-limiting resistor. This is a BIG MISTAKE. Very poor design.
The current limiting resistor is a very low value and this gives the circuit a very small supply-voltage HEADROOM. This is something the designer knows nothing about.
The text does not say if the project is operated via a timer to raise and lower the brightness of the LEDs according to the time of day or if a photocell is included to adjust the illumination. Since there is only one input and one output, a very small micro is needed.
Just another JUNK project with very poor design-features and a price that would scare almost all electronics engineers.
If you can't produce a project like this for $20.00 or less, you are in the wrong market. No-one is going to buy a lamp for $219.00
They are trying to sell their projects to desperate final-year electronics engineers who can't design their own project. They think EVERYONE'S a FOOL.

Here's another ($229) design from Edge FX:

It connects 4 parallel phones to the phone line and only allows one phone to receive the incoming call. The circuit has been copied from a project on the web and the mistakes in the circuit have been copied without any corrections. Note the 220k across the LED on the opto-coupler. It should be 100R

Click for full size image

It is not obvious from the diagram, but the relay contacts are normally closed  so that all the phones are on the phone-line when the incoming signal (ring signal) is received. This is an AC signal and will not have sufficient current to illuminate the LED in the opto-coupler.
When one phone is lifted, the corresponding LED in the opto-coupler illuminates because the phone draws the phone line down from 50v to about 12v and takes about 20mA to 40mA. The resistor across the LED should be about 100R so that some of this line-current flows through the resistor.
The transistor in the opto-coupler turns on and produces a LOW on the corresponding input to the microcontroller.
The micro turns on the other three relays so that other phones are removed from the phone-line.
When the phone is put down, the relays are de-energised.
The project above has LEDs corresponding to each relay and there is no need for the 7-segment display.  

This circuit can be created with gating diodes and a few components. Here is the circuit:

Click for FULL SIZE image

See full details on: http://www.electronicsmaker.info

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