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We start another page of faults, hints and corrections to circuits found on the internet.
This time we study the circuits found on RED CIRCUITS website.
This website is owned by Flavio Dellepiane.
To see all his circuits: to   simply increment the page number from 1 to 165.
When asked to explain some of the following faults, he just said "Goodbye."
He has since replied again by email but failed to explain any of the mistakes below. He did, however, say he was an amateur and I am not sure if he designed any of the circuits himself, but he did say he included component values on the diagrams he uses himself.
It is a pity he did not continue this trend when presenting the diagrams on the web, as this would have made my investigation much easier. 

I know it's embarrassing to be shown-up, but the only way to strengthen the web is to weed out the faults and scams.
A lot of the scams have been exposed and the links have been closed, and yet the magnetic motor, based on Tesla's concepts, still fills my email in-box. See: Tesla's Secrets.
Scams on eBay have however been eliminated and this has made purchasing things in the web totally reliable. 

Now to Red Circuits.
Many of Flavio's circuits contain extra components that have no function.
One of the points I have made in the past is to remove each component and see if it alters the operation of the circuit.
Next, increase and decrease the value to see if it has any effect.
This serves two purposes. It lets you know if the value is critical and if it is needed. Some circuits are so badly designed that they stop working when the rail voltage drops a small amount. Others fail to work if one or more components are substituted.
But adding extra components is a mystery. I cannot work out why the designer added the parts. Maybe he added them when designing the circuit and failed to remove them at the end.
Let's see . . .

You can instantly see that Flavio is not an electronics engineer. He lists the components on the circuit as R1, R2 etc and provides a list.
A circuit without component values is totally useless.
An electronics engineer can "see" the circuit working when component values are included. It's a bit like showing a photo with the faces "painted-out" and placed below the photo.
I have absolutely no idea what his circuits are doing until I fill in the component values.

Halloween Flashing-eyes Badge

Two faults with this circuit.
The current-path though the two transistors consists of junction-drops. The path is as follows: Emitter-base of Q2 then collector-emitter of Q1. The first junction-drop is 0.7v. The second drop is normally 0.2v. The supply is 3v. There is no resistor to take up the excess voltage.
What happens is this: Q1 does not turn on fully and the voltage across collector-emitter is 2.3v.
Since Q1 is turned on very lightly, the collector-emitter current will be less than 1mA.
Q2 will amplify this current 100 times and the current through the LEDs could be as high as 50mA each.
There is no current-limiting for the LEDs.

Bike Light-1

What is the purpose of D1?  Remove it and see if the circuit still works. It will change the setting of the LDR (R1) but the circuit will work when it is removed. Bike Light-2

The main fault with this circuit is the driving-power of the CMOS 7555 IC. It will only sink 30mA when the supply is 3v.
This gives about 5mA for each LED.
The circuit should have a driver transistor on each output.
Bike Light-2 is quite useless.

Capacitive Sensor

The purpose of D2 and D3 is unknown. They are not needed.

Fridge Alarm

This circuit is badly designed. The two diodes pull the 100k down and this takes a small amount of current when the alarm is not being used.
By re-arranging the parts, the circuit will not take any current in this section.

Note: The first circuit has an AND gate made up of the two diodes and 100k resistor. The lower circuit has an AND gate made up of a 10k resistor and diode. The two resistors connected to the base produce a voltage divider so the base sees only about 0.35v in the LOW state. It is turned by the top 10k and when pin 6 is HIGH, the diode does not remove (or "short" or "deck") the "turn-on voltage" from the chip.
Pin 3 is the highest division for this chip (it divides the clock by 16,384. This is 128 x 138) and the timing components on pins 9 and 10 allow pin 3 to go HIGH after about 20 seconds.
Pin 6 has a lower division (it divides the clock by 128) and it goes HIGH/LOW 128 times before pin 3 goes HIGH. But nothing happens during this first 128 outputs as pin 3 is not HIGH.
The circuit requires pin 3 to go HIGH to pull the base of the transistor HIGH and turn on the piezo buzzer.  But this only happens when pin 6 goes HIGH and the result is a beep from the buzzer. Pin 6 goes HIGH/LOW 128 times during the next 20 seconds and this makes the buzzer create a "beep-beep-beep" sound.

Police Siren

The only problem with this circuit is the 100n driving the base of the output transistor.
A 100n will not provide the energy-transfer needed from the output of the chip to the base of the transistor.
The 100n will only supply short pulses of waveform and the output signal will be very weak and not of the required shape.
The 100n should be replaced with 220R to 470R.

Signal Tracer:

The only fault with the circuit is the placement of R3. It should be connected between base and positive rail so the output transistor is not taking any current when the circuit is switched on. When a signal is detected, the first transistor will activate t he second transistor via C3.

LED Flasher

Here is a circuit that does not work.
Maybe it worked for Flavio. But it doesn't work for me. I tried changing the value of the 100k pot and altered the supply voltage. But it never made a flash.
This circuit reminds me of some very early multivibrator circuits. The resistance values had to be accurately selected and when the battery voltage reduced, the flasher stopped working!
A circuit must work the first time it is assembled and the parts must not be critical. Especially when we have reliable flasher circuits that work from 3v to 9v and down to 2v.

Touch Switch

This clever circuit has one problem. It is not reliable. The operator must have static or mains hum on his/her finger to activate the circuit.
It may work the first time due to static build-up on the operators finger but if this charge is not present, the circuit will not activate.
The solution is to provide two touch pads to create a "turn-ON' circuit as shown in the diagram below:
One advantage of this circuit is the current drops to zero when not activated as both transistors are turned off.

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. He designed a similar faulty circuit and we covered it on Page 12. He never tests anything.


Transistor T2 will never turn off. It is being turned ON by R3. The circuit is trying to turn it off via T1 and D1. But the voltage drops across T1 is 0.2v and 0.7v across D1. This produces a voltage drop of 0.9v
T2 requires a voltage drop of less than 0.7v to turn it OFF. It will NEVER turn OFF.
 All his circuits are faulty. Why doesn't D. Mohan Kumar test his circuits before putting them on the web?  Simply remove D1 and the circuit will work.

Here's another disastrous design by D.Mohan Kumar
The circuit has NOT been tested as it DOES NOT WORK in the way it has been described.

Here is his wording and my corrections:

Here is an ideal Mobile charger using 1.5 volt pen cells to charge mobile phones while travelling. It can replenish cell phone battery three or four times in places where AC power is not available.
Most of the Mobile phone batteries are rated at 3.6 V/500 mA.
They are rated as 3.6v / 500mA-Hr. 

A single pen torch cell can provide 1.5 volts and 1.5 Amps current.
Penlite cells (or any dry cells) are not rated as maximum current capability. They are rated in capacity (amp-hours or mAHr) and a penlight cell has about 1 amp-hour when discharged at about 10mA. 

So if four pen cells are connected serially, it will form a battery pack with 6 volt and 1.5 Amps current.
This should be: 6v and 4 amp-hours. This is equal to 6wattHrs

When power is applied to the circuit through S1, transistor T1 conducts and Green LED lights. When T1 conducts T2 also conducts since its base becomes negative. Charging current flows from the collector of T1.
To reduce the charging voltage to 4.7 volts, Zener diode ZD is used.
I have no idea what he is talking about????

Resistor R4,and R5 allows 20 mA charging current.
This is NOT true. Battery voltage = 6v. Voltage emitter-collector of transistor = 0.3v.  Battery voltage rises to more than 3.6v during charging, but say it is 3.6v. Voltage across 270R + 27R = 2.1v   Current = 7mA.

If more current is required, reduce the value of R4 to 100 Ohms so that with in 20 to 30 minutes battery will become fully charged.
This is NOT true. Current through 100R + 27R = 16mA.
Battery capacity is 500mA-Hr  Time to charge = 500 / 16 = 31 hours !!!!!!!

D. Mohan Kumar is a University Professor. How can he get calculations so wrong??????
How can he design such a useless circuit?

What is the circuit supposed to do?   It does not turn off when the phone battery is charged. The zener diode simply takes over and wastes energy until the 4 penlite cells are flat.

No-one sells SK100 transistors. Use BC 557 or BC327
What is the function of the 27R resistor?   It is not needed.
What is the function of the 100u.  An electrolytic across the power supply is only needed when the circuit contains AC signals.
What is the function of T1?   It is not needed.
What is the function of T2? It is not needed.
As you can see, the circuit is an absolute DISASTER.

Another D.Mohan Kumar disaster:

Don't worry if you can't see all the component values. DO NOT BUILD THIS CIRCUIT. It is a DISASTER.

This is the worst battery charger circuit I have seen.
The transformer is 18v RMS and this will convert to at least 22v after rectification.
The battery voltage is about 14v when charging and this means the difference in voltage is 8v.
There is no dropper resistor to drop the 8v and this means the transformer will deliver more than 2 amp and it will burn out in less than an hour.
You simply cannot connect a transformer to a battery without some form of current-limiting. On top of this, the 1N4007 diodes are 1 amp and they will burn out in about 10 minutes.
Why is it that D. Mohan Kumar keeps presenting circuits he has not tested?
What is wrong with the Professor????
Why do professors wear slip-on shoes?  Because they haven't learnt to tie shoe-laces.
"Professor, you have one black shoe and one brown shoe"
"That's funny, I noticed another pair just like this in the cupboard."

Another untested D.Mohan Kumar circuit:

The 15v regulator will NEVER charge a 12v battery. 1.7v  is dropped across the LED and 0.7v between the base and emitter. This leaves 12.6v 
The battery voltage rises to over 14v during the charging process and you need a voltage higher than this to create a current-flow.
What is the purpose of the 15v regulator?
This is another disaster by
D. Mohan Kumar.

I am waiting for D. Mohan Kumar to write up another "disaster."  You can learn more from a faulty circuit than a circuit that works!  I have learnt more from D. Mohan Kumar than anyone else. I can see his incorrect reasoning and many other beginners will have the same faulty reasoning.
This is something text-books should have done 50 years ago. They should "pull you up" and prevent faulty reasoning.
It is not good enough to provide only the correct reasoning because a beginner may think there is also an alternate approach and the faulty concepts are not addressed. 

This circuit has 3 minor faults that need addressing:

The TOUCH PADS work on a voltage-dividing arrangement. The voltage on the input (made up of inputs 1&2) needs to be higher than 66% of rail voltage for the gate to see a HIGH. (This is because the chip is a Schmitt Trigger and it works on 1/3  2/3  detection).
The resistance of a finger can be about 100k. This gives the input as 430k:1M  or 70% of rail voltage.  This is very close to 66%  and the touch switch may not work every time. The 330k needs to be reduced to 47k to produce reliable triggering.
The resistance of R9 puts a voltage of 0.65v on the base of Q3 when using 6v. 
This is barely sufficient to turn the transistor ON. Use 10k for a reliable operation.
When R9 is 10k, D2 can be removed and D3 replaced with 10k.
I would make R8  = 47k to signify it is just a value to keep Q3 turned off.
The TIP121 is a Darlington transistor and it is used in an emitter-flowerer arrangement with the 10k pot.
Adjusting the pot will reduce the speed of the motor.
This creates a big problem.  Many times a motor will not start when the speed is reduced. The circuit should be designed with PWM using the 4093 chip and a few extra components. This will reduce the heat dissipated by the TIP transistor and guarantee reliable start-up.

Drawing Circuits
Always draw a circuit so it is easy to read. This includes simplifying a circuit and making sure the project works by assembling the components and testing it with slightly different values to make sure you have selected the best values.
In this "Mistake" we have selected a simple component:  A BRIDGE RECTIFIER and shown how it can be drawn incorrectly.

This makes it easy to see it is a rectifier and you don't have to work out what it is doing.
The following diagram shows a bridge rectifier with 4 individual diode symbols and a bridge symbol with a single diode to indicate the direction of the 4 diodes. 

The following diagrams are hard to read and three are drawn incorrectly:

Here's a circuit that shows the designer has no idea what he is designing:

The yellow signal wire to a servo does not require any current. It is purely a signal line and requires a fairly high amplitude. The 2N2222 buffer transistor is not assisting in any way AT ALL. It is reducing the amplitude of the signal by about 0.5v and the signal does not have to be buffered.
It can be connected directly to pin 3 of the 555.

Here's an over-designed circuit:

The value of R1 is too low. If R2 is turned fully, pin7 will allow about 100mA to flow through the pot and it will be damaged. R1 should be more than 1k.
C1 should be reduced to 47u and R2 and R3 increased to 100k. This will gave the same timing with an easy-to-get 47u.
R4 is not needed.
The BC338 transistor can be a BC547 device to show a high current transistor is not needed.
R10 is not needed.
TIP142 should be drawn as a Darlington transistor.
R9 is not going to provide any protection for the LEDs because the  Darlington transistor requires very little base current and if the 100k pot is turned fully the LEDs will see nearly 11v.
The circuit is very dangerous and should be adjusted very carefully.

Here we have another poorly designed circuit by Mr Giorgos Lazaridis  of PCB Heaven:
He tried the circuit with white LEDs and used two 1.2v cells. This gives a supply of 2.4v.
The problem is:  White LEDs require 3.2v to 3.5v for correct illumination. Some white LEDs will operate on a lower voltage but the brightness will be very low.
The second fault in the circuit is the 1R resistor. This is the current-limit resistor and prevents excess current flowing through the LEDs.
When the supply is increased, this resistor will need to be increased.
Refer to our article in
30 LED Projects for an explanation on how to determine the value for this resistor.
The BC327 is a fairly high-current transistor and using this type infers that a high current will be flowing in this particular part of the circuit. This is not so and the transistor should be given as BC557.
However the BC547 transistor needs to be a fairly high-current device and should be stated as BC337.
The circuit should be supplied via 4v5 (three cells) and R5 should be 8R2 or 10R. 

An emitter-follower design is very bad as you are losing over 2v in the base-emitter junctions. This means the supply should be 2v higher than the voltage of the relay for guaranteed operation and the final transistor has about 1.5v between collector and emitter. This will add to the losses and heat up the transistor slightly more than a common-emitter design. The second circuit shows the improved design.

Here's another circuit from  D.Mohan Kumar:
The circuit is badly designed as the LED will only allow 20mA and we don't now the current requirement of the buzzer. But the main issue is the over-design of the circuit. The same result can be achieved with fewer components, less cost and it will consume NO CURRENT when sitting around.
It gives an audio visual indication when the water level rises towards the top of the tank so the water pump can be switched off to prevent overflow. Probes A and B are placed at the top of the tank and when the water reaches the probes, the buzzer is activated.

Water Level Alarm

Here is the improved circuit:

Water Level Alarm

Here is a toggle circuit that does not work. Pin 2 detects less than 33% of rail voltage and pin 6 detects more than 66% of rail voltage.
These two pins are being held at 50% of rail voltage via the two10k resistors. The problem is this: The 100k resistor from the output of the chip will not raise or lower the two pins by the 15% needed to change the state of the circuit. If the two 10k's are changed for 100k and the 100k changed for 10k, the circuit will work.
However we have a time-delay created by the 1u and if the switch is pressed for too-long, the circuit will change back to it original state.

An improved design places the switch as shown below: The 1u is now charged to about rail voltage or about 0v via the action of the output of the chip and this voltage is immediately delivered to pins 2&6 via the switch. If the switch is kept pressed, the ratio of the resistors will not cause the chip to change states, as discussed above.
Simply changing the position of the switch turned a faulty circuit into a success.

One further problem with the 555 is this: Pin 3 does not go below about 0.6v and the 2N3904 transistor does not fully turn off.
Fit a 2k2 between the base and 0v to reduce the voltage on the base to fix this problem.

Some readers have asked for the circuit to turn-on in RESET mode.
Pin2 detects a LOW to make the output HIGH. If a 100n capacitor is placed across the upper 10k, the 555 will turn on in RESET mode.

Here's an example of "clear thinking."
In other words, look at the problem and see if it can be solved very simply.
Here is the answer from a reader with "mathematical understanding."
Using the node method, find the voltage across Rx in terms of Rx.
Use E^2/Rx and find the term for power across Rx in terms of Rx
Find the first derivative of the power with respect to Rx and solve for its zero value.
When Rx = 18 ohms, the maximum power is obtained.

The above reasoning is entirely WRONG.

The circuit is very complex until you realise one thing.
Firstly, remove the 6R, 18v and 12R.
Maximum power in Rx is achieved when Rx is equal to 3R.
To see if this is true, simply change Rx to 2R or 4R and work out the watts dissipated.
Another way to look at it is this: When Rx is zero, the watts dissipated is zero, when Rx is very high, the watts dissipated is very small.

Make Rx = 3R and see what happens:
The voltage across Rx will be 18v.
Now put-back the 6R, 18v supply and 12R load resistor.
The voltage across Rx is 18v. This means no current will flow though the 6R resistor because it has 18v on each end. 
Current through Rx = 36v/6R = 6 amp
Wattage dissipated by 3R = V2/3 = 108watts    or 
I2xR  =  62 x 3  = 108watts
We have turned a seemingly complex question into a very simple answer. 

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

The 100R resistor on the input has 12v - 7.2v = 4.8v across it and 48mA will flow through the resistor and 7.2v zener.
Suppose the power supply delivers 20mA to a load. The 7805 takes 10mA, making a total of 48+20+10 = 78mA
The voltage across the 100R will increase to:  V = 0.078 x 100 = 7.8v   The input voltage will be: 12v - 7.8v = 4.2v    The power supply cannot even deliver 20mA!!!!!!!!
This is another of Professor D.Mohan Kumar's disasters. He should stop putting rubbish like this on the web. He doesn't know a thing about electronics.

Here's a circuit from  D.Mohan Kumar and other Indian websites, that does not work. They copy each other and never build the circuit. They obviously think "it'll work" because I am a Professor!

The output of the UM66 is not sufficient to turn on the transistor. It may work for some transistors and it may work when the load is very high. But it doesn't work for the circuit above.
The solution is to self-bias the output transistor and couple it to the chip with a capacitor.
Always build EVERY circuit before putting it on the web. You never know what fault will come out of even the simplest circuit.
The UM66 series of chips have a range of tunes. The UM66T19L outputs "For Elise." The other chips are listed below. When ordering, make sure you ask for the exact song you require.


Here is the list of different chips and their content:

Here's a simple circuit driving 16 white LEDs from the mains.

Here's three technical mistakes from a chief electronics editor:
1. He says:
"The 48V zener is there to clamp the supply to the LED chain to +48V maximum."
Not true.
If it actually did that, the LEDs would not work or be very dull. The REAL purpose of the zener is to discharge the capacitor.
2. He says:
The current through the LEDs is 25mA.
Not true.  The current though the LEDs is about 7mA
3. He says:
"With the 22uF present during the negative half cycle, the 22uF supplies a lower holding current to the LED's so they do not go completely off."
See discussion below.
He also falsely thought the LEDs would flicker at 25Hz on a 50Hz supply.

The current is limited by the 0.22u capacitor and it is a capacitor-fed power supply that is commonly known as a CONSTANT CURRENT SUPPLY. In this case the current is limited to 7mA.
At 7mA the voltage across a white LED is 3.3v. At 3mA the voltage is 3v.

Without the electrolytic, the LEDs produce a noticeable flicker.
This is because they are being fed pulses during the positive half-cycle and turn-off completely during the negative part of the cycle.
The voltage only has to drop from 16 x 3.3 = 53v to 16 x 3 = 48v and the LEDs do not illuminate.
This means they are only illuminating during the top 10% of the voltage-curve and when this is analysed on a width-basis, it represents about 50% ON-time.
You can observe this by setting up the circuit and shaking it from side-to-side and see when the LEDs are illuminated.
When the electrolytic is added, some of the current charges it during the peak of the waveform and as the waveform reduces, it delivers this energy to the LEDs to keep them illuminated during the dip in voltage.
The electrolytic would be much more effective if it were connected to the supply voltage with a 470R connected to the LEDs. This would allow it to be charged to a higher voltage so it can deliver its energy during the remainder of the cycle. 
In fact the circuit will not work with 16 LEDs as they drop 16 x 3.3v = 53v.  The zener limits the voltage to 48v and the LEDs will be very dull.
The zener should be replaced with a 1N4004 diode.
It's a very poorly designed circuit with at least 3 faults.
See: "30 LED Circuits" eBook for a number of correctly-designed circuits.

This circuit produces a constant 350mA to illuminate a 1 watt LED. The circuit is over-designed.


The circuit can be reduced to 2 components:

The 7805 can be converted into a content-current device by connecting a resistor as shown above.
We will take the operation of the circuit in slow-motion to see how it works.
As the 12v rises from 0v, the 7805 starts to work and when the input voltage is 4v, the output is 1v as a minimum of 3v is lost across the 7805. The voltage rises further and when the output is 5v, current flows through the 15R resistor and illuminates the LED. The LED starts to illuminate at 3.4v and the voltage across the 15R at the moment is 1.6v and the output current will be 100mA. The input voltage keeps rising and now the output voltage is 7v. The current through the LED increases and now the voltage across the LED is 3.5v. The voltage across the 15R is 3.5v and the current is 230mA.
The input voltage keeps rising and the output voltage is now 8.6v The current through the LED increases and the voltage across the LED is now 3.6v. The voltage across the 15R is 5v and the current is 330mA. The input voltage keeps rising but a detector inside the 7805 detects the output voltage is exactly 5v above the common and the output voltage does not rise any more. The input voltage can rise above 13v, 14v  . . . . 25v or more but the output voltage will not rise.
If the output voltage rises, more current will be delivered to the LED and the voltage across the 15R will increase. The 7805 will not allow this to happen.
The LED will have 3.6v across it. The 15R will have 5v across it and the output will be 8.6v. The input voltage will have to be at least 12.6v for the 7805 to operate.

Here's another over-designed circuit by Professor D.Mohan Kumar.
The 555 is consuming 10mA all the time and this is very wasteful when using a 12v battery.

The transistor circuit consumes less than 0.1mA when the relay is not activated. It is also simpler and cheaper to construct.

Another over-designed circuit by Professor D.Mohan Kumar.  SOLAR NIGHT LIGHT

1. What is the purpose of D1?   It is not needed.
2. What is the purpose of C1?   It is not needed.
3. If the LED is 1w and has about half rail voltage across it, R2 will also have half rail-voltage across it and it needs to be 1 watt also.
4. Turn VR1 to zero ohms and it will burn out via the emitter-base junction of T1.
5. A 4.5Ah battery needs to be charged at about 400mA.  400mA flowing through R1 will produce a voltage drop of 4v.   The 6v solar panel will never charge the 6v battery.

This is just a junk circuit that will never work.

Here's an interesting circuit. Let's look at the design.
The first two transistors are connected to form a PNP Schmitt Trigger. This give a fast action to the relay. But this is not needed. The action of the relay turns ON and OFF in the second circuit without any problem and it does not need a diode across the relay because the transistor is turning OFF slowly. 

Firstly, you don't need 3 transistors to operate a relay from an LDR (Light Dependent Resistor).
When light falls on the LDR, the relay turns OFF.
Here is a circuit that turns OFF a relay when light fall on the LDR and requires only 2 transistors:

Let's look at the first transistor in the 3-transistor circuit:
When no light falls on the LDR, the transistor is OFF.
The voltage divider made up of the  7k5, 1k5 and 2k2 puts 4v on the base of the second transistor. This turns ON the second transistor and the emitter will be about 4.6v. This will turn ON the third transistor and also the relay.
As light falls on the sensor, it forms a voltage-divider with the 22k.
The emitter of the first transistor is 4.6v and when the base is 4v due to the resistance of the LDR, the first transistor starts to turn ON. This increases the voltage across the 2k2 resistor and has the effect of turning OFF the second transistor. The voltage across the 1k5 reduces and turns ON the first transistor slightly more. After a short period of time the first transistor is turned ON FULLY and the second transistor is fully tuned OFF. This action occurs without the LDR changing resistance. This is the action of the Schmitt Trigger.
The circuit does work but is over-complex for the requirement.

Here's a circuit from Electronics Maker, an Indian magazine. It looks fairly workable at first glance, but let's look into it in more detail:

The transformer is not referred to as "Secondary: 0-18v." It is simply 240v:18v. An 18v secondary will normally be 5v higher than the rating to account for voltage-drop under load. This means the output will be 23v and when this is rectified, it will produce a voltage of 23 x 1.4 -2v = 30v.
The 68ohm resistor will drop 30v - 18v = 12v. The current though the 68R = 175mA.
This current will also flow through the 18v zener.  The wattage dissipated by the zener will be:
18v x 0.175 = 3.15 watts.  The circuit specifies 1.5watt.  The zener will be damaged !!
The 2N3055 has a gain of about 50.
Suppose you turn the pot to produce 9v. It will be at approx 235ohms.
When the transistor is delivering 1A, the base current will be 1,000/50 = 20mA. This current will flow through the 235 ohms of the pot and the voltage drop will be: .02 x 235 = 4.7v.
The output voltage will DROP 4.7v !!
This is a typical JUNK CIRCUIT. It has not been tested and no-one from the magazine has picked up the mistakes.
For a 1Amp Power Supply, it would be much better (and cheaper) to use a 3-terminal regulator such as LM317. It is adjustable from 1.2v. It has almost no drop in output voltage and the ripple is about 5mV.
Here is a kit using the LM 317 regulator. You can see the two different inputs (power plug and flying leads), WO4 bridge rectifier and pot to adjust the output voltage. The regulator will need a larger heat-fin if you want to deliver more than 200mA.

Here's a circuit from someone who knows nothing about designing. It has a number of mistakes.

The condenser microphone is actually called an electret microphone and it will be totally damaged in the circuit above.
It is connected directly across the supply with a base-emitter junction of the fist transistor. It needs a 10k resistor in series with the microphone.
It is not a good idea to connect the microphone directly to the input of the transistor, however it can be done. The transistor must be only lightly turned-on so the signal can be amplified.
This is very hard to judge and if  a series resistor is adjusted from 10k to 1M, you will see the change in output signal.
This is a huge range and will depend on the type of microphone. That's why you cannot present a circuit with such a wide range of values.
The 4k7 resistor is not doing anything and can be removed.
The 1k and 470R resistors are too low.  They should be 22k to 47k.
Pin 2 of the 555 is not being held HIGH via a 100k so it can be taken LOW via the signal from the second transistor.
The whole circuit is a bad design. It is taking 10mA when sitting around and does not offer an ON-OFF feature but simply a delay when a clap is detected.     The circuit should be avoided.

Here is a 2-transistor Switch-Mode Power Supply from Roman Black.  He has shown the operation of the circuit using 4 diagrams but they do not actually explain HOW THE CIRCUIT WORKS.  In fact the diagrams are hard to decipher and misleading. He is a correct description of the operation of the circuit.

The circuit shows 2 transistors, an inductor L1 and high-speed diode D1.
Here's how the circuit works:

The circuit does not turn on until the output is slightly less than 5v.   The voltage on the base of  Q2 is 5v6 and the emitter is slightly less than 5v. This produces a slightly higher base-emitter voltage and causes Q2 to turn ON. This turns on Q1 via the 2k7 resistor.
Current flows through inductor L1 and it builds up a magnetic flux. This magnetic flux produces a voltage across the inductor that is almost equal to the difference between the 5v output and the supply. This allows Q1 to turn on. The inductor can keep producing this "back voltage" until the the core becomes saturated. At this point the back voltage suddenly ceases and the right plate of the 1n drops a small amount. This is transferred to the base of Q2 to turn it off slightly.
This action continues between the two transistors until they are both fully turned off.
The large purple arrow is correct.  
The thin purple arrow is vague and the description of the yellow arrow is a mystery.


The arrows above are vague and misleading and incorrect. See Phase 2A below for correct description.


When the two transistors turn off, the magnetic flux in the inductor collapses and produces a voltage IN THE REVERSE DIRECTION to the applied voltage. This means the left side of the inductor produces a voltage that is LESS THAN 5v.  This pulls the voltage on the base of Q2 lower and turns off the transistor even MORE. At the same time it charges the 1n and C1 is discharged slightly.
If the voltage on the left side of L1 reaches -0.6v, the high-speed diode "flips-over" and starts to conduct. This stops the left side of the inductor creating a higher negative voltage and now any energy produced by the inductor is passed to the 5v regulated output. This is how the energy from the inductor adds to the output.

PHASE 3: OFF (Delay Period)

I don't know what is trying to be explained in Phase 3. I don't know what the author means by "Delay Period" He thinks the delay period is set by C1, C2. This is not so. The releasing of the energy from the inductor sets the time for this part of the cycle.
Both C1 and C2 are charged by a very small amount by Rz.


The energy in the inductor is very soon passed to the output and the voltage across L1 reduces. The voltage on the right-hand side plate of C2 rises from -0.6v and the energy in C2 charges C1 and turns on Q2.
The yellow base-current path does not pass through R1 but the emitter-base junction of Q1 passes about 1mA and R1 passes about 0.5mA.

As you can see, arrows on a diagram are misleading and meaningless. The author did not have an understanding of the operation of the circuit and this confused me greatly.  
I could not follow the operation. That's the purpose of a language. It explains things clearly and accurately.

I did not receive a reply until a reader posted this discussion on an electronics website.
Roman replied on the website:
"I had just chosen to ignore it, believing my description was fairly accurate and easy to understand."
Things cannot be "fairly accurate" They are either accurate or not.
Roman replied to some of my comments and got himself further entwined in inaccuracies. That reinforced my comment that he did not know how the circuit worked.
He said: "The turning off phase involves the discharge of caps. As the buck turns off, the inductor pin point A slams below 0v, this causes C1 to discharge and the +ve voltage in C1 to cause a current through C2, as shown by my blue arrow."
The "turn off" phase is a transitory phase where C1 energy is dumped through C2 causing the voltage on point Z to drop sharply as C1 energy is dumped through to the load.
This is entirely untrue. 
The timing of the "Off Period" is controlled by the energy in the inductor.
Refer to Phase 2A where I have shown the high-speed diode sits below the 0v rail and will have -0.6v on the cathode.
The right lead of the 1n will drop from about 12v (in one direction) to -0.6v (in the other direction) and this 13v is called "energy" and it will discharge the 4n7 by an amount equal to the ratio of the two capacitors. That's why the 4n7 drops about 3v.   The voltage on C1 will go below the zener voltage and the 10k will provide a small current to "top up" C1 and C2. This means that the energy in the two capacitors will be slightly more than at the beginning of this part of the cycle and when the energy from the inductor has been fully released, the voltage on the cathode of the high speed diode will start to rise due to the voltage on the electrolytic on the output and the resistance of the inductor. As the voltage rises, the energy in C2 will be passed to C1 and the voltage on the base will rise to 5v6. The energy will do two things. It will pass into the zener diode and also into the base of Q2 to turn it ON.
This turns on Q1. This raises C2 to turn on Q2 further.  The small amount of energy provided by the 10k Rz resistor provides the energy into the base of Q2 to start the next cycle.

Roman further states:
It is common in SMPS buck ICs to use a "timed off" period, usually a monostable. I managed to mimic this effect by the combination of C1 and C2 and the energy dump as explained in Phase2. Once C1 has dumped and is at 3v, there is a clearly defined "delay period" while C1 charges again (via RZ) and during this delay period the buck Q1 remains off. This gives a big drop in frequency and improves efficiency for a number of reasons, and also greatly increases stability.
Roman is mixing up the defined "timed off" period in a buck IC circuit with the 2-transistor circuit above. The capacitors C1 and C2 have no effect on the frequency. The frequency is generated by the time it takes to release the energy from the inductor during the "turn off" period.
He says: "The "turn off" phase is a transitory phase where C1 energy is dumped through C2 causing the voltage on point Z to drop sharply as C1 energy is dumped through to the load." This is incorrect. The energy is not passed to the load as the right hand end of C1 is connected to the cathode of the high-speed diode and it is CONDUCTING during this part of the cycle and the energy from C1 is being LOST.
This is a clear factor that he does not understand the operation of the circuit.
I have been criticised for stating this FACT, but how else can you describe an article that is COMPLETELY WRONG!
He believed his descriptions were "fairly accurate." 
You be the judge.

1. He never once mentions the fact that the inductor produces a reverse voltage. This is the crux to the operation of the circuit.
2. He never shows the high speed diode "flipping over" and conducting.
3. He concentrates on current through the 10k entering the load. This is less than 1mA.
4. He thinks the frequency of the circuit is generated by C1 and or C2. This is not so.
5. He says energy from C2 is dumped into the load. This is not so. It is LOST.
6. He says the charging of C2 is via Q1, but initially it is via the electrolytic (at the load). 

In simple terms, the operation of the circuit is this:

The high voltage from the supply is passed to the load via the inductor.
The two transistors turn on very quickly and the current flows to charge the electrolytic on the output. This current passes through the inductor and produces magnetic flux.
This magnetic flux produces a "back voltage" that allows Q1 to turn on fully. At a point in time the core of the inductor becomes saturated and the "back voltage" ceases. At this point  the voltage on the collector reduces and this is passed to the base of Q2 via the 1n capacitor.
The two transistors turn each other OFF and the magnetic flux collapses and produces a voltage in the opposite direction.
The rest of the cycle has been covered above.

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