I am not against anyone building any of the circuits on my website … and not buying a kit. In fact, if you build the project yourself, and use junk-box components, it will cost you less and you will learn 10 TIMES MORE than buying a kit.
How do you think I learnt so much?
Kits were not available in my day.
And printed circuit boards had hardly been invented.
Everything was hand-wired on tag strips and then punched board came on the market.
Then we got that terrible strip-board.
And finally expensive Printed Circuit Boards.
But if you make your own Printed Circuit Board and use salvaged components (and get the circuit to work), you will have learnt things from the BASIC LEVEL. Not the UNIVERSITY LEVEL where you pretend to make a project and come out of a 6 year course without having to touch a soldering iron . . and have the attitude that every circuit you design will work first first-go.
All the successful people who have gone though my training have build dozens . . if not hundreds . . of circuits and it is only by construction that you gain an understanding of being able to look at a circuit and see what is happening.
Here is the metal detector NAIL FINDER project from: Hashem Pakdel
Here is a Power Supply project from October 2019 issue of DIYODE magazine:
Let’s see how many faults and bad-designs you can get in a single project.
From first impressions, the project looks ok. But when you look deeper, it is riddled with faults.
The most glaring mistakes are these:
The bridge rectifier diodes are 1N4004 for a 1.5 amp power supply. These are 1 amp diodes.
The second mistake is the rotary switch. It is designed for currents up to 100mA and the circuit requires 1.5amp to flow though the tiny contacts. They will be destroyed in 10 minutes. The rotary switch needs to be replaced with a flying lead and machine pins. At least they will not burn out.
The 2,200u electrolytic used in the project is 50v. You can see it is very small and will have a very small ripple current capability.
When you use a high voltage electro, the ripple current capability is reduced. The manufacturers cannot get high voltage and high ripple current into the same package.
The current through the 0.82R and 1R will produce about 1.5watts of heat and this heat will flow through the leads and if the pads on the circuit board are not large, the solder will melt and the resistor will drop off.
The 4 diodes in the bridge are Schottky 3 amp diodes and normally they have a low voltage drop across them. But I have used these diodes and found the voltage-drop at high current to be nearly 0.9v. This means nearly 1.4watts will be dissipated in two of the diodes when DC input is applied and the pads are so small and isolated that the diodes will drop off the board.
When 19v DC is applied (as suggested in the article) and the output is 5v @ 1.5amp, the losses in the project will be 18 watts.
I have not worked out where the losses will occur but it looks quite impossible for the heatsinks in the photo above to dissipate this wattage.
The small heat fin is 2 watts, the medium fin is 5 watts max and the large fin is 5watts because it is in an inefficient place. The heat is generated UNDER the transistor not on the top and gluing a fin to the top will be very ineffective. The bridge is 3 watts.
The 5k, 1k pots and 270R resistor on the second regulator will have the capability of producing an output from 1.25v to more than 27v and thus the project is using less than half the rotation of the pot to get to 12v.
The 10R resistor is 5watt. But the regulator will have to deliver only a small current as the MJ2955 is doing all the work.
But here is the major fault with the circuit.
There is no electrolytic on the input of the second regulator. The regulator needs something to “push against.” Something rigid and stable so it can produce a constant output voltage as the current-requirement of the load, varies. This circuit has no electrolytic and the output variations will be enormous. The circuit is worthless.
Nothing is marked on the top of the board. You would have absolutely no idea how and where to place the components.
Normally, a LED has the voltage dropping resistor on the anode, so the cathode is connected to the 0v rail. This just makes it easier to service.
I explained to the author of this project that it is a “dog-breakfast” with the components all over the board and not following the layout of the circuit.
The project shows a complete lack of understanding of electronics and current. The 10R resistor will pass about 60mA and 0.25watt will be suitable. The 0.82 and 1R can be much smaller. The rotary pot will burn out. The 5k pot will use only a portion of its rotation, the PCB has no component markings … just to mention a few.
If you look at Talking Electronics projects, you will see the layout of the components follows the circuit and that makes it easy to service. This project is just a jumble - a jumble of mistakes.
Here are a few points on what to expect from this power supply.
A LM317 is capable of delivering 1.5amp but the input voltage must not be higher than 12v for an output voltage of 5v so the voltage across the regulator will be 7v and the current will be 1.5A, making a total loss of 10.5watt. If the losses are higher than this, the regulator will get too hot, even though you may have a very large heatsink, and it will close down and reduce the output voltage.
When operating correctly, like this, the output voltage will drop (fluctuate) about 5mV to 25mV when the current fluctuates, from say zero current to 1 amp, due to the regulator driving say a 5watt amplifier or a resistive load that is being turned on and off.
The output will only have this low voltage-drop (fluctuation) if the input voltage is stable and does not drop below 10v as the regulator needs 3v or more across it AT ALL TIMES to provide the feature of a STABLE OUTPUT VOLTAGE.
If the voltage across the regulator falls below 3v, the fluctuations on the output become enormous, so it is the 2v (up our sleeve) on the input that we are relying on to maintain a low voltage-drop on the output.
Now the 2,200u on the input has the ability to store energy and if the input voltage is AC, it will be getting 100 pulses of energy each second.
You have to remember, the AC on the input is rising and falling and some of the time it is not present AT ALL. That’s why the 2,200u is added to the circuit to deliver during these parts of the cycle.
But the 2,200u really has very little energy and if you are taking 1.5amps, the voltage across the electro will drop about 5v during part of the cycle. That’s why, to get a minimum of 10v for the regulator, you need to charge the electro to 15v.
All this applies to a 5v output.
You can see, the 2,200u is really quite useless but when combined with the 3-terminal regulator, the regulator improves the performance of the electro about 1,000 times. It’s called an “electronic improvement,” or “electronic improver.” It reduces the fluctuations (called “dips”) that will be present on the electro by not allowing the peaks on the electro to be passed to the output and thus the output “thinks” the electro has say “10v across it at all times.” In other words the output thinks the electro has a constant, fixed, rigid voltage across it at all times.
For any voltage above 5v, the same amount must be added to the voltage across the electro. Thus for 12v out, the voltage across the electrolytic must be 22v.
Finally, the output voltage from a transformer has a rating. Say the rating is 14v @1.5amp.
This is an AC rating and the output will be 14v when 1.5amp is flowing.
But the secondary winding of the transformer has resistance and when current is flowing, there will be a voltage drop in (across) the secondary winding.
This drop may be 5v, so the manufacturer adds extra turns so the voltage is 14v AC when 1.5amp is flowing. This means the output voltage will be 19v on NO LOAD.
When this winding is connected to a bridge, the voltage gets converted to pulses of DC and stored in the electrolytic. The conversion from AC to DC increases the voltage 41%, minus about 2v drop across the diodes in the bridge.
These voltages must be taken into account when designing a power supply and some of them are only known when you test a prototype.
The 220u on the output has almost no effect (no improvement) as we already mentioned a 2,200u on the input sees a 5v drop.
If the regulator is not delivering a quality current and a voltage-dip of less than 25mV, the 220u on the output will make no improvement. In fact it will not alter the 25mV.
Here’s another way to look at the circuit.
Suppose we have an electrolytic charged to 25v and it dips to 20v when delivering a current (due to a bridge and AC input).
If we were able to “tap” it at say 18v, (or any voltage below 19v) the output current would be perfect and no ripple-current would be present and no voltage-dips would occur. But we cannot do this by simply connecting a wire to the electro, so we need a 3-terminal regulator. The regulator is, in effect, picking off the voltage of the electro at say 18v and delivering it to the output.
The battery pack has 6 cells in parallel and 3 cells in parallel. What do you think is going to happen when you use the battery???
The amp-hr of the battery will be limited by the 3 cells. This means the battery will only deliver about half the available amp-hr.
And the 4 pin plug will only handle 100mA. The battery will deliver 30 amps !!!
Pity the artist did not understand what he was doing!!!!
The circuit works perfectly but it contains a technical fault that needs to be covered.
The emitter-collector junction of the BC557 and the base-emitter junction of the 2N2222A are directly across the supply.
If, and when these junctions are fully saturated, they will try to have a voltage across them of 0.2v and 0.7v If this were to occur, a very high current would flow and both transistors would be destroyed.
The only thing preventing this is the bottom 2M2 resistor. It limits the current through the base-emitter junction of the BC557 to about 1.5 microamps. If the transistor has a gain of 300, the emitter-collector current will be 450uA (less than 0.5mA) and the transistor will not be destroyed.
This is the same current that will flow in the base-emitter of the 2N2222A and is sufficient to turn on the transistor and drive the circuit, but not get damaged.
The circuit does not have a timing capacitor and relies on the time taken to saturate the inductor and then release its energy to the LEDs.
I posted a comment about the 555 Metal Detector article and it was immediately removed. That’s what they do to all my comments.
The circuit is a copy of one on my website but it has 3 mistakes. The 2u2 is up-side down, the 100u to the speaker should be 10u and the 1k pot will burn out. The coil has almost no details. Who winds a coil by saying the wire is 57 metres long!! Why use 51k when a beginner has enough trouble reading a 47k resistor.
Here is a DELAY CIRCUIT from a design engineer in India:
It has a number of faults and will not work at all.
The circuit is far too complex but looking at the base of T1, it will be at 0v when T2 is turned ON via the delay on its base.
The emitter is connected to the base of T5 and the emitter of T5 we will assume to be at 12v for the moment.
This means the base of T5 will be 12v and the emitter of T1 will be 12v.
This places the base-emitter of T1 in a reverse voltage situation between the base and emitter and this voltage can on be about 5v before the transistor starts to leak. This means the reverse voltage on the base-emitter of T1 will zener at 5v and the emitter of T1 will not go lower than 5v. Pin2 of the 555 needs to go below 4v for the 555 to start timing and must go above 8v for the chip to stop timing.
The circuit is such a mess that I don’t know what will happen.
The second problem is the 1,000u on pin7 of the 555. Getting pin7 to discharge this capacitor will put a very high current through the discharge transistor and damage it.
The third problem is the output of the opto-coupler being able to turn off the BC558. The load on the output of the coupler is 12mA and I don’t know the voltage across the output with this current. It may not be able to go as low at 0.5v
The reason is this: As you load the output of the transistor device, the transistor cannot fully saturate and the voltage between the collector and emitter will rise from 0.3v to 0.4v or 0.5v or even 0.6v
It depends of the quality of the transistor and the current is designed to handle. When you are relying on unusual features like this for a circuit to work, you have to test a number of devices to see if the whole batch will work as required,.
The fourth problem is the 7 minute timing. Delays above 3 minutes are very unreliable with high value electrolytics and high resistances. You are charging the 1,000u at less than 3uA and the leakage current for this type of electro can be 3uA so it will never charge. This circuit was designed to go into an air conditioner for a life of 20 years and and you can see the problems that will start to show after 12 months and the company will go bankrupt.
The second relay and opto coupler are not needed.
And improved design will eliminate most of the components. It can be designed much simpler.
Here is another circuit from D.Mohankumar. https://dmohankumar.wordpress.com/
None of his circuits work and he has 2 million Indians going to his website each month and seeing faulty circuits that don’t work and never will work. It is no wonder the poor Indians are hopeless at understanding electronics when they see this RUBBISH.He says: Here is a simple 3.7 volt Lithium-Ion battery level indicator. Green LED lights when the battery is full with 3.7 to 4.2 volts. When the battery voltage decreases below 3 volts, Green LED turns off and Red LED lights. Just connect it to the 3.7 volt battery charger.
There is no current-limiting resistor for the green LED. Imagine what would happen to the green LED when the battery reaches 4.2v.
A green LED drop 2.4v The diode drops 0.7v and the base-emitter junction drops 0.7v That is 3.8v A fully charged cell is 4.2v and after testing the circuit for 1 minute the green LED burnt out. It got so hot on the high voltage that the crystal overheated and died. I connected another green LED and the battery voltage dropped enough so the LED did not burn out. The circuit is badly designed and I suggest you do not build it. However the red LED did come on at 3v - - so just the green LED section has to be fixed.
AUTOMATIC LED LIGHT
Another failure from D.Mohankumar.
Basically the circuit will not work because the 100R resistor will allow 60mA to flow and produce good brightness.
The way the transistor works is this: It effectively reduces the 100k by a factor of 100 due to the gain of the transistor and this means it adds a 1k resistance to the 100R and less than 6mA will flow. The brightness will be a lot less than expected and you will wonder why. You have to design a circuit so that it works 100% better than expected to take into account all the differences in the features of the components.
SIMPLE FIRE ALARM
Another failure from D.Mohankumar.
The circuit will not work because the LED and resistor will not get any voltage. A photo diode has a very high resistance and putting 470R in series has no technical understanding. Turning on a mechanical buzzer slowly like this does not work. Just try it yourself and see how the circuit fails. Another untried circuit from Professor Mohankumar who has NO understanding of electronics.
Here is a simple circuit that not been understood by any of the readers or authors:
The zener is designed to limit the current when the switch is turned ON and the mains voltage is at a peak. In this condition the current will be very high and we know a LED can be destroyed INSTANTLY.
Let me say the circuit is badly designed because it is too close to destroying the LED when the conditions are such that the capacitor is uncharged and the switch is turned ON when the supply is at a peak. The current through the LED at this peak will be 90mA when the characteristic voltage of the red LED is 1.8v and the zener is 2.7v This leaves 0.9v for the voltage across the 10R resistor. This will allow 90mA to flow.
A much better arrangement is to use a 100R resistor and the 4mA will flow normally and when a peak occurs the current will be 10mA.
This is a much better design.
The secret is understanding the resistor can be increased considerably without affecting the current.
Here is a zener regulator circuit.
The first thing you have to accept is this: The voltage across the zener will be constant when the current taken by the load changes.
Of course the voltage will change, but if you don’t accept the variation, don’t use a zener regulator.
The voltage will change because the zener gets hotter when it is passing more current and you need to provide good heatsinking.
But if you don’t accept CONSTANT ZENER VOLTAGE don’t argue with the following description.
To start with, all zener diodes have a tolerance of +-5% so a 9v1 may be 500mV less or 500mV more. This means the voltage has a range of 1v !!
That is called the initial setting of the output of the zener regulator. But when the circuit is in operation, the output voltage will change by only a few millivolts or maybe 100mV
On top of this, every 9v1 zener will produce a different characteristic voltage across it when taking 20mA so this is a start to understanding the wide range of output voltages from a batch of 9v1 zeners.
Suppose you set up a circuit with 20mA flowing through the zener. As you put a load across the zener, the current flowing through the load will be taken from the zener-current. The current through the supply resistor WILL NOT CHANGE. It will simply be split into two paths.
You can keep “robbing” the current from the zener until only about 2mA flows through it and the output voltage will remain constant.
If you did the same thing with a VOLTAGE DIVIDER made up of two resistors, the output voltage will drop 2, 3 or 4 volts. The zener circuit may drop 100mV or 200mV. It has much better features. A zener regulator will be about 50 times better than a voltage-divider. For instance a voltage divider may change 100mV and a zener circuit will only change 2mV.
In reality the zener will go colder when the load increases and the natural characteristic voltage across it will alter and drop slightly. That will be main contributing factor to the change in output voltage. Good heatsinking will reduce this.
Here is a circuit from April 2021 of DIYODE Magazine.
The author is trying to explain how to design a SERIES REGULATOR circuit.
But he is missing the hidden facts and disasters of using a transistor such as a BD139. He does not understand the BASICS of electronics and that’s why he has made so many mistakes in this articles.
The basics of a POWER TRANSISTOR is this: The gain of these devices changes according to the current flowing. It may have a gain of 250 when 15mA flows, and a gain of 100 when 150mA flows but the gain will only be 40 when 500mA flows and less than 25 when 1 amp flows.
Let us say the transistor has a gain of 40.
This means the current into the base must be 13mA.
This means 13mA must flow through the base-collector resistor. And then you need another 10mA to keep the zener in regulation.
When the output is delivering no current, the current through the zener will be 23mA. As the output current increases, the transistor “robs” the zener of current and theoretically when the output current is 500mA, the transistor will use 13mA and 10mA will flow through the zener. The zener will stay in regulation until the current through it drops to about 1mA but when it drops to zero, the output of the supply will drop.
The voltage across the base-collector resistor is 6.4v so the value of resistance must be: 6.4/0.023 = 280R
But at 1 amp the gain of the transistor is about 25 and the base current must be 1,000/25 = 40mA plus 10mA = 50mA
The base resistor must be: 6.4/0.05 = 130R
The wattage dropped in the zener will be 0.05 x 5.6 = 280mW and it will get slightly hot.
The circuit above will “fall over” when more than 200mA flows and the output voltage will dip and drop appreciably. AND YOU WILL WONDER WHY!!!!
You must test everything before you put it in a magazine.
I don’t know what the 330u is doing as you need another 1,000 on the output for 1Amp and the 100n does nothing. He took none of the essential calculations into account when he decided the resistor should be 820R.
It is obvious the author has no idea how the zener works in this circuit as his calculations were totally incorrect.
The zener works like this:
The value of the resistor is worked out by allowing 10mA for the zener and the maximum current required by the transistor. Say it is 40mA. The total current through the “supply resistor” needs to be 50mA. This current will flow ALL THE TIME.
When the output of the SERIES REGULATOR is zero, all the current we are talking about will flow through the zener. As the output current increases, the transistor will take (rob) the required current from the zener. When the output current of the SERIES REGULATOR is a maximum it will take 40mA from the zener and the zener will be left with 10mA.
Here is a circuit from June 2021 of DIYODE Magazine.
When some LEDs receive very bright illumination they produce an output voltage accompanied by a very small current.
In the following circuit, the OPTO-COUPLER has a bank of diodes on the output that produces about 10v and a current of 10 microamps. This is not obvious from the diagram and you have to reference the component date to see its features. The top of the stack of diodes is positive but it not obvious that the opto coupler is connected around the wrong way.
This is the part of the circuit we are discussing:
The circuit has been rearranged so you see the MOSFET with a positive voltage on D and a lower voltage on S. In addition you can see the negative output of the opto-coupler is connected to G and this will not turn on the MOSFET.
Here is the correct connection to the OPTO-COUPLER.
This modification is supposed to speed up the signal from the opto-coupler.
I cannot see how it works, and I say it does not work, so I have sent the request to Scott Williams, the technical editor of DIYODE for a clarification.
He has not replied to me.
Here is how the circuit works. Let us suppose the 10v from the opto-coupler rises slowly as the circuit is supposed to turn on at some special point.
When the 10v is say 3v, the base will be 3v and the emitter will be 2.5v
For the transistor to turn ON, the emitter has to be higher than the base.
When the voltage rises to 10v, the base will be 10v and the emitter will be 9.5v
So the transistor NEVER TURNS ON.
However the 10v from the opto coupler goes through the diode and into the Gate of the MOSFET and turns it ON.
The transistor does NOTHING.