METAL DETECTOR MkII
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
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:
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
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
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
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 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
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
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
The fourth problem is the 7 minute timing. Delays above 3 minutes are very
unreliable with high value electrolytics and high resistances. You are
charging the 1,000u at less than 3uA and the leakage current for this type of
electro can be 3uA so it will never charge. This circuit was designed to go into
an air conditioner for a life of 20 years and and you can see the problems that
will start to show after 12 months and the company will go bankrupt.
The second relay and opto coupler are not needed.
And improved design will eliminate most of the components.
It can be designed much simpler.
Here is another circuit from D.Mohankumar.
None of his circuits work and he has 2 million Indians going to his
website each month and seeing faulty circuits that don't work and never
will work. It is no wonder the poor Indians are hopeless at
understanding electronics when they see this RUBBISH.
He says: Here is a simple 3.7 volt Lithium-Ion battery level indicator. Green LED
lights when the battery is full with 3.7 to 4.2 volts. When the battery
voltage decreases below 3 volts, Green LED turns off and Red LED lights.
Just connect it to the 3.7 volt battery charger.
There is no current-limiting
resistor for the green LED. Imagine what would happen to the green LED when the
battery reaches 4.2v. A green LED drop 2.4v The diode drops
0.7v and the base-emitter junction drops 0.7v That is 3.8v A fully charged
cell is 4.2v and after testing the circuit for 1 minute the green LED burnt out.
It got so hot on the high voltage that the crystal overheated and died. I
connected another green LED and the battery voltage dropped enough so the LED did not
burn out. The circuit is badly designed and I suggest you do not build it.
However the red LED did come on at 3v - - so just the green LED section has to
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
Here is a simple circuit that not been understood by any of the readers or
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
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
Before we leave this topic, here is a diagram from DIYODE magazine:
How can one end of a supply rail be 12v and the other end 9v
???? It is blatant, GLARING MISTAKES, like this that are not
corrected and the editor is saying I am: "nit-picking."
If I was a beginner, I would think it is possible to achieve the result above
AND HOW MISLED I WOULD BE.
Here is a circuit from April 2021 of DIYODE Magazine.
The author is trying to explain how to design a SERIES REGULATOR circuit.
But he is missing the hidden
facts and disasters of using a transistor such as a BD139. He does not
understand the BASICS of electronics and that's why he has made so many
mistakes in this articles.
The basics of a POWER TRANSISTOR is this: The gain of these devices
changes according to the current flowing. It may have a gain of 250 when
15mA flows, and a gain of 100 when 150mA flows but the gain will only be 40 when
500mA flows and less than 25 when 1 amp flows.
Let us say the transistor has a gain of 40.
This means the current into the base must be 13mA.
This means 13mA must flow through the base-collector resistor. And then you need
another 10mA to keep the zener in regulation.
When the output is delivering no current, the current through the zener will be
23mA. As the output current increases, the transistor "robs" the zener of
current and theoretically when the output current is 500mA, the transistor will
use 13mA and 10mA will flow through the zener. The zener will stay in regulation
until the current through it drops to about 1mA but when it drops to zero, the
output of the supply will drop.
The voltage across the base-collector resistor is 6.4v so the value of resistance
must be: 6.4/0.023 = 280R
But at 1 amp the gain of the transistor is about 25 and the base current must be
1,000/25 = 40mA plus 10mA =
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.
SOLAR CHARGER CONTROLLER
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
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.
HIGH SIDE MOSFET
Here is a stupid circuit from EngineerGarage.com
They are trying the explain High-Side connection of MOSFETS. But if the 5v
supply is capable of delivering current to the load, what is the point of
including the MOSFET !!!!!
Here is a circuit with a number of problems. Apart from the fact
that is overly complex, the question is: why would you want to produce 330v DC
for a LED lamps when most lamps operate on 5v from a USB connection??
And 240v LED globes work on 240v AC !!
The other question is: why select an expensive transistor? The only two
sources on the web are $7.50 each or 10 for $30.00!!
The circuit is designed for 7watt output, so why use a transistor capable of
handling 5 amps?? But if you design the circuit to pass 5 amps, the
voltage across the transistor will be 1v and the losses will be 5 watts and yet
the transistor is the size of a BC547 and the data sheet says the transistor
will handle 1.5 watts. I would like to see a BC547 dissipating 1.5
watts !!! Maybe it will dissipate 300mW.
The whole idea of selecting a common transistor is to allow the hobbyist to
build the circuit from readily-available components.
The circuit is only suitable for 4v operation
To build this circuit you will need to know a lot about ferrite core
transformers. The primary winding is 16 turns for 4v. Making it 4
turns per volt, The secondary is 1360 turns, making the output 340v AC. I
don't know why you want 330v DC. What LED lamp operates on
The feedback winding Nf appears as an air core on the circuit, whereas it is
wound on top of the primary.
Removing the air gap is not a good idea. Without an air gap the
transformer can be saturated and the energising current will simply go into
heating the winding.
Where are you going to buy the wire with all the different diameters?
Finally the 10k pot is in a low-impedance part of the circuit and the value is
only suitable for a low power inverter. The value is too high for a
Here is a basic run-down on its performance.
The circuit takes 2 amps. The voltage across a transistor is 1v. This
gives the transformer 3v. The gain of the transistor is 100. The base
current will need to be 20mA. The output of the feedback winding will be 3v
minus the diode drops of 1v. You have 2v remaining. The pot resistance
plus the 68R must be 100 ohms. The pot must be 32 ohms !!! How
are you going to select 32 ohms from a 10k pot !!!! The whole
circuit is so impractical !!!
It can be simplified by cross-coupling the bases of the two transistors
to the primary winding and using base resistors.
The whole idea of putting a circuit on the web is to help hobbyists produce a
worthwhile project as cheaply as possible with readily-available components.
There are a lot of cheaper, simpler designs, that I would recommend.