# Questions about a simple variable power supply

#### samy555

May 11, 2010
63

My questions are:
1- Why the 12V zener?
2- If i replace the IRF740 MOSFET by any 400v/10A BJT transistor, what would happen?
note: I know how to connect the BJT.
3- why C2 is only 10uF dislike C1?
4- from:
http://avecircuits.blogspot.com/2012/07/0-300v-variable-high-voltage-power.html#.UI3WD2dQaoY
I found a similar circuit:

Author said:
The degree of R3 gets from testing in this circuit, which depending on the gain of the transistor or the hFE value, so you may need to tune the value of R2
How R3 value depends on HFE?
thank you alot.

#### duke37

Jan 9, 2011
5,364
1. The 12V zener is to limit the gate/source voltage to +12V and -0.7V
2. The fet will take very little base current so the voltage across R1 will be close to zero. If a bjt is used, then there will be a base current and a voltage drop across R1. You may need a Darlington to work properly.
3. C1 is large enough to keep the voltage up between input pulses. C2 is there to smooth the voltage if the load takes pulses of current.
4. I do not see R3. You should put a 1K resistor in the base of the BC337 for safety.
5. A moderate capacitor from the gate of the fet to -ve will reduce the output ripple greatly.

Keep your fingers out, it would be nice to have another post from you.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393

1. The reason for limiting the gate-source voltage of the MOSFET is that the MOSFET gate has a very high impedance and can easily be driven to any voltage, but if the gate-source voltage exceeds about 20V (positive or negative) the MOSFET will be damaged. The zener restricts the gate-source voltage and protects the MOSFET.

2. What duke37 said.

3. What duke37 said.

4. I assume you're talking about the 3.3 ohm base-emitter resistor. This is not related to the gain of the transistor; it relates to the base-emitter voltage needed to bias the transistor into conduction, which is about 0.6~0.7V. Ohm's Law tells you that with R=3.3, this voltage will be reached when the load current is about 180 mA. At this point the transistor will start to conduct and will suck bias away from the MOSFET, thus limiting the current.

I also agree with duke37 that you should add a resistor in series with the base, to prevent damage to the transistor if the output accidentally gets shorted.

Regarding adding a capacitor from the MOSFET gate to the negative rail, I suggest adding a resistor of ~10k in series with R1 on the right side of it, and connecting the capacitor from the centre point of those resistors. Also I would limit the capacitance to around 0.1 uF. This is to prevent possible damage to the zener or the MOSFET.

Neither of these designs have any kind of power limiting for the MOSFET. The MOSFET will dissipate power according to P = I V, where I is the output current and V is the voltage ACROSS the MOSFET. Assuming 230VAC input, and output set at say 150v and 180 mA load current, the MOSFET will dissipate 30 watts (from (320 - 150) x 0.18) and will get extremely hot and destroy itself unless it is properly heatsinked.

IMPORTANT! Your first circuit rectifies the mains voltage directly, and the output is therefore HALF LIVE. This applies even to the negative output connection. BOTH of the output points are a clear and present electrocution hazard. The second circuit at least has an isolating transformer at the input.

IMPORTANT! Both of your circuits can produce up to 340V DC at significant current. They are a clear and present electrocution hazard. Please be careful!

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
In addition to all of the above, I would caution that 350V or so across a potentiometer is asking for trouble. Also you're going t have to be particularly brave to touch the metal shaft which may not be that well insulated...

#### samy555

May 11, 2010
63
Thank you, duke37, KrisBlueNZ and steve
1. The 12V zener is to limit the gate/source voltage to +12V and -0.7V
2. The fet will take very little base current so the voltage across R1 will be close to zero. If a bjt is used, then there will be a base current and a voltage drop across R1. You may need a Darlington to work properly.
What is the similarity between Darlington and MOSFET?
3. C1 is large enough to keep the voltage up between input pulses. C2 is there to smooth the voltage if the load takes pulses of current.
4. I do not see R3. You should put a 1K resistor in the base of the BC337 for safety.
5. A moderate capacitor from the gate of the fet to -ve will reduce the output ripple greatly.
moderate like 100u/25volt?
Keep your fingers out, it would be nice to have another post from you.
I'm never built a circuit before knowng expert opinion like you, thanks.

1. The reason for limiting the gate-source voltage of the MOSFET is that the MOSFET gate has a very high impedance and can easily be driven to any voltage, but if the gate-source voltage exceeds about 20V (positive or negative) the MOSFET will be damaged. The zener restricts the gate-source voltage and protects the MOSFET.

2. What duke37 said.

3. What duke37 said.

4. I assume you're talking about the 3.3 ohm base-emitter resistor. This is not related to the gain of the transistor; it relates to the base-emitter voltage needed to bias the transistor into conduction, which is about 0.6~0.7V. Ohm's Law tells you that with R=3.3, this voltage will be reached when the load current is about 180 mA. At this point the transistor will start to conduct and will suck bias away from the MOSFET, thus limiting the current.
Is it suck current or voltage?
I also agree with duke37 that you should add a resistor in series with the base, to prevent damage to the transistor if the output accidentally gets shorted.

Regarding adding a capacitor from the MOSFET gate to the negative rail, I suggest adding a resistor of ~10k in series with R1 on the right side of it, and connecting the capacitor from the centre point of those resistors (Are you mean a Low pass filter?). Also I would limit the capacitance to around 0.1 uF. This is to prevent possible damage to the zener or the MOSFET.

Neither of these designs have any kind of power limiting for the MOSFET. The MOSFET will dissipate power according to P = I V, where I is the output current and V is the voltage ACROSS the MOSFET. Assuming 230VAC input, and output set at say 150v and 180 mA load current, the MOSFET will dissipate 30 watts (from (320 - 150) x 0.18) and will get extremely hot and destroy itself unless it is properly heatsinked.
How to add a kind of power limiting for the MOSFET?

IMPORTANT! Your first circuit rectifies the mains voltage directly, and the output is therefore HALF LIVE. This applies even to the negative output connection. BOTH of the output points are a clear and present electrocution hazard. The second circuit at least has an isolating transformer at the input.
Is this isolating transformer like the one in the front end of a SMP power supply?

IMPORTANT! Both of your circuits can produce up to 340V DC at significant current. They are a clear and present electrocution hazard. Please be careful!

In addition to all of the above, I would caution that 350V or so across a potentiometer is asking for trouble. Also you're going t have to be particularly brave to touch the metal shaft which may not be that well insulated...
Thank you all for a dvice, I know that.

still one point: all modren power supplys found in computers, TVs and monitors are all using the same front end, i.e. 220v mains, RF filter, rectifier bridge, big electro cap then to a mosfet through a primary windings of a high frequency transformer. I want to say that is actually used in a real life
thanks alot

#### duke37

Jan 9, 2011
5,364
A mosfet is controlled by the voltage at the gate, the current is essentialy zero if the voltage does not change. If the voltage changes, it is necessary to charge and discharge the input capacitance.

A bjt is essentialy controlled by current into the base so should be connected to a low resistance source. The output current is the transistor gain times the input current. transistor gains may be as low as 10 up to 500 or so.
Putting another transistor in front will give a gain of gain1 * gain2 so the input current will be low. A Darlington transistor consists of the two transistors in one package, gains of well over 1000 is possible.

A capacitor to the gate of the fet or, better, to a tapped resistor could be 100nF 400V. A low voltage electrolytic will soon go pop.

You have not said what this circuit is to power, do you mean to use it to provide a power supply similar to that used in a computer? If so, you do not need the regulation, the switch mode supply will control the output voltage even if the input voltage varies.

Your second circuit has an isolating transformer running at mains frequency (50Hz?). A switch mode supply has a small high frequency transformer which gives isolation at the correct voltage.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Is it suck current or voltage?
It sucks current, which decreases the gate-source voltage and makes the MOSFET "turn OFF more". As the load current increases past the limit value, the transistor starts to conduct and starts to suck bias away from the MOSFET. The MOSFET conducts less, and the output voltage drops. Assuming the load is resistive, this causes the output current to drop as well, until an equilibrium is reached. Thus, the output current is limited.
(Regarding adding a capacitor from the MOSFET gate to the negative rail)
Are you mean a Low pass filter?
Yes. I would describe it as smoothing, but low-pass filtering is the same thing.
How to add a kind of power limiting for the MOSFET?
It's not easy to add. Ideally the circuit needs to limit the instantaneous power dissipated in the MOSFET, and limit its temperature. I haven't seen any circuits for this that I could point you to. Anyone? Anyone? Bueller?
Is this isolating transformer like the one in the front end of a SMP power supply?
No. Switching power supplies don't have a transformer at the input. Maybe you're thinking of the common-mode choke at the input of a switching supply? That is part of the interference reduction circuit. It's not an isolating transformer.
still one point: all modren power supplys found in computers, TVs and monitors are all using the same front end, i.e. 220v mains, RF filter, rectifier bridge, big electro cap then to a mosfet through a primary windings of a high frequency transformer. I want to say that is actually used in a real life
No, that's different. Switching supplies DO have isolation, from the live-side circuitry (all the stuff that's powered from the bridge-rectified mains voltage) to the secondary side.

Everything that's galvanically connected to the mains is LIVE and must be properly insulated, in order to pass safety regulations. Circuitry that's connected to bridge-rectified mains voltage is HALF LIVE. If you connect a light bulb from the negative rail (after the bridge) to mains earth (which is connected to mains neutral), the bulb will light up. And so will you, if you touch anything that's fully live or half live.

The output(s) are isolated from the input, and the 0V rail for the outputs can either be floating relative to mains, or linked to mains earth.

The output terminals of your circuit are just marked + and - and there's no indication of what they are to be used for. If you're bringing them outside the fully insulated circuitry and making them available externally, you MUST use an isolating transformer, as in the second circuit. Even then, there is potentially a very high voltage ACROSS the terminals, which is an electrocution hazard, but at least the actual terminals themselves are not live relative to mains earth.

The safety regulations are not put in place without good reason. You can, and will, electrocute yourself if you touch the negative or positive output terminal of your first circuit, if your body has a path to earth (which it usually does, more or less).

#### samy555

May 11, 2010
63
Thank KrisBlueNZ you from my heart, you are a wonderful man

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Thank KrisBlueNZ you from my heart, you are a wonderful man
Aww thanks
I should have explained in my previous post: the reason why we're all concerned is that we don't know what you intend to use this power supply for. The output is potentially very dangerous. If you intend to use it for experimentation, where you will be connecting wires and components to it, it's a real danger. Normally, everything that's galvanically (directly) connected to the mains needs to be fully insulated and enclosed. But it looks like you want to bring those output terminals out to circuitry or components that you might be able to touch. That's why we're concerned.

#### samy555

May 11, 2010
63
I know that the circuit is very dangerous
I do not intend to implement
I love to discuss the theory of work
I would love to know more about circuits and their components, this is my hobby.
Accept my greetings and respect

Nov 2, 2012
3
good job guys

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