# Questions about a simple variable power supply

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by samy555, Oct 30, 2012.

1. ### samy555New Member

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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:
How R3 value depends on HFE?
thank you alot.
2. ### chemelecActive Member

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HFE is essentially DC Gain.
So Depending on the DC Gain of a Different Transistor, that resistor might be incorrect
3. ### ChrisP58Active Member

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The purpose of the zener is to limit the gate-source voltage from reaching the failure level. Most mosfets have a G-S voltage limit of 20VDC, so the zener could actually be anyting from about 6V to 18V

The mosfet could be replace by a bjt, but you will need to come up with a driver circuit since a BJT needs real current into the base, whereas a mosfet needs only effectivly zero gate current.

In the second circuit, Q2 is acting as a current limiter. When the current through R2 (~200mA) generates enough voltage to turn Q2 on, it reduces the mosfet gate voltage, causing the output voltage to drop. As I see it, the hfe of Q2 is a non issue, since almost any gain will be able to shunt the current coming from R1. Where is R3?

Be aware, that this circuit will deliver that 200mA into a dead short. At that time the mosfet will be disipating a lot of heat.

Vin - Vout * Iout, 300V - 0V * 200mA, 300 * 0.2 = 60 watts

Now that assumes that the input is 300V, if the output is to be regulated at 300V, then the input voltage will need to be higher. So the actual disipation will be higher that 60W.

Also, be aware that this circuit has no voltage feedback from the output. The control voltage is derived from the unregulated input voltage, so any change in input voltage due to line fluctuations or loading conditions, will affect the output voltage.
Last edited: Oct 31, 2012
4. ### samy555New Member

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Thank you chemelec for fast reply

Why 12v? why not 9 or 6 volts?

In TVs they use a BJT in the horizontal circuit while using MOSFET in the power supply, can you tell me please how BJT would require a Higher wattage Control?

I thought that the resistance 3.3 Ohm is calculated by dividing the VBE voltage(0.6-0.7V) by the maximum current required.
5. ### simonbrambleActive Member

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The Zener is to protect the Gate Source voltage, not the gate drain voltage. You can use a 9V or 6V Zener, as long as it is lower than the gate-source breakdown voltage of the FET and larger than the gate source turn on voltage. A bipolar transistor requires current into the base whereas a MOSFET does not. The 3.3 Ohm resistor is purely dependent on the Vbe of the transistor. If you put a current of x through a 3.3Ohm resistor, you get a voltage drop of 3.3 * 'x'. If you scale the resistor right, you can get the transistor turning on at a certain current.

BTW - if you only want a low output voltage (less than, say 30V), I would use a transformer in front of this circuit. this will give you isolation as well as enable you to use lower voltage (and cheaper) component. Just stick it in front of the circuit above and feed the ouptut of the transformer into the L and N terminals above
6. ### spicaNew Member

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Hi everyone

Chanced upon this discussion about 300V DC Variable Power Supply while researching on the IRF740 circuit.

Thank you all for explaining the circuit as I have exactly built the circuit with current limiter part. But the thing is if I need a higher current, do I just simply remove the Q2 and R2??? Or Can I lower the value of R2 so that I can get around 500 to 600mA out of this circuit? ChrisP58 was right too about poor load regulation of the circuit. I am having the same problem in my application as well in which the output voltage just drops upon encountering low resistance load!

Can anyone help advise on this problem?

Thank you very much.
7. ### dr pepperWell-Known Member

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Yes, lowering r2 will increase the o/p current limit however at 600ma you'd be working q2 hard, and q1, might be better to replace q2 with something with a higher current rating, maybe something with a heatsink bolt (to220).

The reason regulation isnt brill is largely down to the fact the reference voltage to the gate of q1 from r1 comes direct from the dc rail, when you load the circuit the dc rail will drop also lowering the reference voltage going inot r1, you'd be able to improve this by regulating this volatge, which is a little tricky at high voltage, maybe a resistor/zener/cap would do this trick.
I'd suggest a resistor in series with the lower leg of vr1 to restrict the lowest volatge to a sensible value, if you tried to run the supply at 5v at max current Q1 would become an electric fire.

I'd like to build something like this, I mess with nixie tubes and vacuum tubes, it'd be handy.

Someone might be along soon with some more ideas
8. ### spicaNew Member

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Hi dr pepper

Regarding this fire hazard "if you tried to run the supply at 5v at max current Q1 would become an electric fire.", I am very concerned about it. Do you mean even after modifying it as per your advice, this still poses a threat somehow?
9. ### dr pepperWell-Known Member

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Yes, if you have a supply capable of 300v output you need a dc rail of at least that voltage, so suppose you have an output voltage of 5v at a max current of 600ma, that means that either accross the o/p transistor or across the current limit transistor you'll have 300v - 5v = 295v, power dissipation would be 295v x 600mA = 177 watts, this electronic wise is a lot of heat and would require a massive heatsink around 0.2 degrees per watt.

My term 'electric fire' was a similie, I meant the output transistor would behave like an electric fire.

This supply would be lot more efficicent if it was a switcher, but that would make things a lot more complicated.
10. ### audioguruWell-Known MemberMost Helpful Member

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The Engrish text in the blog is horrible, isn't it?

I agree that this extremely simple circuit has NO voltage regulation. The output voltage depends on the input voltage, the output current and the temperature of the Mosfet because the Mosfet is a simple follower.
11. ### WTP PepperActive Member

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I agree. the Vds of the MOSFET (which in this case controls the output voltage) is also dependant of the Ids current.

The MOSFET is being used as a variable resistor to control the current and not as a current controlled device. Zeners have horrible temperature characteristics which is why they can be crudely useds as temp sensors.
12. ### audioguruWell-Known MemberMost Helpful Member

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The zener diode has nothing to do with voltage regulation. It simply limits the max VGS voltage so the Mosfet does not exceed its max VGS rating of about 20V.
13. ### spicaNew Member

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I c, dr pepper. Seems I need a real big heatsink!
14. ### spicaNew Member

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Haha, right...

Anyway, do you or anyone else have any idea how to improve the circuit??

Thanks a lot for all the advice.
15. ### audioguruWell-Known MemberMost Helpful Member

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If you want voltage regulation then replace the simple circuit with a more complicated circuit that has voltage regulation.
Maybe you could use an LM317HV regulator IC with a variable regulated output from 1.2V to 57V then amplify it to 6.5V to 311V with a power amplifier circuit.
16. ### spicaNew Member

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Thanks a lot for your advice, audioguru. Do you have any circuit idea in mind for the power amplifier stage as I am not very familiar with it.

Regards

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