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Questions about a simple variable power supply

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samy555

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**broken link removed**
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:
https://avecircuits.blogspot.com/2012/07/0-300v-variable-high-voltage-power.html#.UI3WD2dQaoY
I found a similar circuit:
**broken link removed**
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.
 
**broken link removed**
My questions are:
1- Why the 12V zener?
Answer: To Prevent Excessive Gat to Drain Voltage

2- If i replace the IRF740 MOSFET by any 400v/10A BJT transistor, what would happen?
note: I know how to connect the BJT.
Answer: Mosfetsare Voltage Controlled and More EFFICIENT than Transistors.
Transistors are CURRENT CONTROLLED and would require a Higher wattage Control and a Different value

3- why C2 is only 10uF dislike C1?
Answer: It is just for Stability, Noyt needed for Filtering out Ripple.

WEITHER you use a TRANSISTOR OR MOSFET, It Needs a GOOD HEATSINK on it!

4- from:
https://avecircuits.blogspot.com/2012/07/0-300v-variable-high-voltage-power.html#.UI3WD2dQaoY
I found a similar circuit:
**broken link removed**
Author said:

How R3 value depends on HFE?
thank you alot.

HFE is essentially DC Gain.
So Depending on the DC Gain of a Different Transistor, that resistor might be incorrect
 
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.
 
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Thank you chemelec for fast reply

1- Why the 12V zener?
Answer: To Prevent Excessive Gat to Drain Voltage
Why 12v? why not 9 or 6 volts?

Mosfetsare Voltage Controlled and More EFFICIENT than Transistors.
Transistors are CURRENT CONTROLLED and would require a Higher wattage Control and a Different value
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?

HFE is essentially DC Gain.
So Depending on the DC Gain of a Different Transistor, that resistor might be incorrect
I thought that the resistance 3.3 Ohm is calculated by dividing the VBE voltage(0.6-0.7V) by the maximum current required.
 
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
 
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. :)
 
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
 
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

Hi dr pepper

Thanks a lot for your advice. :)

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?
 
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.
 
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.
 
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.

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.
 
Zeners have horrible temperature characteristics which is why they can be crudely useds as temp sensors.
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.
 
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.

Haha, right... ;)

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

Thanks a lot for all the advice.
 
Anyway, do you or anyone else have any idea how to improve the circuit??
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.
 
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.

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
:)
 
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.
Some IC regulator datasheets show how to add a power transistor to increase the output current from the IC regulator. Then it can still regulate the voltage down to +1.2V.
 
Hey
Know it's a long time since any have written in this posting, but have a question.
Have tried with this

jb13514723291-1.jpg

The only thing I did not take in the experiment was the fuse and C2, and C1 is 2 * 4700uF/450V parallel connected. Would use it as a power supply 0-300V up to 9A for testing. Although I am not much to mess with high power :woot:

I have set the IRF740 on an old CPU cooler and supposed that I would use two or more mosfet connected in parallel so that the load is less per unit.

The problem I have encountered is that the MOSFET burnes. Have so I can regulate the AC which I have put on the primary side before the bridge, better to gradually regulate the the primary side up 220AC rather than 311DC instantly :( Have had set a chain of lights I had lying with 3'40 watt bulb 220AC, and measured secondary after DC regulation, and as soon as I reach a certain voltage it burns and opens fully. Have also tried to regulate DC down while I regulate the AC up, same result. measured DC to around ½ watt when it burns.

Have tried to switch D2 out with at 10V, the result is the same. Any idea of what is going wrong?
 
Speedy, your circuit has no voltage regulation and its zener diode does not do anything.
If the input is 311VDC and the output is 5V at 9A then the Mosfet will heat and burn with (311V - 5V) x 9A= 2754W. The chip in the Mosfet will be at its max allowed temperature if the heatsink is perfect (impossible because it will heat up) when the power is only 125W.
 
Yes, know that everything depends on the heatsink why I also wrote about parallel connect several, and also came to think it was due to the approximately 110/120 watts which was the reason for the burned, so even it has good heatsink, I will to morrow try to give it a heatsink on the top side, to see if it helps. And whether it is that which is the cause, could manage more if it is what is the cause.

Did answer a little too quick with a yes, you say that the limitation of voltage plus the current through the circuit allocates watt in the circle, which I do not really get to fit, because it would burn faster at low setting than at full load where it dedicates consumption through it. So should it be understood so that if we say that it burned at 200DC 500ma which is 100 watts and 55 watts before the limitation, then this should be 155 watts?
 
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