50-500 VAC from a 15VDC

Hello,:razz:

:shock: I am looking to generate 50-500 VAC from a 15VDC Power supply. The component I wish to use :shock:

Hi,
I think you got it wrong. D1+D2 voltage should equal 15v. Because at first, with no load connected, you will have an open output of very high voltage. Forgetting that, what is your current requirement. You're going to have a VERY HOT transistor if you draw any reasonable current. If for example, you need 100mA, you have a power dissipation of (at 220v line) 29.6W in the transistor for an output power of 1.5W. That's an efficiency of 4.8%. That means, in your circuit, you're wasting 95.2% of energy input. That's a huge amount of power to lose, don't you think?

---------- Post added at 20:19 ---------- Previous post was at 20:16 ----------

What you need for an efficient design, is a multistage circuit, where you first chop the very high voltage into a more reasonable voltage and then chop that further using PWM. eg. First using PWM to step down 500v to 200v, that to 80v, that to 50v.
Go through this thread, which contains something similar:
https://www.edaboard.com/threads/193441/

Hope this helps.
Tahmid.

with the circuit provided you can't produce 50-500v , do you want 15V output or 50-500v output?, if it is 15V remove D1,D2 and Place D4 there,

/********* it is not isolated from mains ************/

thank u both of u.
i think i am worng direction. pl suggges me right direction.
isolated from mains no prablom.but i can't used bridgh rect. only hulf wave rect can use.

remove D1 and put D4 instead of D1 at the same direction , the output volt will be 15V - (.7 may be ) depending of transistor , since it is half wave it is better to replace c1 c2 for little higher values,
if your current requirement is small better add a capacitor before 1N4007

Hi sahu,
Can you please tell us what is your current requirement. If you state it, then it will be much easier to help you make a proper circuit for your required purpose.

Tahmid said:

Hi sahu,
Can you please tell us what is your current requirement. If you state it, then it will be much easier to help you make a proper circuit for your required purpose.

Click to expand...

my current requirement is max 500ma.

it is not good to design a liner regulator with 500v input for 15v 500ma output, the remaining 485V* 500ma = 242VA is wasted as heat, better go for smps type design

Hi sahu. First of all ! Take a look at your post "50-500 VAC from 15VDC"! It means you have 15Volts direct current(VDC) source. and you want to produce 50-500 volts alternating current(VAC)! Later you state the opposite ! The schematic you provide, is parametric stabilizer and stabilize at (Vzenner)D1+D2=180+180=360 Volts ! The input capacitor filter is useless- 100nf, assuming 50/60Hz, what to say about the rest... Galvanic isolation means going through a transformer, not rectifying ! The best approach is buck converter.Although I would not recommend YOU do any of that!!! I must warn you that This kind of voltage is VERY DANGEROUS! You may get electrocuted! Please, stick to voltages lower than 30VAC or VDC until you gather enough experience and knowlege. If any of this helped you in any way please click:Helpful?

Hi,
If you require 15VDC from 50-500VAC, you could first rectify the AC to DC and then use a buck converter. You can't use a simple zener based linear regulator. At 500VAC input (707VDC) input and 15V output, at 0.5A output, you dissipate [(707-15)*0.5] = 346 Watts (!!!) for an output of 7.5W.

If you require 50-500VAC from 15VDC, then you can first boost the voltage using boost converter or push-pull converter and then use a full-bridge converter to convert the DC to AC.

However both techniques are dangerous as they deal with very high voltages. So please be careful when dealing with such circuits.

Hope this helps.
Tahmid.

Tahmid said:

Hi,
If you require 15VDC from 50-500VAC, you could first rectify the AC to DC and then use a buck converter. You can't use a simple zener based linear regulator. At 500VAC input (707VDC) input and 15V output, at 0.5A output, you dissipate [(707-15)*0.5] = 346 Watts (!!!) for an output of 7.5W.

Tahmid.

Click to expand...

he is using half wave rectification , it won't be 707V

Ok, didn't see that. I thought he was going to use full-bridge rectification. Anyway power losses are still over the top and beyond practical limits.

Tahmid.