Build a DC to DC Converter Accepting 300-900V Input

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Discussion Overview

The discussion revolves around the design and construction of a DC to DC converter that can accept high input voltages (300V, 600V, or 900V) and output a voltage that is 1.25 times the input voltage. Participants explore various approaches, safety considerations, and the practicality of using a disposable camera flash as the input source.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Experimental/applied

Main Points Raised

  • One participant requests information on commercially available converters or guidance on building a custom solution for the specified input and output voltages.
  • Another participant suggests using a small battery instead of a complex converter, questioning the necessity of the high voltage input.
  • Some participants inquire about the application and safety requirements, suggesting alternatives like a resistive voltage divider and linear regulator for achieving the desired output.
  • Concerns are raised about the feasibility of using a boost converter due to the short duration of the high-voltage signal from the flash unit.
  • Participants discuss the potential dangers of working with high voltages and emphasize the importance of safety precautions.
  • There is a suggestion to consider using a voltage doubler circuit to achieve higher voltages before rectification, although this approach is also cautioned against.
  • Some participants propose dropping from 600V to 375V as a simpler alternative, discussing resistor values needed for this method based on output current requirements.
  • A question is raised about effectively grounding high voltage circuits, with a suggestion to use a grounded metal case.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to achieve the desired voltage output, with some advocating for simpler methods while others propose more complex solutions. There is no consensus on the optimal design or method, and safety concerns are a common theme throughout the discussion.

Contextual Notes

The discussion includes various assumptions about the input source and output requirements, as well as the limitations of certain components in high-voltage applications. Participants highlight the need for safety measures when working with high voltages, but specific safety protocols are not detailed.

Who May Find This Useful

This discussion may be useful for individuals interested in high-voltage electronics, DC to DC converter design, and safety considerations in experimental setups.

soho1123
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i want to have dc to dc converter which accept a Vin (300,600,900V) dc as input and is able to give 1.25Vin dc as output. any commercially avaible converter?or how can i build one?

thanks
 
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It would require several stages of (expensive) down-conversion. Why not just use a small battery?

- Warren
 
soho1123 said:
i want to have dc to dc converter which accept a Vin (300,600,900V) dc as input and is able to give 1.25Vin dc as output. any commercially avaible converter?or how can i build one?

thanks

Could you tell us more about the application? That is a pretty strange reqirement. Can you just use a resistive voltage divider, and then a linear final regulator for the 1.25V output? What is the output current requirement? What are your safety isolation regulations for this setup?
 
i am using disposable camera flash as my input .it approxmately gives 300v dc .so i don't want to use some other input source.output current can be 1 to 2 milli Amps.i want to have circuit which takes for example 300V dc from camera flash as input and gives me out 375 v dc..

i want it to be cost effective too.

output =1.25*input

where input can be either 300 ,600 or 900V
 
Why would you want to derive some signal from the flash tube itself? Why don't you just grab a low-voltage control signal off the PCB?

It sounds like you're trying to make this (whatever it is) a lot more complicated than it ought to be.

- Warren
 
soho1123 said:
i am using disposable camera flash as my input .it approxmately gives 300v dc .so i don't want to use some other input source.output current can be 1 to 2 milli Amps.i want to have circuit which takes for example 300V dc from camera flash as input and gives me out 375 v dc..

i want it to be cost effective too.

output =1.25*input

where input can be either 300 ,600 or 900V

Ah, Lordy. chroot and I both misunderstood your original post (OP) where you said "1.25Vin". We thought you meant 1.25V output, so your question made no sense to us.

What you are really asking is to get 1.25 * Vin as your output, hence 300Vdc input and 1.25 * 300 = 375V output. Okay.

First of all, be careful with those high voltages. Have you worked with high voltages like that before? Do you understand what kind of safety precautions and procedures you should use?

To go from 300V to 375V, you would use a boost topology DC-DC converter. The simpler/cheaper ones do not run at that high of a voltage, though, so you would probably need to build your own out of high voltage discretes. You can learn more at the following websites:

http://en.wikipedia.org/wiki/Boost_converter

http://www.national.com/analog/power/simple_switcher_lm258x

.
 
Last edited by a moderator:
ya i know its dangerous .i am taking all the required precautions...thanks ..i'll try these links ...i'll post again if i face any problems...
 
It's not going to work, berkeman. The high-voltage signal on the flash lasts only a millisecond at full power, and perhaps only a tenth of a millisecond at moderate power.

There's no way a boost converter is going to be able to turn on, stabilize, and produce anything meaningful in that short a time period.

soho1123, it's a very bad idea to try to connect extra circuitry at the flash tube, and none of the small integrated switchers that berkeman is discussing are going to be able to tolerate or produce hundreds of volts.

You should really re-think your entire approach and find a low-voltage control signal to trigger your circuit, whatever it might be.

- Warren
 
The high-voltage signal on the flash lasts only a millisecond at full power, and perhaps only a tenth of a millisecond at moderate power.


The high voltage in a flash unit takes a few seconds to develop as the small charger circuit slowly charges up a large capacitor.

After that,the charge stays there until you fire the flash.

The real problem is that, for small flash units, the charging circuit really struggles to recharge the capacitor in a reasonable time and there isn't much power to spare.

I wouldn't recommend doing this, but you COULD get to the AC voltage in the flash before it was rectified and install a voltage doubler. Usually there is only a half wave rectifier there.
This might give you something over 600 volts which you could divide down with a resistive network.

ya i know its dangerous .i am taking all the required precautions

Despite the small size of these flash units, the voltage across the big capacitor is quite lethal and there would be no second chances with it.

You seem to be saying that you have a source of 300 V 600 V and 900 V. Why not drop down from 600 volts to 375 V? Easier than developing 375 from 300.
 
  • #10
Why not drop down from 600 volts to 375?

how do you recommend to do that?
 
  • #11
Why not drop down from 600 volts to 375?

how do you recommend to do that?


Depends on the current, but if it was 2 mA you could use a resistor.
(600 - 375) volts / 0.002 amps =112 K 0.45 watts

If it was 1 mA the resistor would be
(600 - 375 ) volts / 0.001 amps = 225 K 0.22 watts.


At these voltages it might be better to use larger resistors (like 1 watt) and use two or three of them to make up the total series resistance. Resistors can break down due to voltage limits before they overheat due to power.
 
  • #12
can anyone tell how to effectively ground these high voltage circuits
 
  • #13
You would normally have high voltages in a metal case that was grounded via a 3 pin plug to the mains.
This ground should be adequate.
 

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