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Voltage divider.

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StormInABottle

Thu Jul 18 2013, 12:21AM

Registered Member #9252 Joined: Fri Jan 04 2013, 06:27AM
Location: Andromeda
Posts: 253

My rectified mains give off exactly 300 volts. I need 150 volts. So what can i do to divide the voltage. Note that i will be pulling about 3kw.

HV Enthusiast

Thu Jul 18 2013, 01:50AM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
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Since you have to ask, there is really only one solution.

Transformer or Variac - PERIOD

Or if you are well experienced with power electronics and switching converters, you could design and build your own switching supply such as a non-isolated buck etc....

But honestly, for this, all you need it a good variac or step-down transformer.

StormInABottle

Thu Jul 18 2013, 02:04AM

Registered Member #9252 Joined: Fri Jan 04 2013, 06:27AM
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Posts: 253

I was talking about things more like this.

klugesmith

Thu Jul 18 2013, 02:35AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

ExplodingLikeYeah! wrote ...
I was talking about things more like this.

OK, then here's the least-lossy solution. The figure is from your wikipedia reference.

V_in = 300 volts.

V_out = 150 volts.

R2 = 7.5 ohms. That's your 3 kW load, drawing 20 amperes.

R1 = 7.5 ohm 3 kW resistor. Chosen to drop 150 volts at a current of 20 amperes.

Note that the output voltage will be higher than 150 V whenever your load draws less than 3 kW.

[edit] How much variation is allowed?

As with any voltage divider, you can make it less sensitive to load variations by using smaller value resistors.
Suppose you can tolerate a +/- 5 volt variation in V_out, as load varies between 3000 watts and zero.
Then a simple voltage divider can serve you well.
R2 = 1.0 ohms, here not counting your load.
R1 = 0.936 ohms
The no-load output is 155 volts at 0 amps.
The full-load output voltage is 145 volts at 20.7 amps (3 kW).

Is that what you wanted?

Ben Solon

Thu Jul 18 2013, 02:44AM

Registered Member #3900 Joined: Thu May 19 2011, 08:28PM
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Posts: 600

Power is I^2R, and you want to draw 3kW at 150v. That's 20 amperes drawn. Kirchoff's current law states that there must be 20A flowing through R1, plus the current flowing through R2.

Your load if resistance is 7.5 ohms(which it must be to get 20A from 150v), R2 must also be 7.5 ohms. add up those parallel resistance and you get 3.75 net. In order to divide by two, R1 must be equal to R2(net 3.75 ohms). Congrats, you now have 3kw applied to your load, but 9kw being dissipated by your divider network.(40A*40A*3.75ohms=6kW, 20A*20A*7.5ohms=3kW ).

So if you want to draw 12kW from the wall to apply 3kW to your load, then by all means use a resistive divider

That's 40A from your 300V source by the way.

edit: or just throw a 7.5 ohm resistor in series with your load, the simpler solution slipped my mind. you still dissipate 3 extra kW in that shunt.

StormInABottle

Thu Jul 18 2013, 02:46AM

Registered Member #9252 Joined: Fri Jan 04 2013, 06:27AM
Location: Andromeda
Posts: 253

There is a capacitor solution right? I was very lucky to get a 40 watt resistor.

Ben Solon

Thu Jul 18 2013, 02:50AM

Registered Member #3900 Joined: Thu May 19 2011, 08:28PM
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Posts: 600

ExplodingLikeYeah! wrote ...

There is a capacitor solution right? I was very lucky to get a 40 watt resistor.

capacitors don't pass dc current. you can make an ac divider that way(the series tapped caps in a half bridge inverter work like that.)

StormInABottle

Thu Jul 18 2013, 02:59AM

Registered Member #9252 Joined: Fri Jan 04 2013, 06:27AM
Location: Andromeda
Posts: 253

Well. Can i put the capacitor divider behind the rectifier?
If the output is going to be AC then i need 107 volts. As you can guess. The input of ac is 220. My meter reads it as 210 though. For 3kw. What can the values be?

Ben Solon

Thu Jul 18 2013, 03:28AM

Registered Member #3900 Joined: Thu May 19 2011, 08:28PM
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Posts: 600

Yes, but Iâ€™m not going to calculate that for you. They have to be large enough to pass enough energy per cycle at 60hz to equal 3kW E = .5*CV^2, and be able to handle all the peak current. I=C*(dV/dT). You'll have a poor poor factor this way.

I back what evr says. If it's for a single application, a buck converter is optimal, if you want to easily change the voltage and use the converter in another project, a variac is your best bet.

StormInABottle

Thu Jul 18 2013, 01:40PM

Registered Member #9252 Joined: Fri Jan 04 2013, 06:27AM
Location: Andromeda
Posts: 253

It's for a single project. And a buck converter seems very easy to make.

The pictures at the theory of operation. Also i read that the switch that is on the schematic is actually an igbt or a mosfet.

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