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How to power 2 low power loads from 1 HV source

Frankie232

Jul 20, 2010
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hello,

i am trying to power two loads from the output of a dc-dc converter (please see diagram). how can i do this?


many thanks
 

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Resqueline

Jul 31, 2009
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What's the nature of those loads, and how accurate does the voltages need to be?
 

Frankie232

Jul 20, 2010
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Sorry for the misunderstanding. I am actually trying to power a system (a colloid thruster) that requires 2 separate voltages.

the voltages should be as accurate as possible.

thanks
 

Resqueline

Jul 31, 2009
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Ok, so you're looking for sub-uA currents. Is the control voltage for output adjustment or is it just on/off?
Aren't those voltages supposed to have positive in common (nozzle)? The Extractor electrode meant for the smaller voltage and the Accellerator electrode for the larger?
I figure both electrodes will vary their current with the flow rate?
If efficiency is not an issue I'd use simple resistor dividers. A number of resistors in series means ordinary types can be up to the task (200 or 500V each).
The higher the divider current is relative to the electrode current variation - the higher the accuracy will be.
 

Frankie232

Jul 20, 2010
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The control voltage is for output adjustment, and you are correct about the voltages. Efficiency is very much an issue. I am actually trying to find out if it will be more efficient (in terms of power) to use 1 DC-DC converter instead of 2.
 

Resqueline

Jul 31, 2009
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I'd expect the converter to use at least 100 times more than the 1-3mW that the accelerator electrode might draw, so spending 30-40mW on a divider would make sense.
The accellerator would be connected directly to the output, and the extractor would be connected 1/3rd down the resistor divider (consisting of at least 9 resistors).
The control voltage would then have to be adjusted to produce a 3300V output.
How would you like the control voltage to be derived btw.? Via a CPU & software or by direct analog (potentiometer) control?
How large resistor values are you able to get hold of? (The target would be around 33M ohms.) Do you have a datasheet or a type # for that converter?
 

Resqueline

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Ok, so the converter is capable of 1W output (200uA), drawing 100mA unloaded & 250mA fully loaded. That means a resistor divider won't affect efficiency very much.

Since 1100V is 1/3rd of 3300V the number of resistors has to be dividable by 3, and since each (1/2W) resistor can tolerate 500V you have to use at least 7, hence 9.

If we choose to run the converter at as much as 1/10th power it'll draw 115mA and output 20uA at 3300V. R=3300V/20uA=165Mohm (/9=18Mohm each).
A 20uA divider current related to 0.3uA electrode current is a factor of 67, ensuring a voltage stability of 1.5% (16.5V) on the 1100V.
Choosing 33Mohm instead will reduce the divider current to 11uA, giving a stability factor of 37 (2.7%, or 30V).

Now, I don't know how big an engine you're going to make, but I saw a figure of one drawing up to 300nA, which I used as a basis for my calcul-/ estim- ations.

But I came to think; will you be using the converter & 4300V for driving something else besides? If so it won't be quite as easy as this, but it'll still be within reach.
 
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