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Author Topic: I have a problem with the numbers.. Unusual for me.  (Read 1113 times)
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I am building a bench supply.  I have a 36 volt 9 amp supply that I was going to use as a voltage source.  But it doesn't have negative voltage.  I came across this circuit for creating negative voltage:


So that's good.  I can use fixed voltage regulators for many voltages but it would be nice to have an adjustable voltage as well.  So I am looking at this from a data sheet for the  LM79xx:
Here is the link to the data sheet:http://www.hep.upenn.edu/SNO/daq/parts/lm7915.pdf
and this is the part I am having some difficulty with:


Look at the bottom of the page.  There is a set of values for R2 at different voltages.  Using these values I need to find R1 but the numbers I am getting are impossible.  So I suspect I don't grok what Vset is and the only place you will find Vset in the data sheet is right there.  I am using 36 volts for the supply, that is what I thought Vset would be, and I want a negative 15 volts out.  How is this done?  I keep coming up with a negative value for R1 and unless things have changed a lot I don't think you can get a negative resistance potentiometer.
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Sorry, joseph_m, but that circuit, as drawn, is dead in the water.
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The circuit is a voltage divider where the voltage on R2 is a set voltage (whatever the regulator's fixed output).  So if you're using the -5V supply, Vset is -5V. 
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Run the number with Vset at -5 volts and that is what you want Vout to be and you get zero for your resistance of R1.  Run the numbers you will see what I mean.
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Sorry, joseph_m, but that circuit, as drawn, is dead in the water.

Well that's helpful.  Want to elaborate on that?  It came right out of the data sheet.
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Run the number with Vset at -5 volts and that is what you want Vout to be and you get zero for your resistance of R1.  Run the numbers you will see what I mean.
Huh?  Why would you be building a variable output circuit when the fixed regulator already has the voltage you want.  Of course you get 0ohm, because you wouldn't need a divider to -5V out of a -5V supply.

Sorry, joseph_m, but that circuit, as drawn, is dead in the water.
Well that's helpful.  Want to elaborate on that?  It came right out of the data sheet.
The first circuit.  Where did you get it?  
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The first circuit will (might) output +15V with respect to its GND on the +15V side, OR -15V wrt GND on the -15V side, but there isn't any way that +15V will be 30V above -15V output, which is also necessary for a bipolar supply.
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@joseph_m

That first schematic in your post ? What the ?  smiley-eek  Really ?

I agree with Runaway Pancake here.

Let see... you have 36 V and you want to divide by 2 ? Correct ?   Like 36 / 2 = 18 V.

I just have a schematic that do just that. I included in this post. 
   






* Voltage_spitter.jpg (37.27 KB, 1024x663 - viewed 18 times.)
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