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design +10v to -10v variable power supply??

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settra

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hello forum! i am trying to design, a bias circuit for an application. the circuit i am using this far is like that:
iBEDFvn.jpg


where, REF102 is a precicion voltage reference of 10v and ina 105 is a differential OP-amp. those components, are fixed, and are the becouse of their low noise... (i will be dialling with pA currents...) i will be using this, to bias a photoCell... i have to be able to bias it from -10v , up to +10v.

0 to 10v, is easy, with a potentiometer.. but what about -10v, to 10v?? how would i go, doint it without the use of a switch?? any advice is welcome!!
 
Well presuming you've got the above circuit to work and thus have a stable +10V and -10V - it really is an extremely simple problem.

Your potentiometer, instead of connecting between +10 and GND, connect between +10 and -10V. assuming your circuit is referenced to ground and not -10V for some strange reason...
 
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Any opamp will not produce a negative output voltage unless it has an additional negative supply voltage.
If your INA105 opamp has an additional -15V supply then its output can be -10V.
 
hello forum! i am trying to design, a bias circuit for an application. the circuit i am using this far is like that: View attachment 82641

where, REF102 is a precicion voltage reference of 10v and ina 105 is a differential OP-amp. those components, are fixed, and are the becouse of their low noise... (i will be dialling with pA currents...) i will be using this, to bias a photoCell... i have to be able to bias it from -10v , up to +10v.

0 to 10v, is easy, with a potentiometer.. but what about -10v, to 10v?? how would i go, doint it without the use of a switch?? any advice is welcome!!
Here is how I would do it. Turning pot from one end to the other goes from -10V to +10V, but it does need dual supplies. Plot shows output vs pot position.
 

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audioguru. it seems strange to me too... BUT if you look at the ina105 (precision unity gain ammplifier), at the figure 7, it has the exact same schematic.... what do you think??

mikeMI. can you explain a bit more plz?? what do the V1 V2, U3 , represent in regards to my schematic??
also, if i connect the -10, +10 , with a potentiometer... wouldn't that make it look like just a 0,20v ???
 
hello forum! i am trying to design, a bias circuit for an application. the circuit i am using this far is like that: View attachment 82641

where, REF102 is a precicion voltage reference of 10v and ina 105 is a differential OP-amp. those components, are fixed, and are the becouse of their low noise... (i will be dialling with pA currents...) i will be using this, to bias a photoCell... i have to be able to bias it from -10v , up to +10v.

0 to 10v, is easy, with a potentiometer.. but what about -10v, to 10v?? how would i go, doint it without the use of a switch?? any advice is welcome!!

Hi

Check figure 9. of the REF102 data sheet. It shows how to get the -10v using the INA105.

eT
 
audioguru. it seems strange to me too... BUT if you look at the ina105 (precision unity gain ammplifier), at the figure 7, it has the exact same schematic.... what do you think??
Look at figure 1 in the datasheet. Every circuit has a positive 15V and negative 15V supply.
You CANNOT get a negative output if it doesn't have a negative supply.
 
... what do the V1 V2, U3 , represent in regards to my schematic??
also, if i connect the -10, +10 , with a potentiometer... wouldn't that make it look like just a 0,20v ???
V1 is the +10V REF102 chip you talked about. It makes a precise +10V out of a not so precise V2 +15V supply, which also supplies +15V to the opamp. V3 is the -15V supply which the opamp needs so that it can pull its output to -10V.
 
listen to AudioGuru you can not create -10V from an inverting op-amp without a negative voltage rail.

wouldn't that make it look like just a 0,20v
Depends, if the rest of your circuitry is ground referenced then then the voltage from the pot will look like a +10 to -10 variable signal. If the rest of your circuitry is -10V referenced then yes this will look like a 20V to 0V variable signal.

I think first you want to work out how your going to make -10V...
Perhaps you should tell us what your trying to do and we can help you find the correct circuit. try to answerer the following questions as a good start:
  • What input voltage / current is available
  • What will the output be used for
  • How accurate do you want the voltage to be ... this will effect the pot used
 
well, ok. i suppose that indeed it can't produce the negative, if i dont supply a -15v as well.. but this is a bit irelevant, becouse, we can consider , that i will ahve a psu that gives both +15 and -15v... so there. -10 and +10 fixed :D

misterbenn. i am trying to build , a photocurrent sensor. that part is the part that is used to bias the photocell from -10v to +10... :)
Mike, i will take a look an i will tell you what i understood :D
 
MIkeMI. in ytour circuit, i cant seem to understand, why it can supply bias, up to 10v.... in a differential amplifier, the output should be R1/R2(V2-V1).. IN our case r2/r1 = 1.. BUT v2 will never be higher than v1, so it could produce positive bias...
what am i missing??
 
MIkeMI. in ytour circuit, i cant seem to understand, why it can supply bias, up to 10v.... in a differential amplifier, the output should be R1/R2(V2-V1).. IN our case r2/r1 = 1.. BUT v2 will never be higher than v1, so it could produce positive bias...
what am i missing??

You dont have the correct equation for this differential amplifier.

Consider this: Since the left end of R1 is tied to +10V, if the non-inverting input is at oV (wiper at the bottom), the output is
10*(-R2/R1) = -10V.
The gain from the non-inverting input to the output (if the left end of R1 is at 0V) is 1+R2/R1. We can use superposition to determine the effect of both inputs at the same time, namely the output is Vpot*(1+R2/R1)+10*(-R2/R1) = 2*Vpot-10 .

Try it for these three cases:

1. Pot wiper Vpot= oV: 2*0 - 10 = -10
2. Pot wiper Vpot= 5V: 2*5 - 10 = 10 - 10 = 0
3. Pot wiper Vpot= 10V: 2*10 - 10 = 20 - 10 = 10

Now look at the plot of output vs postition of pot wiper, again.
 
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