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-1.25 volts from positive supply.

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yafch

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Hi,
can any one suggest me an easy way to get -1.25v(20mA) from a -5v power supply.
regards.
 
A resistor.... What are you trying to power?
 
i need to use an operational amplifier in an application. supply voltages for the amplifier are +5v and -1.25. 20mA is just enough.i can use a little more voltage like -3 to -3.5, but for precision -1.25v is required.
 
Can you link the amplifier you're using?
 
here it is.
 

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Google for: 1.25V shunt regulator

Ken
 
yafch I meant the opamp itself, there are a lot of unconnected points in that circuit, the reason I'm asking is a opamp's performance should improve with a lower negative voltage, unless it's required for some other portion of the circuit to function.
 
If you want your LM317 adjustable regulator to go down to 0V then why not use the simple circuit in its datasheet?
Change the values of the resistors for an LM317 instead of the LM117 shown, change them for a different max output voltage and change the 680 ohm value for a different negative supply voltage which does not need regulation.
 

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I agree. I would change the point that's labled as -1.25V to -5V. Use a resistor and 1.25v shunt regulator/reference in series from the -5V to common. And move the bottom for R2 to the junction of the resistor and shunt regulator.
 
op-amp itself is common LM358. see the diagram i attached with the second post. it has a current limiting feature.

it may work without a negative supply for op-amp. but don,t you think that current limiting feature may not work properly under short circuit conditions.
 

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Oh. The opamp makes an adjustable current-limiting circuit for the LM317 regulator.

If the circuit does not have a -1.25V supply then its minimum output voltage is +1.25V which might cause a much higher output current into a short circuit than its setting.
 
You said you had a -5 volt supply, use that instead of the 1.25....
 
The -5V would be for the opamp and transistor, the -1.25 is needed to provide a negative reference to the regulator's divider R1/R2 so the output can be brought to 0.0V.

ken
 
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