Voltage converter

[ACDC]

Can anyone suggest a method or methods for converting a incoming 0-5 vdc control voltage
so there is the opposite voltage direction. When input is 0 output is 5, input 5 output 0.

[Robert Reed]

How about an Op-Amp. Plus 5 volt bias on the positive input. 0-5 volts applied to the negative input.Out put will drop from 5 volts to zero volts as the negative terminal rises by subtraction .The gain of the circuit will have to be more than one, but both input voltages can be adjusted thru resistive divider networks to get your desired results.

[Externet]

If the control voltage is only either 0V (low) or 5V (high) and never stays in between, one transistor in common emitter or one optocoupler or one logic inverter IC does the inversion.
I cannot verify this link in the compfuser I am now, but should explain it :
---> http://www.falstad.com/circuit/e-rtlinverter.html

Edited: Yes, that is it. Click on the left H or L. The output inverts.
Miguel

[MrAl]

Hi,


Here is a circuit that will do what you want.
The power supply has to be around 7v but can be higher too up to 30v.
The 5v supply is a reference or a good regulator.
The 0.2v supply can be 0.2v to 15v as long as the positive supply is limited to 15v.
The 0.2v supply is a negative supply, but it is only needed if the output has to get right
down to zero volts and 0.1v for example isnt good enough for the low end.

[ACDC]

Thanks ALL, appreciate the information and direction.

(This is for a voltage that will vary between 0 and 5.)

Now I just need to find the box that has those op amps
in and build a test circuit.

[haklesup]

Sounds like you want an inverting amplifier with gain of 1 and supply voltages close to 5V so the Op amp needs rail to rail operation. Higher operating supply voltage will give you a bigger selection of op amps without the rail to rail requirement.

[redrocker]

The describing equation for the OP's circuit is simply:

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Vout = 5 - Vin.

The Kirchoff's current equation for the inverting node of an inverting amplifier is:

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I1 = (Vin / Rg) = -I2 = -(Vout / Rf)

This assumes a 'virtual ground', or zero volts, at the inverting input, because the non-inverting input is held at ground. However, if a bias, such as 2.5 volts, is applied to the non-inverting input, as in MrAl's circuit, the above equation becomes:

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I1 = ((Vin - 2.5) / Rg) = -I2 = -((Vout - 2.5) / Rf)

Under the special circumstance where Rf = Rg, this equation reduces down to the aforementioned describing equation. Hence, MrAl's circuit mathematically implements the equation. Nice job, MrAl.

Furthermore, the OP might be interested in this simple technique for obtaining the -0.2 V bias for the negative voltage rail (which I lifted from another forum):

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 +------------ VCC 
_|_ 
 - 
 | 
 | 
 +----|<|----- GND 
 | 
 | 
 +------------ V- 

The diode could be Schottky type. Also note that National Semi makes the LM7705 IC specifically for this function.