I have a 5V power supply and an input voltage that’s between 0V-5V. I need to create a circuit where, after feeding the input voltage X (X is between 0V-5V), the output voltage is 5-X (5 MINUS X) V. Would this be possible to achieve without any ICs?
Edit: The output is supposed to be used as the inverting input of an Op-Amp. Some give is okay (~300mV).
If you're OK with the output only swinging to within 150mV of each rail, you can try this IC-less circuit.
It's a current mirror where the input current flows through R4 and Q1 and is duplicated in R5 and Q2 which provides an output inverted with respect to input.
If the signal were connected directly to R4 the current through Q1 would crap out when its base reached 0.6V, so instead R1-R2-R3 attenuate the signal and center it around 2.5V such that the negative peak never turns off Q1. R5 being higher than R4 adds gain to compensate for the attenuation.
Unfortunately I only have a LM358 which is not Rail-To-Rail.
Then you can add an NPN transistor to achieve nearly a rail-to-rail output;
Note that the + and - inputs of the op-amp are reversed to make up for the fact that the NPN transistor inverts the signal. The output ranges from 4.8 V to about 0.2 V. You may need to add a small capacitor across the feedback resistor to avoid oscillations. Also, add a 100 nF capacitor across the 5 V supply close to the IC.
Would this be possible to achieve without any ICs?
Yes. But it would be much more complicated.
With a IC: It requires an op-amp with rail-to-rail output and 4 resistors.
Without an IC: You could recreate the op-amp functionality with discrete transistors, which would add at least 5 transistors, a capacitor, and a few more resistors. For example, this one.
With minimal components and no ICs, you could use a relay and 2 capacitors. It's ugly, but workable.
Don’t expect much accuracy and you must give up about 0.5V on both sides (Vcc and Gnd) from 0-5V range.
Make R6 variable from 220 ohm trimmer to set some bias current when a 0V is on input.
Blue is V2 , a green is Q1 collector.
Note: The Q2 compensates the Q1 BE offset (0.7V) and increases the input impedance.
If your 5 V supply and 0 V to 5 V signal don't need to share a common ground you can connect them in series opposing (in series with positive to positive) and take the output from across a load resistance. The signal then subtracts from the supply voltage, inverting the signal.
This is about as simple as it gets, but again, it relies on the signal voltage being floating. You'll also have to take into account the signal source impedance as it will affect the output voltage. The load resistor should be at least two orders of magnitude higher (x100) than the signal source impedance. So if the source impedance was 100 Ω the load should be at least 10 kΩ.
In the schematic E1 is just there to give a ground referenced version of V2 so it's easier to chart, you would leave it out in hardware.
You can try use this circuit:
simulate this circuit – Schematic created using CircuitLab
It is very important that Rc is big enough so that Q1 is in saturation.The output impedance(under no load) of this circuit is Rc||Re and since typically Rc>>Re it will be a little bit smaller than Re.
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