operational-amplifier
I have a +3.5V/-3.5V AC signal (audio frequency range) that I need to convert to 0V/5V range so that I can use it as input for an ADC.
I was trying to use a non-inverting summing opamp, but I can't figure how to calculate the resistors.
I know that I need a gain of 7.14 (5V/7V) + 2.5V offset, but I don't understand how to determine the resistors for the non-inverting topology (assuming 5V as voltage reference for the offset).
can anyone help me understand how to calculate that? thank you
There is a calculator on this website here. Here are the results for the two scenarios of: -
I've arbitrarily chosen R1 to be 1K and I knew from experience that V2 needs to be half the range of the output requirements so I set that to 2.5V. By simple examination of range of input voltages to range of output, the gain has to be 0.25.
As you can see, both scenarios produce the same result of R2 = 250ohms, R3 = 1kohm and R4 = 250ohms.
They are regulators as you thought.
To properly control the MOSFET on each regulator, the gate needs to rise several volts above the source, probably over 6 volts greater than the source. If the source voltage is 3.3 volts the gate might need to rise as high as 9 or 10 volts when the MOSFET is driving a heavy load current. This means the op-amp MUST be powered from a supply greater than 10 volts.
The non-inverting input is the demand voltage for the regulator - for the 3V3 supply the non-inverting input will be 3.3 volts and ditto the 2V5 supply.
The MOSFET is rated at 20 amps and this means the supply current to the circuit might be in the realm of 5 to 8 amps - try and find a linear voltage regulator that would fit the bill an be a low drop-out type - realistically with a drain voltage of fractionally above 3.3 volts the output would still be in regulation. This is of course due to the op-amp being powered from 12 volts.
As to whether it is a bad design, I'd say not BUT I don't know the full spec of the circuit.
Best Answer
This is how it goes with a simple voltage divider:
simulate this circuit – Schematic created using CircuitLab
Just in case you don't have a 10.5VDC supply, you can cap couple it:
simulate this circuit
The actual values would depend on the lowest frequency you need to measure.