# Analog divider using OpAmp

My teacher was teaching me about analog multiplier and divider using opamp. Following is the circuit is made us write.

simulate this circuit – Schematic created using CircuitLab

Could someone please explain me how is it working and where it is used. A little detail and further reading source is appreciated.

The circuit is based on (and can be seen as) a simple inverting opamp circuit. In general:

Acl=-Hf/(1/Aol+Hr) and for Aol infinite (ideal opamp) Acl=-Hf/Hr with

Acl: Closed-loop gain, Aol:Open-loop gain

Hf: Input signal portion available at the inv. input (Vout=0);

Hr: Output signal portion available at the inv. input (Vin=0).

In our case: Hf=1/2 and Hr=Vr/2.

Therefore: H(s)=-1/Vr

Special case: Vr=1 with H(s)=-1 (as known from simple inverter circuit).

The circuit shows an inverting opamp configuration, thus both the terminals of the opamp are at the same potential(ie 0 V). Now the current through the resistor connected to the source Vz is I = Vz/R. Since the opamp input draws no current ideally, this current through the resistor R connected to on of the inputs of the multiplier resulting in the input voltage of -Vz at the multipliers input.The product of -Vz and Vy is then multipliers output which gts inverted due to the opamp configuration.

• Multiplier direction is opposite to what you've taken. I've updated the diagram. How is it working as a divider? – Ansh Kumar Oct 2 '15 at 17:10