# Electrical – How to amplify the 0-60mV voltage drop of a current (sense) shunt to 0-10V

amplifiercurrent measurementplc

I want to measure the current of my system with a shunt.
My system has a maximum current of 10 A.
How can I change the 0-60mV signal from the shunt to a 0-10V signal for my PLC?

What you need is called "non-inverting operational amplifier".

And it looks like this:

The voltage amplification will be \$1+\frac{R_2}{R_1}\$, in this case \$R_2\$ is \$10 kΩ\$ and \$R_1\$ is \$60 Ω\$.

But I won't assume that you got \$60 Ω\$ laying around, but perhaps you got a potentiometer laying around that you can tune.

In that case it would simply look like this:

As you can see it says \$2.4 V\$ rather than \$10 V\$, that is because in the simulation there's only 20 steps, or is it 10, doesn't matter. I can't make it to \$10 V\$ in the simulation, but in reality you can easily do that.

Here's a link to the simulation if you want to... play around.

## EDIT after Transistor's comment

If you got an op-amp that can't output negative voltage supply, then these circuits, updated of the two above, will most certainly work out for you. Notice how the inputs to the op-amp have switched and that there is an Resistor-Transistor inverter as the output. If the input is \$0 mV\$ then it outputs about \$10 mV\$ (not perfect, but it beats \$1.5 V\$). if the input is \$60 mV\$ it outputs \$10 V\$.

Notice how I changed the feedback to \$600 Ω\$ and \$100 kΩ\$ instead. 10 times higher than before. That is because I chose \$10 kΩ\$ pull up resistor to make sure that the transistor can pull the resistor down and there won't be excessive power being lost. Had I chosen \$1 kΩ\$ resistance instead it would've been \$P = \frac{V^2}{Ω} = \frac{12^2}{1000}=144mW\$. Rather than \$144 mW\$ it's now \$14.4 mW\$.

And here's the other circuit that now contains a \$100 kΩ\$ potentiometer

And here's the link for those if.. you by any chance want to mess around.