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

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?

• Your education is mainly learning how to learn. This means learning where to look and find the answers easily. ( like a search tab or a search engine with the right key words. electronics.stackexchange.com/questions/333592/… Commented Oct 9, 2017 at 20:37
• Is the shunt in the low side (ground) or on the high side (supply) of the load? Commented Oct 9, 2017 at 21:14
• @peufeu Good point, but I'd go further. What if neither end of the shunt is near ground or a high side? It could actually be "anywhere" in the circuit.
– jonk
Commented Oct 9, 2017 at 22:21
• This is a design site and engineers are eager to show you stuff you can make, but I would suggest simply buying a signal conditioner module that provides galvanic isolation suitable for the maximum expected voltages. You are using a ruggedized PLC for a reason, why compromise it with homemade barnacles? There are also more ways than I can count that this could end in tears without a LOT more information than has been provided. Commented Oct 9, 2017 at 23:05
• +1 for @SpehroPefhany also LEM makes a whole line of isolated hall effect current sensors which are nice when you don't want to think about voltages... Commented Oct 9, 2017 at 23:21

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.

• And the op-amp needs to be able to work with inputs and output down to negative supply rail if you have a positive-only supply. Commented Oct 9, 2017 at 19:58
• @Transistor Your comment confuse me. Commented Oct 9, 2017 at 20:04
• At I = 0 the non-inverting input will be 0 V and the output should be 0 V. For a single-ended supply (I said "positive-only" because I was concerned that the OP would not understand "single-ended") you need an op-amp that can input and output 0 V. Many standard op-amps can only swing the output down to 1.5 to 3 V from the negative supply rail. See electronics.stackexchange.com/questions/327109/…. Commented Oct 9, 2017 at 20:06
• @Transistor huh, that's the first time I've heard of op-amps not being able to reach the negative rail. I thought all op-amps could do that and it was some op-amps that couldn't reach the positive rail. Thank you for elaborating. Commented Oct 9, 2017 at 20:13
• Questions are regularly posted about "why my opamp output won't go below X.X volts". Be aware that on the rail to rail op-amps the pull-down or pull-up is often very weak as you approach the rail so you can't source or sink much current. "... and it was some op-amps that couldn't reach the positive rail." Most of the output stages are somewhat symmetrical so you'll have the same problem at both ends. There are those that are optimised to reach the negative rail as this is the most common problem due to negative being circuit GND. Commented Oct 9, 2017 at 20:18