Current measuring with shunt resistor

Thread Starter

Peter523

Joined Aug 15, 2021
97
Hello,

i would like to construct a circuit measuring the current passing through the load of a buck converter i designed. So i constructed the circuit shown below (from internet sources) in PSpice: initial.png
This circuit operates normally giving desired result(the voltage on the output of LT1498 gives 1:1 the current passing through the load and RShunt).

But after changing the topology(for some reasons) of the buck converter the circuit doesnt operate correctly as it doesnt measure correct current.
The new topology is :final.png

Could anyone help on this?

Thank you.
 

Ian0

Joined Aug 7, 2020
3,760
The voltage on the op-amp's inputs is outside it common mode input range.
The voltage on its inputs is about 8V, whereas it common mode input range only extends as high as its positive supply (3.3V)
You can use a specialist shunt amplifier such as AD8418 which has a common mode input range which extends from -1V to +70V even though the device runs off a 3.3V supply
 

Thread Starter

Peter523

Joined Aug 15, 2021
97
Yes , with 5V power supply to the initial opamp i used the result is correct. But i would prefer ,if possible, a solution with 3.3V max supply. Unfortunately there is not AD8418 ,that @Ian0 proposed, available in PSpice to test its operation.
 

Thread Starter

Peter523

Joined Aug 15, 2021
97
Another thing that i observed is that powering the opamp with higher voltage e.g. 5V the accuracy of the measurement is being a little reduce. For example, for current 3A the initial circuit measured ~2.9A but this may measure ~2.65A.
 

DickCappels

Joined Aug 21, 2008
7,981
Another solution is to change R142 and R131 to reduce the maximum DC voltages at the inputs to the LT1498. That will reduce the gain quite a bit, and that loss of gain can be made of by increasing the shunt resistance and/or following the differential amplifier (LT1498) with a gain amplifier and/or scaling in firmware.

1631465184110.png
 

Ian0

Joined Aug 7, 2020
3,760
Another thing that i observed is that powering the opamp with higher voltage e.g. 5V the accuracy of the measurement is being a little reduce. For example, for current 3A the initial circuit measured ~2.9A but this may measure ~2.65A.
If it does, then you still have a problem. Provided that the input common mode voltage is not exceeded, it will be accurate.
 

Ian0

Joined Aug 7, 2020
3,760
increasing the shunt resistance
Most commercially-available shunts are designed to produce an output voltage between 50mV and 100mV. It seems as though it is that way because it works.
So, when choosing a shunt resistor, then it's probably not a bad idea to design it to drop 50mV or so. Then it doesn't need a ridiculous amount of amplifier gain which risks offset errors and noise
 

MrAl

Joined Jun 17, 2014
8,555
Hello,

i would like to construct a circuit measuring the current passing through the load of a buck converter i designed. So i constructed the circuit shown below (from internet sources) in PSpice: View attachment 247755
This circuit operates normally giving desired result(the voltage on the output of LT1498 gives 1:1 the current passing through the load and RShunt).

But after changing the topology(for some reasons) of the buck converter the circuit doesnt operate correctly as it doesnt measure correct current.
The new topology is :View attachment 247756

Could anyone help on this?

Thank you.
It looks like the choice of op amp does not match the application very well.
The op amp has reasonable input offset voltage for a lot of applications but not really for precision DC applications like this one.

For one thing, you dont need 10MHz bandwidth to measure DC. In fact, a lower bandwidth op amp would most likely be a better choice because a lower bandwidth device can be found with a much lower input offset voltage spec leading to less error in the DC measurement. The chopper stabilized op amp type can have extremely low input offset voltage (because that is one of the main target design goals for that kind of op amp) and so there is much less error in the DC measurement. These kinds of op amps may have a bandwidth less that 1MHz, in fact possibly much less, but when it comes to measuring DC they provide the gain needed as well as the lower error in measurement.

Of course it does in part depend on what kind of accuracy you are after too, but consider that with the current choice of op amp even without any current flowing it may appear that the current through the load is almost 0.5 amps, and that's without any current at all. That's a lot for a circuit that has to measure from 0 to around 3 amps. If it was from 0 to 30 amps that would be different, but for measurements from 0 to 3 amps an error of 0.5 amps is somewhere around 15 percent, which although not the end of the world, why not just select a better op amp more suited to the application. But that's not the worst part of it. If the actual current is 1 amp, then the error could end up being as high as around 50 percent which is most likely unacceptable.

It makes sense to look for an op amp that fits the application better and that would be a chopper stabilized type which can help create a circuit that is much more accurate.

Something should be said about the sense resistor itself too and the wiring to that resistor.
The wiring should be such that the current handling wires go to the resistor terminals as expected, but the sense wires also go directly to the terminals of the resistor. These sense wires can be of much lighter gauge wire because they dont handle any current, but they must connect directly to the terminals of the resistor.
The value of the sense resistor also comes into question however. 1 mOhm is just 0.001 Ohms and that means you will have to measure just 1 millivolt for a 1 amp load current. That's a very very low voltage to have to measure even with a good op amp. That means just 0.003 amps for a full load current, which is also very low. Better would be a 0.01 Ohm resistor which still does not drop much voltage and interfere with the normal circuit operation but of course it depends what you need here. However, 0.01 Ohms will give you a starting voltage of 10 times higher than with a 0.001 Ohm sense resistor so the gain of the circuit can be set 10 times lower. That could make a big difference in the measurement accuracy and noise.
There are also special types of resistors made for current sense like this, and they have four terminals not two. You may want to look into that also. If efficiency is not as much an issue then you can go to high side current sensing where the feedback for the buck still comes directly from the output and so the sense resistor effectively does not drop any voltage at all.

As the output is from a buck circuit which is a switching regulator the output will be jumping up and down at regular intervals. This means the output current will not be perfectly constant. That means some sort of low pass filtering would also be a good idea. Many measurement circuits like this dont have to be in a feedback loop though, if they do, then you have to be more careful about how you filter the signal because then the circuit becomes more advanced with a more complicated output response. This means you'd have to study the effects of added feedback carefully and do some careful tests.

The most important point here though is that the op amps choice should match the application better so you get reasonable measurements. If for some reason you need the super speed of 10MHz bandwidth then we'd have to talk about this more but that's not usually the case with DC measurements, and in fact often slower is better because it averages over a longer time period which makes for more consistent measurements.
 
Last edited:

dcbingaman

Joined Jun 30, 2021
498
Hello,

i would like to construct a circuit measuring the current passing through the load of a buck converter i designed. So i constructed the circuit shown below (from internet sources) in PSpice: View attachment 247755
This circuit operates normally giving desired result(the voltage on the output of LT1498 gives 1:1 the current passing through the load and RShunt).

But after changing the topology(for some reasons) of the buck converter the circuit doesnt operate correctly as it doesnt measure correct current.
The new topology is :View attachment 247756

Could anyone help on this?

Thank you.
If for some reason you need to use that op amp and you need to power it from 3.3V you can always do this:

1632006953072.png
This will reduce the gain by a factor of 2. If you want to recover the gain just increase R132 and R127. For less loading effects you may want to use 10 ohm resistors in place of the 100 ohms I choose here. This will increase the common mode voltage from around 3.3 to 6.6V.
 

dcbingaman

Joined Jun 30, 2021
498
Hello,

i would like to construct a circuit measuring the current passing through the load of a buck converter i designed. So i constructed the circuit shown below (from internet sources) in PSpice: View attachment 247755
This circuit operates normally giving desired result(the voltage on the output of LT1498 gives 1:1 the current passing through the load and RShunt).

But after changing the topology(for some reasons) of the buck converter the circuit doesnt operate correctly as it doesnt measure correct current.
The new topology is :View attachment 247756

Could anyone help on this?

Thank you.
This is in LTSpice and it is far superior for high side current measurements. Reduces parts count and increases simplicity. This uses an INA250A1 shunt current monitor. The shunt is built into the chip and it can handle up to 36V common mode.
 

Attachments

Top