How to measure current with opamps?

[pvh1987]

I know that current can be measured with an opamp (configured as a current-to-voltage converter) so that the output voltage is proportional to the input current. It is easy to find theoretical information about this, but I wonder how it works in a practical situation.

Here is a theoretical drawing from wikipedia:
**broken link removed**

The idea is that I want to make a digital precision ammeter that will measure current in the 0-100 mA range. I suppose that means a suitable current-to-voltage converter and an ADC. I have worked with the latter before and I don't have any question about ADC or DAC's - right now it is all about the current-to-voltage converter.

Please don't recommend using precision shunt resistors instead - I don't know where to get them and they are probably very expensive. If you insist on using shunt resistors, please read my questions anyway and tell me why

So my questions is:
1. How do I connect the opamp to measure current? I suppose it has to be connected in series like an "analog" ammeter, but is it across the inverting input and output or across the inverting input and ground?

2. What parameter in opamp datasheets states the maximum allowed input current?

3. Do I need power opamps or can I divide the input current across several opamps and sum the output voltage?

4. What about the voltage in the circuit being measured? Example: Measuring current flow through a BJT with VCC=15 volts and the ammeter/current-to-voltage converter is connected to supply (VCC) and the collector of the transistor. Does the VCC have to be within the limits of the opamp's supply?

5. Can I calibrate the current-to-voltage converter with a constant current source or sink? Like an IC precision voltage reference can be used to calibrate a voltmeter? If yes - how? Is there any cheap or reasonable priced IC precision current sources/sinks available? I'm asking this question because I probably want it to be sure that 100 mA => 10 V output, 10 mA => 1 V output or something like that.

Those questions have been on my mind for quite some time. I really hope that I can get some answers on this forum - any help/advice is appreciated

 

[audioguru]

The circuit you have simply short-circuits the circuit you are measuring. The max output current of an opamp is low so the current being measured is also low. A power opamp can be used for higher short-circuit current. But usually a circuit is not short-circuited.

Usually current is passed through a low value resistor then the voltage across the resistor is measured for the current to be calculated with Ohm's Law. An opamp can be used to amplify the voltage so that an extremely low value resistor (called a shunt) can be used so there is a very low voltage drop.

 

[MikeMl]

The current to voltage converter you show in your post has some serious restrictions:

1. It can only measure current to GROUND. (i.e. where the non-inverting input of the opamp is connected to). It cannot make a "floating" measurement.

2. The opamp has to sink an equal amount of current as you are measuring. (to measure 100mA, the opamp has to sink 100mA, requires a POWER opamp).
Try to build a I/V that measures 10A!

3. The opamp requires dual (split) power supplies.

Get a shunt!

 

[crutschow]

As noted by the previous posters, an op amp current-to-voltage converter is not suitable for your requirements. Shunt resistors are what you need. They are low value resistors, (typically 1 ohm or less depending upon the current measured) and should be available at any large parts supplier that sell resistors (for example can you buy from Farnell?).

 

[MrAl]

Hello pvh,

To use it like that, you would connect from the inverting terminal to the non inverting terminal, with the non inverting terminal being common to both the input and the output. If your non inverting terminal does not connect to ground then you have to measure that voltage and subtract that from the output voltage to get the adjusted output voltage which would then still represent the input current. The non inverting terminal still has to be common to the input however.

The max input current will be small, it's the max input voltage that you have to think about.

As others have pointed out, you need a power op amp. The only problem here is that you may not be able to find one that also has very low input offset voltage. The input offset voltage affects the measurements so you want that to be as low as possible. With this in mind, if you are measuring a unipolar input current you might get away with a PNP drive transistor on the output. If not, you may need two transistor. This way you can get both low input offset and 100ma output drive.

Vcc (input) does not have to be within the Vcc of the op amp as long as you use a protection diode(s) on the input. This would protect the input of the op amp long enough to allow the op amp to respond normally. If you dont use a protection diode(s) the voltage may be too high.

You should be able to calibrate it yes. Apply a known current at the input, however you choose to generate that. It would be a good idea to have a meter you can compare readings with. If not, you might use a precision reference with a precision resistor to generate a precision current. When you do it this way you may want to consider the input offset voltage which introduces an error.

If you want 10v out for 100ma in then you use a 100 ohm resistor in the feedback.

As i was saying before, input offset voltage affects the measurement so you want a op amp that has very low input offset voltage, and since internal gain also affects the measurement you want high gain. Most op amps should provide a gain of at least 100000 so i think you'll be ok there. The gain affects the input offset by the equation:
Voffset=Iin*R/Aol
where
the additional input offset is Voffset,
Iin is the input current with Voffset already present,
R is the feedback resistor (100 ohms here for example),
Aol is the open loop gain of the op amp near DC (typically 100000 or more).

 

[Jaguarjoe]

Current sense amplifiers are tailor made to do this kind of stuff along with a shunt resistor. Every major op-amp maker has a line of CS amps. Linear Technology has their LT6100 series. They can be modeled in LTspice so you don't have buy parts and assemble, just immagineer away on the tube. LTspice is a free download.

Accurate shunt resistors are not that expensive. As Crutschow mentioned, all the big suppliers have them, Mouser, Newark, Digikey etc. There are also websites that have calculated the resistance for various short lengths of copper wire for use a shunt.

LT also makes a 2 terminal programmable precision current source, the LT3092 will do up to 200ma.

This will help:

https://www.electro-tech-online.com/custompdfs/2011/03/an105.pdf

 

[KeepItSimpleStupid]

I've successfully built a I-V converter to work in 4 ranges to +-100 mA FS with a 10 V output. Yea, it's tough. It's tougher to make it a 4 terminal device. Hall effect current sensors might be better for you. See Electronic Components Distributor | DigiKey Corp. | US Home Page.

1. The connection is between the inverting input and ground.

2. Vout is -IRf so, I and Rf and the max output current of the OP amp determine the max output. You can use a buffer such as the LT1010 inside the feedback loop. A cap is usually placed across R in the feed back loop to prevent oscillations.

3. You can use a power OP amp, but Vos and Ib are usually higher. You can place the power OP amp or buffer within the feedback loop.

4. This is harder to answer. Usually these are placed in the ground lead of the device being measured, but it doesn't have to be. The problem is that your output would float above ground. This could be an issue. Just how far it can be floated depends on lots of other factors. For higher currents, you can put back to back protection diodes between the inverting and non-inverting inputs. Vcc (measured device) does not have to be within the limits of the circuit being measured.

5. Calibration. It's best to make the calibration using fixed resistors. 1% or better resistors can be used. You can also decide to select others in series for a better calibration. Potentiometers in the FB loop are possible to some extent. FETS in the feedback loop have also been used in some designs. You can also follow the OP amp with a second gain stage, but this introduces errors as well. Your calibration source is just that, a calibration source. You would need to have a NIST traceable one.

Usually with I-V converters, one is interested in measuring extremely low currents. Ib and Vos are the most critical parameters in this case. Electrometer amplifiers can have an Ib around 100 fA. Other parameters generally suffer. One is drift. Vos is temperature sensitive. Zeroing of the amplifier is also necessary. If your not going too low in current then possibly a chopper-stabilized amplifer can be used.