# AC Current source

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1. Jun 8, 2015

### SanDiegan

Hi everybody,

As part of my research at UCSD, we are trying to measure Hall resistance of some materials. And to do that we need to send an AC current. So I wondered how we could easily build an ac current source, probably based on op-amp. I know there is the Howland circuit (current pump), but I am not sure it can be used to get an AC current as well.

The specifications of the current we need are the following: constant ~µA RMS, frequency range ~1kHz, sine wave. The impedance of the material we are testing is about ~mΩ.

Yours

Last edited: Jun 8, 2015
2. Jun 8, 2015

### meBigGuy

Do you have a sine wave source from a signal generator. If so, a bipolar current source like the howland is the correct answer.

You can cheat, in that a 1V sine wave source in series with a 1 megohm resistor driving a milli-ohm load is very nearly a 1uA constant current source to 1 part in 10^9. (but maybe you need better than that?)

3. Jun 8, 2015

### SanDiegan

We can have a sine wave source, but I cannot find any information anywhere clearly stating the Howland circuit can be used to generate an AC current.
And actually, we would like to avoid using a resistor, but yes that would have been an easy solution!

4. Jun 8, 2015

### meBigGuy

The howland circuit will generate a current that follows the reference voltage. In order to generate AC without coupling capacitors you need to have a bipolar power supply for the opamp (like +5 and -5 volts). Since the opamp is supplying/sinking the current, it can go positive and negative.

You probably want to use a high quality cmos opamp to avoid bias currents.

I'm surprised that I can't find a google image of the simplest possibility (if your load need not be grounded).

Opamp drives the load in series with a sense resistor to ground. Sense resistor/load node feeds back to - input of the opamp. Reference voltage is applied to + input.
So the reference voltage appears across the sense resistor, causing a reference current through the load.
Again, this requires a + and - supply referenced to ground.

Sometimes op-amps have crossover distortion when they drive high resistances. You can use a resistor from the output of the opamp to the negative supply to "bias" the output into class A operation.

5. Jun 9, 2015

### SanDiegan

Okay now I understand how it could generate an AC current. Does the attached picture correspond to your circuit / can it be used in my case?
And last question how do you choose the power supply for the op-amp and the resistors? I know how to choose R3,R4,R5 (cf. the circuit) since they set the current amplitude, but what about the others?

Thank you!

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6. Jun 9, 2015

### tech99

The resistance under test is only milliohms, so it is very easy to obtain constant current by just placing a resistor in series with either a signal generator or a simple oscillator (such as a Wien Bridge type). For instance, a resistor of 100k would enable a 1 volt SG to pass 10uA through your device under test, and would swamp out the milliohm fluctuations of the device itself. I suggest also placing a capacitor in series to eliminate the chance of a DC component.

7. Jun 9, 2015

### meBigGuy

The circuit you drew cannot generate an AC signal in the load. The transistor will only source current to the load. It cannot sink current (draw current out of ground through the load). Take out the transistor and connect the opamp directly to the R4 R5 junction. If you need the load grounded, just follow the howland design notes. http://www.ti.com/lit/an/snoa474a/snoa474a.pdf. Otherwise try the simple circuit I described.

8. Jun 10, 2015

### Jeff Rosenbury

Naively I would go with tech99's solution as simple and effective.

However, I don't know what Hall resistance is? I understand the Hall effect. Are you trying to measure the resistance of the material side to side with a working Hall effect sensor?

If tech99's solution is too simple for your needs, you might consider a current mirror as a current source. These can give multiple, identical (with hand matched betas) currents. I could see where multiple, identical current sources might be useful with Hall measurements. You could for example insert a current in a working circuit and extract the same current from the other side of the sensor. (You may need to build some sort of photo transistor current mirror if different voltages are a problem.)

9. Jun 10, 2015

### SanDiegan

Hi All,

Thank you for all your contribution. I am going to try few circuits based on op-amp, and also with only resistors in series first before trying anything more complicated. We want an AC current just to be able to modulate it and filter the signal with the lockIn in order to cancel the noise.

@JeffRosenBury The Hall resistance is simply related to the Hall voltage you can get according to the current you send. Actually to analyse and compare the materials, the Hall resistivity is more relevant (just have to take into account the dimensions of the sample).

10. Jun 10, 2015

### meBigGuy

Like the one in post #2?

11. Jun 11, 2015

### Jeff Rosenbury

Perhaps.

Sorry, I read your first sentence and wandered off to look up the Howland. I should have kept reading.