VCCS howland circuit - no current

Thread Starter

yonigargi

Joined Nov 10, 2021
16
Hi
i'm trying to do research in the bioimpedance field.
i've built this design of VCCS based of modified howland circuit.
i'm using a signal generator - 1VPP, 1KHZ sinusoidal signal.
By using oscilloscope at the tissue electrodes i'm getting an augmented sinusoidal signal (VPP 1.9v).
But when i'm connecting a multimeter in order to check the microamperes at these electrodes i'm getting a zero result.
it doesn't change when changing the frequency or vpp of the signal generator.
how can it be? there is a sinusoidal voltage but no current.... I know I should get around 39 microampere by this test.

I would be grateful for your help.
Thanks
 

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crutschow

Joined Mar 14, 2008
34,470
What kind of multimeter?
Most won't measure such a low AC current.
What's the sensitivity and series resistance of the multimeter in the AC current mode for that sensitivity setting?
 

Thread Starter

yonigargi

Joined Nov 10, 2021
16
What kind of multimeter?
Most won't measure such a low AC current.
What's the sensitivity and series resistance of the multimeter in the AC current mode for that sensitivity setting?
hi, thanks for the answer
i'm using
AstroAI digital multimeter TRMS 6000
it is capabe of measuring microamperes
 

crutschow

Joined Mar 14, 2008
34,470
As Irving noted, the AC current response drops off above 400Hz (from manual below).

Which terminals are you using for the current measurement and are they connected in series with the signal generator?

1636582729801.png
 

Thread Starter

yonigargi

Joined Nov 10, 2021
16
the multimeter two probes attached to the two tissue electrodes. the electrodes are not connected to tissue at the measurement time.
 

Thread Starter

yonigargi

Joined Nov 10, 2021
16
Another question...
When i measure voltage with oscilloscope connected to the tissue electrodes (not connected to tissue) i get a signal with 180 degree phase shift. why is that?
Thanks
 

Irving

Joined Jan 30, 2016
3,897
The gain of your circuit is 1 so with a 1v p_p input signal the current generated across the meter is 1/51k or 20uA p-p or 7uA rms. At 1kHz that probably won't register on the meter. Try it on a DC signal, you might just see something.
 

Thread Starter

yonigargi

Joined Nov 10, 2021
16
The gain of your circuit is 1 so with a 1v p_p input signal the current generated across the meter is 1/51k or 20uA p-p or 7uA rms. At 1kHz that probably won't register on the meter. Try it on a DC signal, you might just see something.
Thanks a lot.
Do you have an explanation for the 180 degree phase shift?
 

Thread Starter

yonigargi

Joined Nov 10, 2021
16
Good question. What do you think the gain should be and why?


Relative to what? I think I know but can you work it out?
The gain of the INA128P is 1+50kohm/Rg which is about 2. The gain for the howlland circuit is 1. that's why it is close to 2 (1.9), why isn't it 2 though, because of wires resistance maybe?
I don't understand the 180 phase shift compared to the original signal.
 

Irving

Joined Jan 30, 2016
3,897
Ah, I was referring to the input to R1, not the INA218. The gain from there is 1 assuming R1 = R3 and R2 = R4. The gain of the INA128 is 1+50/51 = 1.980 +/- 0.024% but what's the tolerance on your 51k resistors and your 10k resistors? 1%? so the total tolerance of the circuit end-to-end gain is around +/-2% or 1.98 +/- 0.04, ie 1.94 - 2.02

180deg phase shift: look very closely at the way you are using the INA128...

edit: corrected tolerance calcs.
 
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Thread Starter

yonigargi

Joined Nov 10, 2021
16
Ah, I was referring to the input to R1, not the INA218. The gain from there is 1 assuming R1 = R3 and R2 = R4. The gain of the INA128 is 1+50/51 = 1.980 +/- 0.024% but what's the tolerance on your 51k resistors and your 10k resistors? 1%? so the total tolerance of the circuit end-to-end gain is around +/-2% or 1.98 +/- 0.04, ie 1.94 - 2.02

180deg phase shift: look very closely at the way you are using the INA128...

edit: corrected tolerance calcs.
You are amazing! Thanks a lot!
 
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