Isolation Amplifier Problem

Thread Starter

ilginsarican

Joined Jul 13, 2017
137
Hello,
I use ACPL-C87B-000E as an isolation amplifier.
I am trying to measure ( OUT1_V_ISO) and isolate it.
The range of this point is 0 -1,5VDC.
The problem is: Even if I dont apply any voltage to OUT1_V_ISO, I measure 1,76V on VOUT- pin (6) of U14.(The reference is GDN2).
And I measure zero voltage on VOUT+ (pin7).
According to the datasheet Unity gain 1 V/V.
I would be very grateful if you could give any ideas.
Thanks,
 

Attachments

Deleted member 115935

Joined Dec 31, 1969
0
Hello,
I use ACPL-C87B-000E as an isolation amplifier.
I am trying to measure ( OUT1_V_ISO) and isolate it.
The range of this point is 0 -1,5VDC.
The problem is: Even if I dont apply any voltage to OUT1_V_ISO, I measure 1,76V on VOUT- pin (6) of U14.(The reference is GDN2).
And I measure zero voltage on VOUT+ (pin7).
According to the datasheet Unity gain 1 V/V.
I would be very grateful if you could give any ideas.
Thanks,
not an answer,
but please

never draw a circuit with ground up, and Vcc down

Fig 19 in the data sheet you look to have re used is a good example,

Why have you gone for an amplifier with only a +v supply, where as the data sheet says use an amplifier with + and - supplies.

Looking at the app note for the evaluation board,
they include a Vref ( U5 ) that is used to shift the offset voltage, down to a virtual ground,
you seem to have missed that off.
 
Last edited by a moderator:

Thread Starter

ilginsarican

Joined Jul 13, 2017
137
not an answer,
but please

never draw a circuit with ground up, and Vcc down

Fig 19 in the data sheet you look to have re used is a good example,

Why have you gone for an amplifier with only a +v supply, where as the data sheet says use an amplifier with + and - supplies.

Looking at the app note for the evaluation board,
they include a Vref ( U5 ) that is used to shift the offset voltage, down to a virtual ground,
you seem to have missed that off.
Actually I am trying to design single ended isolation amplifier circuit, because of that I did not use - supply.
Here is an example:
https://www.arrow.com/en/reference-...on-amplifier/a85e21201447fb83b20e3fe054bd8b5e
1631212465182.png
 

MisterBill2

Joined Jan 23, 2018
9,543
The connections on the input side of the isolation amplifier in post #3 do not look right. What I see is two connections to the one side , I am guessing that is both a signal line and a power connection. It looks like power and signal are connected, and that is wrong.
To measure motor current with only one polarity of signal voltage, put the shunt in either the driver low feed or the driver high feed, so that it is always the supply current through the shunt, always the same direction. Of course, this is presuming that the driver is a full bridge circuit.
 

ronsimpson

Joined Oct 7, 2019
1,549
I am using a number of different isolation amplifiers. They mostly work the same.
When U14-2 is "0V" then U14-6 & U14-7 should be 1.76V. (the same voltage) I think you have one problem there!
UH1A pin 1 = 0V. R247 & Re51 make a voltage divider that you do not want! Remove the bottom 10K.
I am using a R-R input, R-R output op-amp. Yours might not pull to ground well.
In my case I am looking at positive voltages.
When "VIN" goes up 1V, Vout+ goes up 0.5V and Vout- goes down 0.5V, OP-amp out goes up by 1V and "AC_OUT1" goes up by 0.5V
When "VIN" goes down 1V, Vout+ goes down 0.5V and Vout- goes up 0.5V, OP-amp out goes to 0V and "AC_OUT1" goes to 0V.
If the op-amp had a -5V supply you would see the output go to -1V and AC_OUT to to -0.5V.
1631228859030.png
Fix the short that pin-7.
Next reverse pin-7 and pin-6 to get a positive output when a negative is input.
BUT before you do any of that!!!! are you certain it works with negative input voltage.
----------------------------------------------------------

1631229876923.png
 

Thread Starter

ilginsarican

Joined Jul 13, 2017
137
I am using a number of different isolation amplifiers. They mostly work the same.
When U14-2 is "0V" then U14-6 & U14-7 should be 1.76V. (the same voltage) I think you have one problem there!
UH1A pin 1 = 0V. R247 & Re51 make a voltage divider that you do not want! Remove the bottom 10K.
I am using a R-R input, R-R output op-amp. Yours might not pull to ground well.
In my case I am looking at positive voltages.
When "VIN" goes up 1V, Vout+ goes up 0.5V and Vout- goes down 0.5V, OP-amp out goes up by 1V and "AC_OUT1" goes up by 0.5V
When "VIN" goes down 1V, Vout+ goes down 0.5V and Vout- goes up 0.5V, OP-amp out goes to 0V and "AC_OUT1" goes to 0V.
If the op-amp had a -5V supply you would see the output go to -1V and AC_OUT to to -0.5V.
View attachment 247614
Fix the short that pin-7.
Next reverse pin-7 and pin-6 to get a positive output when a negative is input.
BUT before you do any of that!!!! are you certain it works with negative input voltage.
----------------------------------------------------------

View attachment 247615
Hi, I noticed that when I measure VDD1 , there is voltage that decreases and goes to zero.
Then I removed R233, VDD1 is ok now.
U14 pin-7 and pin-6 are 1,2V, they are same finally.
UH1A pin-3 is 0,6V because of resistor divider.
I was hoping UH1 opamp subtraction circuit worked and zero volt output on UH1A pin-1 but I measured 0,9V UH1A pin-2 and 0,6V UH1A pin-1.
And this does not make sense.
PS: I removed R247 and R251.
 
Last edited:

MisterBill2

Joined Jan 23, 2018
9,543
I suggest searching for "precision Rectifier" circuits to learn how to do such a measurement. They will provide the peak voltage of your AC signal.
 

ronsimpson

Joined Oct 7, 2019
1,549
I think the op-amp is not good at 0V out. I think it can not pull down well. Try a 10k to 1k pull down resistor go ground, Maybe remove the capacitor.
 

MisterBill2

Joined Jan 23, 2018
9,543
For an AC signal from a resistor shunt you could use a small transformer to both boost the voltage AND provide the isolation. Given the very low source impedance of the typical current shunt It should work quite well. So there is another option to consider. It would be quite reliable, very stable, and not require an external power supply.
 
Top