Circuit distorting AC signal from CT

[Red_Micro]

Picture 1 shows a 60Hz signal from a CT only with the burden resistor connected. Picture 2 shows the signal when I connect it to the opamp circuit. Picture 3 shows the op amp circuit. Some points on it:
x1,x2 - z1-z2 are analog switches: SN74LVC2G66DCUR
L1 L2 are two identical secondaries (same turn ratio) from the same CT. One is used for primary sensing and the other is occasionally used for testing.
D12 sets the DC offset to Vcc/2

My guess is that the culprit here is related to the analog switch. Any light on this?

[pardo-bsso]

What makes you think it's the analog switch?
Have you tried replacing it with jumpers?

[coppice]

If you ran this without a burden resistor across L2, didn't it blow up the analogue switch Z1/Z2?

[Red_Micro]

If you ran this without a burden resistor across L2, didn't it blow up the analogue switch Z1/Z2?

But if there's a burden across L1, doesn't this load get reflected across L2?

[wraper]

What makes you think it's the analog switch?

How about what's written in datasheet?

[Red_Micro]

What makes you think it's the analog switch?
Have you tried replacing it with jumpers?

I'm doubtful about L2 being referenced to GND and L1 to Vcc/2. So the switch may be clipping the signal?

[coppice]

If you ran this without a burden resistor across L2, didn't it blow up the analogue switch Z1/Z2?

But if there's a burden across L1, doesn't this load get reflected across L2?

Why would it? L2 is producing a current. Without a load the voltage across it will rise until the limitations of the CT bound it. If you have a really good CT it can arc and self destruct without a burden resistor in place.

[wraper]

It seems that voltage is going below GND and apparently you are powering analog switches without negative rail but simply use GND as negative supply. They do not work with voltages below GND. Opamp inputs do not operate properly below GND either.

[Red_Micro]

It seems that voltage is going below GND and apparently you are powering analog switches without negative rail but simply use GND as negative supply. They do not work with voltages below GND. Opamp inputs do not operate properly below GND either.

The opamp is not intended to work below GND. That's why the DC shift. Perhaps I should reference L2 to Vcc/2 as well.

[NiHaoMike]

Move C21 to the other side of the analog switch, that way the ESD diode in the switch will not clamp the negative peaks.

[Red_Micro]

Move C21 to the other side of the analog switch, that way the ESD diode in the switch will not clamp the negative peaks.

I will try your suggestion. What do you think of my thought of referencing L2 to Vcc/2 as L1? Originally I reference to GND because the test_square signal to the test winding comes from a micro. The current from the micro has to return to the micro through ground and if I reference L2 to the VCC/2 point then the return current may affect the reference. Not sure actually.

[srb1954]

Omit analog switch Z1/Z2 altogether.

When you aren't driving the test winding tri-state the micro output so it doesn't load down the CT winding. You will need some protection diodes on the micro output so it is protected against transients feeding back from the CT.  These diodes will need to have a reasonable current rating (>100mA) as they also have to handle any charging current through the coupling capacitor C21.

[Jay_Diddy_B]

Red_Micro,


**************************************************
This analysis is wrong!!

The REF pin on the LM4041 is referenced to pin 2, the cathode.

Ignore
**************************************************

I think you need to revisit this part of the circuit:






With the component values shown, the LM4041 will try to regulate TP9 at

V_TP9 = 1.225V (1 + R33/R35)   = 1.225 (1+ 30K/10K)= 4.9V

This can't happen because you only have a 3.3V supply.


To regulate mid-rail, 1.65V

Make

R33 = 10K
R35 = 30K

V_TP9 = 1.225 (1 + 10K/30K) = 1.633V


(With the original values, the voltage on TP9 = 3.3 x (10K +30K)/ (10K + 30K + 3.3K) = 3.05V )

In addition this circuit is a shunt regulator, it has limited ability to source current. You need add a bypass capacitor from TP9 to ground.

Regards,
Jay_Diddy_B

[Red_Micro]

Red_Micro,

I think you need to revisit this part of the circuit:

(Attachment Link)




With the component values shown, the LM4041 will try to regulate TP9 at

V_TP9 = 1.225V (1 + R33/R35)   = 1.225 (1+ 30K/10K)= 4.9V

This can't happen because you only have a 3.3V supply.


To regulate mid-rail, 1.65V

Make

R33 = 10K
R35 = 30K

V_TP9 = 1.225 (1 + 10K/30K) = 1.633V


(With the original values, the voltage on TP9 = 3.3 x (10K +30K)/ (10K + 30K + 3.3K) = 3.05V )

In addition this circuit is a shunt regulator, it has limited ability to source current. You need add a bypass capacitor from TP9 to ground.

Regards,
Jay_Diddy_B

I measure 1.65V at TP9 so the LM4041 part is OK. According to the datasheet: VO = VREF[(R2/R1) + 1] or VO = VREF[(R35/R33) + 1]

[Jay_Diddy_B]

Red_Micro,

You are right. I missed that the reference in the LM4041 is with respect to the cathode.

Can you check the voltage on TP9 with a scope?

Jay_Diddy_B

[Red_Micro]

Red_Micro,

You are right. I missed that the reference in the LM4041 is with respect to the cathode.

Can you check the voltage on TP9 with a scope?

Jay_Diddy_B

Voltage at TP9 is a clean DC signal 1.65V. I will try to also upload the graph. But it looks pretty good.

[Jay_Diddy_B]

Hi,

What is the part number of the opamp?

Jay_Diddy_B

[Red_Micro]

Hi,

What is the part number of the opamp?

Jay_Diddy_B

It's OPA317. Good news. I disconnected L2 and now the circuit works well for L1. It seems like the switch after L2 is distorting the signal and that is reflected into L1. Also, I had to lower the bias resistor R31 for D12 because the reference was affected by the load current. By lowering R31 I allow more current. Now I just need to fix the test coil circuitry. I need to find a more reliable way of injecting a current from the micro. I will try moving C12 to the other side of the switch as suggested, above but still unsure if L2 needs to be referenced to GND or D12.  If I'm injecting current from the micro, then it would need to be referenced to ground so it finds the return path.

[Red_Micro]

Omit analog switch Z1/Z2 altogether.

When you aren't driving the test winding tri-state the micro output so it doesn't load down the CT winding. You will need some protection diodes on the micro output so it is protected against transients feeding back from the CT.  These diodes will need to have a reasonable current rating (>100mA) as they also have to handle any charging current through the coupling capacitor C21.

I will try that also. The thing is that when I'm not driving the test winding and I let it open, a reasonable voltage develops across L2. Even L1 has a burden.

[srb1954]

If the two secondary windings have identical number of turns the voltage developed across the drive winding should be the same as that across sense winding in normal operation. Therefore the effect of a load resistor on the L1 winding will also be reflected into the L2 winding. You should never operate a CT without a low-value load resistor on the secondary winding to avoid the generation of high output voltages from the CT.

However, under certain transient conditions you can get spike voltages on the windings caused by the effects of transformer leakage inductance, which is due to imperfect magnetic coupling between all the windings of the CT. In this case the TVS on winding L1 might not full protect circuitry connected winding L2. You should perhaps also put a TVS on the L2 winding to clamp any such spikes.

[Red_Micro]

If the two secondary windings have identical number of turns the voltage developed across the drive winding should be the same as that across sense winding in normal operation. Therefore the effect of a load resistor on the L1 winding will also be reflected into the L2 winding.

That's what I think. But the voltage I'm seeing across L2 when it's open it's much bigger than the one across L1 even with the burden.  I'm sure they are both identical, i.e., same turn ratio. I will upload some graphs.