I have designed a circuit to measure high currents indirectly using a TAK8348 current transformer:
https://cdn.shopifycdn.net/s/files/...-75c4-4b1d-a63b-a792b07066b4.pdf?v=1616374876

This is the circuit that connects to the microcontroler:



The circuit works fairly ok and can measure to a degree of 0.1A accuratly.

The problem occurs when current flows for a very long time (1h+). When the whole thing runs for a long enough time, it heats up just a little bit, and the ADC signal on the pin increases. This causes the measurment to eventualy be off by 1A (shows 31A instead of 30A).

My question is what would cause this drifting. I have eliminted a change in Vref as a possible reason, since i tried powering the whole thing through an external USB charger and got the same result. The microcontroler (ESP32) isnt more then 10 degrees warmer as well, so that is also not the reason.

The strange thing is that the microcontroler also measures voltage, which is much more complex and sensitive circuit, and that stays perfectly accurete even after a long time. So by that i am deducing that the issue might be in the part of the circuit in the picture.

Any recomendations for possible causes or troubleshooting?

 

The diode changes voltage drop with temperature.
Is the CT connected from TAK8348-1 to -2? It is hard without seeing more of the circuit.
I do not understand R59. I think it should be 0 ohm.
Do you have a resistor to ground from point 3P_MEA?

There are ways to compensate for diode drop. Do you want to go down that road?

 

The diode changes voltage drop with temperature.
Is the CT connected from TAK8348-1 to -2? It is hard without seeing more of the circuit.
I do not understand R59. I think it should be 0 ohm.
Do you have a resistor to ground from point 3P_MEA?

There are ways to compensate for diode drop. Do you want to go down that road?

Thank you, I did check the diode datasheet and found that that is something that could cause this effect. I will try to replace it with an op amp circuit that acts as an ideal diode.

As for R59 i do not understand it either. I found this circuit in a reference design and it seems to work nicely and gives a clean 0-3V sine wave for the current range i am measuring. I will try without it as well.

TAK8348-1 and TAK8348-2 are the output pins of the transformer as spacified in thedatasheet. I didnt have a model for the component, so i just named the nets that.

I dont have a pull down resistor on 3P_MEA, but the esp32 has an integrated pull down resistor, which i am utilizing for this

 

I will try to replace it with an op amp circuit that acts as an ideal diode.

You don't need that to cancel the diode drop.
Just put the 100Ω resistor to ground at the output of the diode rectifier.
Since the transformer output is a current, the diode drop will not affect the current through the resistor, and the voltage is then just a function of the current.

LTspice simulation below:

It has a full-wave bridge rectifier, giving a filtered DC output equal to the average value of the sinewave.
Note that the averaged output is less affected by distortion and spikes in the current waveform, than the peak detection you were using, so would generally be more accurate.
Of course the most accurate would be to take multiple samples of the rectified but unfiltered waveform (node 1) with the micro, and calculate the RMS value.

What frequency current are you measuring?

Edit: Changed circuit to full-wave rectifier

 

Use Crutschow’s circuit. That is your best approach.

If you would like to balance the transformer’s volts-seconds for improved accuracy with high crest factors, add another 100 ohm resistor in series with D2.

 

If you would like to balance the transformer’s volts-seconds for improved accuracy with high crest factors, add another 100 ohm resistor in series with D2.

Sorry, I changed the circuit to use a full-wave bridge, so that mod is not necessary.

 

Sorry, I changed the circuit to use a full-wave bridge, so that mod is not necessary.

Thank you for the sugestion. But wouldnt the output voltage of this circuit be very heavily based on the capacitance of C1? If the capacitence differs from component to component , wouldnt that cause every board to show slightly different results?

 

Can you confirm which of the TAK8348 series pulse transformers you are using, there are several in the range.
Also the Vf of D3 will decrease with increasing temperature which I’m assuming would result in a lower current measurement.

 

Whoops, forget that last sentence, brain fog !!

 

Can you confirm which of the TAK8348 series pulse transformers you are using, there are several in the range.
Also the Vf of D3 will decrease with increasing temperature which I’m assuming would result in a lower current measurement.

I am using the TAK8348 -200

 

Thank you for the sugestion. But wouldnt the output voltage of this circuit be very heavily based on the capacitance of C1? If the capacitence differs from component to component , wouldnt that cause every board to show slightly different results?

No, C1 value do not alter output voltage .

I suggest you to use a small Shottky diodes instead of 1N4148. BAT41, BAT85 or similar. They have much lower forward voltage drop. This leads to better accuracy, due to reduced load on the current transformer

 

No, C1 value do not alter output voltage .

I suggest you to use a small Shottky diodes instead of 1N4148. BAT41, BAT85 or similar. They have much lower forward voltage drop. This leads to better accuracy, due to reduced load on the current transformer

Wont the forward voltage change with temperature anyway and cause the signal to change?

 

If the capacitence differs from component to component , wouldnt that cause every board to show slightly different results?

The value of capacitance only affects the ripple voltage, not its average value.

Wont the forward voltage change with temperature anyway and cause the signal to change?

As I previously noted, the output is a current so the forward drop has negligible effect on the output current, and thus voltage across the 100Ω resistor.
Do you not understand what it means to be a current source?

You didn't answer my question about what frequency current you are measuring.

I suggest you to use a small Shottky diodes instead of 1N4148. BAT41, BAT85 or similar. They have much lower forward voltage drop.

I doubt that less than a volt difference between the two different bridge drops will have any significant effect on the transformers output current.

 

I doubt that less than a volt difference between the two different bridge drops will have any significant effect on the transformers output current.

don't be so sure.
Voltage drop on the load resistor for given trafo TAK8348-200 is:
Uload= 25mA*100 Ohm= 2.5 V

Voltage drop on the bridge built on 1N4148 ( based on 10 mA current from the datasheet) :
Udiode1= 2*1V= 2V
Total drop Usum1= 2.5V + 2V= 4.5 V

Voltage drop on the bridge built on BAT85 ( based on 10 mA current from the datasheet):
Udiode2= 2* 0.4V = 0.8V
Total drop Usum2 = 2.5V+0.8V= 3.3V

Difference is 27 percent. Not so neglible

 

don't be so sure.
Voltage drop on the load resistor for given trafo TAK8348-200 is:
Uload= 25mA*100 Ohm= 2.5 V

Voltage drop on the bridge built on 1N4148 ( based on 10 mA current from the datasheet) :
Udiode1= 2*1V= 2V
Total drop Usum1= 2.5V + 2V= 4.5 V

Voltage drop on the bridge built on BAT85 ( based on 10 mA current from the datasheet):
Udiode2= 2* 0.4V = 0.8V
Total drop Usum2 = 2.5V+0.8V= 3.3V

Difference is 27 percent. Not so neglible

But we are only interested in the voltage across the burden resistor, which is unchanged.

 

With the TAK8348-200 and the 100ohm burden resistor R57 you should measure an ac voltage of around 1.5v across this resistor for a current of 30A, ie (30A ÷ 2000) x 100ohms = 1.5V.
What happens to this voltage if you monitor it over a period of an hour or longer?

 

But we are only interested in the voltage across the burden resistor, which is unchanged.

Exactly. Unchanged if an ideal current source is used.
But, in the real world only a real current trafos do exist. And for those trafos best working conditions are if they are shorted.
Of course, we can't to short the transformer by a jumper. A small value resistor to be in the circuit to read the current value. The main word is- small . To get as low as possible load to the current trafo.
Diodes in the bridge act as an additional resistors. That's the point

 

Of course, we can't to short the transformer by a jumper.

You can if you use an op amp connected as a high-current transimpedance amp, which essentially looks like a short circuit at its input.

Unfortunately the transformer spec sheet makes no mention of the maximum output voltage it can generate before it causes degradation of the accuracy.

 

You can if you use an op amp connected as a high-current transimpedance amp, which essentially looks like a short circuit at its input.

Unfortunately the transformer spec sheet makes no mention of the maximum output voltage it can generate before it causes degradation of the accuracy.

Agree.
But, this time, we discuss the circuit with diode bridge and burden resistor.
And, yes, the trafo's spec sheet do not tell nothing about the accuracy. It seems the burden resistor can be as high as you want

 

Exactly. Unchanged if an ideal current source is used.
But, in the real world only a real current trafos do exist. And for those trafos best working conditions are if they are shorted.
Of course, we can't to short the transformer by a jumper. A small value resistor to be in the circuit to read the current value. The main word is- small . To get as low as possible load to the current trafo.
Diodes in the bridge act as an additional resistors. That's the point

I now see the point you were making!