[SOLVED] Current XFORMER not fully reset at high duty cycle

Hello everyone,

I am wondering about resetting an unipolar current transformer at high frequency under worst case conditions as follows:

*CT waveform frequency=200kHz
*Dmax=95% => t_on(MAX)=4.75us and t_off(MIN)=250ns
*Current sense pin not allowed with negative voltage => can not use zener diode instead of the usual "rectifier" diode
*For the reset, less amount of parts possible => I am considering the option of using only a reset resistor (the option of diode+zener, discarded.. for the time being)

Sense voltage set to 2V for a current of 5A.
CT used: CS4100V-01L from Coilcraft

Compute needed data:
Volt*second each cycle under this worst case conditions ~13V*us => increment of mag current=0.65mA
Xformer saturates at 14.9 mA through the secondary mag inductance.

To reset fully the Current Xformer under this conditions, I need 400k as reset resistor (calculated for 5 time constants). This high reset resistance value is not allowed because of the reverse voltage it produces across the "rectifier" diode, also due to noise.
This means, I need to lower the resistance, to say, less than 100k, but that means that the Current Xformer is not fully reset, but not saturated either.

So, I have designed as follows:
1) Set maximum allowed current the mag inductance can reach in steady state under this conditions --> Ipeak=1mA
2) Compute the current it will drop to in steady sate --> I_LOW=Ipeak-increment of mag current --> 1mA-0.65mA=0.35mA
3) Compute time constant needed using the exponential decay equation: tau=-t_off(MIN)/ln(I_LOW/Ipeak) --> tau=2.38*10^-7
4) Compute reset resistor: Rreset=Lmag/tau --> Rreset~84k => can select e.g. 100k

Can anyone see any issue with this ?
Manufacturers do not specify parasitic capacitance, does anyone know typical secondary capacitance in usual CT with 100 turns ?

P.D.:LTSpice sim below

Any opinion/comment is welcome.

for your CT - what is the LV winding capacitance? this will limit the rate of rise of the reset volts.

if you measure the SRF of the sec, then again with 100nF in parallel, you can work out the L and hence the intrinsic C of the winding.

for 2V in sec (+ diode etc) for 4.75uS and Toff = 250nS the reset volts must be 38V if a square wave and say ~60Vpk if a haversine.

you cannot saturate the CT if the pri and sec currents are balanced - excepting if you overshoot the Volt-seconds (298V.uS) - you want to keep the peak flux low - for accuracy.

dB /dt = V / N.Ae so if you know the area of the core (break one open) you can calc peak flux, ideally < 50mT.

The only way to know performance is to test, no resistor might well be the case - a small snubber on the sec might be in order to limit noise - e.g. 47E & 150pF.

the sec side C will limit dv/dt on the sec and so alleviate some noise issues, more so with a snubber ...

- - - Updated - - -

p.s. you can't use a zener instead of a fast rectifiying diode because the zener is very slow to turn off - and so useless as a CT rectifier.

Use a fast 80V schottky ... or 2 x 45V schottkies in series ... 1N4448 at a pinch ...

CT's can also be biased for faster reset per: https://www.ti.com/jp/lit/an/slua174/slua174.pdf

Easy peasy said:

CT's can also be biased for faster reset per: https://www.ti.com/jp/lit/an/slua174/slua174.pdf

Click to expand...

I think the article is wrong regarding the "faster reset" statement (paragraph 2, page 2).

A RL circuit has the same transient response (given by the time constant) with or without voltage excitation. (same if the parasitic capacitance is included, bias voltage source has no effect in the transient, but only in the steady state solution)

The only advantage the bias voltage can add is the fully saturation in the opposite side, hence doubling the Volt*seconds available, but will also add error to the "sensed current".

Make Rr a current source - for speed, and then divide down the output to take away the extra voltage from the current - sorry thought that was obvious.

having two sources of current, the CT and the biasing - will give a faster response though.