# High current/high voltage/high frequency measurement

I have a circuit that I'm having trouble understanding. It measures the current flowing through a conductor. The measured load is 20-100 Amps, 9kHz-34kHz, and approx. 340 volts. The main conductor passes through a toroid that has a 10 turn winding (winding #1). The two ends of the winding extend away from the toroid. One end of the winding loops through another toroid that has a 100 turn winding (winding #2). The two ends of winding #1 are then spliced together. Winding #2 has both ends terminated across a 100 ohm resistor. The voltage across the resistor varies from 0 to 6.6v p-p, dependent on the current of the the measured load. The question I have is that it seems like the voltage would be MUCH higher. Winding #1 should produce 3400 volts, then winding #2 should produce 340000 volts? I know I'm missing something very fundamental. Could someone help me to understand this?

• Draw a schematic with the tool – laptop2d Aug 16 '18 at 15:42
• I think what you need to research is Current Transformers then realize that here, you have two CT's in series. This is common to "step-down" very large currents into much more manageable ones. – rdtsc Aug 16 '18 at 15:48

## 2 stage Current Transformer 1000:1

Typical thru-hole torroidal CT's use a burden of 1 Ohm per turn. The 2nd core has 100 turn and uses 100 Ohms which appears as 10 Ohms burden to the 1st core so no additional R is needed.

The frequency response "can" be quite limited, depending on design and cost.
e.g. one frequency decade 20kHz ~ 200kHz

That may contribute to your error.

simulate this circuit – Schematic created using CircuitLab

This is my attempt to draw a 1 wire CT insulated from secondary (not shunted as shown)

Since voltage drop on primary ~ 0 the voltage is not amplified. This is because the transformer is in series , not parallel with the load.

The burden resistor ( here 100 Ohms) is per Mfg suggested value to create Volts per Amp.

The ten turn secondary winding on the first CT is shorted hence for 100 amps flowing through the main conductor in that CT, 10 amps will flow in the shorted secondary.

But, remember that the voltage across the short length of the main conductor that produces a magnetic field in the CT's core is just milli-volts or less. Most of the 340 volts (99.99%) appears across your primary load and not across a short length of wire going through that CT so there is no possibility of seeing thousands of volts.

So, moving onto the second CT; it has 10 amps flowing through the core primary winding and the secondary is 100 turns feeding 100 ohms. The 100 turns means that this CT's secondary current is 10 amps / 100 = 100 mA.

And, the voltage across the 100 ohm resistor will be 100 ohms x 100 mA  = 10 volts.


You say you are operating in the kHz region and this may account for only seeing 6.6 Vp-p. Leakage inductances could account for a significant reduction in the voltage being only 6.6 Vp-p as opposed to 10 volts (presuled RMS).

• – Andy aka Aug 17 '18 at 17:12