# Buffering the input to a passive RC integrator used to measure magnetic flux density

Note: this is part of a larger question which I was asked to separate into subquestions. Other subquestions: 1 and 2

I'm using the method outlined here to measure the induced magnetic flux density in a toroidal transformer made of Nanoperm (additional datasheets here and here). I am trying to adapt the method to use a passive integrator rather than the op-amp integrator described in the tutorial. (I tried building the active one but it was giving me issues.)

The diagram of my circuit is given below:

simulate this circuit – Schematic created using CircuitLab

where I am varying the input frequency of V1 to be either 50 Hz (as reported in the datasheets to measure hysteresis), 350 Hz, and 380 Hz, and I am varying R2 to be either 5 kOhm, 10 kOhm, 27 kOhm, or 37 kOhm. Note that the signal generator is connected to an audio amplifier capable of driving large currents. A result from one of my experiments is shown here:

For more details about the experiment and results, please see my original larger question.

While the shape of the hysteresis loop looks somewhat similar to that of the loops reported in the data sheet, it is orders of magnitude off in terms of both H and B. I posted a related question about this possibly being attributed to the way I was calculating H and B and/or to the fact that the gain of the RC integrator at the frequencies I'm looking at being much less than 1. I'm also wondering, since I'm using a passive integrator, does the EMF induced in the measurement coil need to be buffered before being fed into the integrator?

• Is R1 << Rp on primary – Sunnyskyguy EE75 Aug 6 '18 at 19:22
• Rp is 200 mOhms, and R1 is 400 mOhms, so no, R1 is > Rp. – Vivek Subramanian Aug 6 '18 at 19:24
• Then reduce by orders of magnitude – Sunnyskyguy EE75 Aug 6 '18 at 20:44
• Doesn’t the impedance of the coil increase when AC signals are passed through it? Because Z = |R + jwL|? – Vivek Subramanian Aug 6 '18 at 20:47
• Yes as long as the IR drop is neglible and V-s integral is driven to saturation – Sunnyskyguy EE75 Aug 6 '18 at 20:58

I'm also wondering, since I'm using a passive integrator, does the EMF induced in the measurement coil need to be buffered before being fed into the integrator?

In short, yes because it eliminates the need to account for the losses in the resistor.

Without having to invoke circuit theory for a lossy secondary, to measure B we need to measure V as accurately as possible, which means open loop.

$$B = \frac{\Phi}{A}$$ and $$V_s = \frac{d\Phi}{dt}*N_s$$

if you draw current from the loop then you need to account for the energy loss, which would mean factoring in a current into these equations because the voltage drop on the secondary would need to be accounted for. Much easier to use a high impedance measurement on the secondary.

In all of the hysteresis measuring curve circuits I have seen only high impedance integrator (which implies active).

• Thanks, @laptop2d. Would one of these work: i.stack.imgur.com/N4RV4.png or i.stack.imgur.com/zpZ9w.png. I ask because I tried using the integrator given in the tutorial (meettechniek.info/passive/magnetic-hysteresis.html, section "Measuring arrangement with an analog oscilloscope") but I was getting really large current draws and eventually gave up. I also tried the op-amp integrator on Wikipedia, which also failed; however, I did not buffer that input to the integrator. Would buffering the input to the integrator, as shown in my second link, make a difference? – Vivek Subramanian Aug 6 '18 at 20:38
• Yeah, I should have posted one of those links – Voltage Spike Aug 6 '18 at 20:50
• Here is another one: info.ee.surrey.ac.uk/Workshop/advice/coils/BHCkt/index.html – Voltage Spike Aug 6 '18 at 20:50
• Thanks so much, @laptop2d. Should R1, R2, and R4 be 68, 1, and 10 ohms, respectively? I just see 68R, 1R, and 10R on my computer. Also how important is the material the capacitors are made of? – Vivek Subramanian Aug 6 '18 at 21:09