# derivative of current confusion

Say you have some B field from an alternating current, B(t)=B0cosωt. And using Faradays law, e.m.f. ε = turns of a coil × the time derivative of the flux of the B field going through the coil area.

$$\\epsilon = -N B \frac{d\psi}{dt} \$$

If my B(t) max has a amplitude of 0.0005 T and a frequency of 500,000 kHz, then does the derivative of B(t) have a 1000000π constant pop out front? This seems wrong, like you can make the ε voltage really high just from the frequency?

What if your B field has a frequency in the Giga hertz range and impedance was knocked out with some capacitors? Then ε would have some 10^9 coefficient out front?

I feel like I'm missing something simple, Thank you

• @S.s. your answer was correct so why delete it? Nov 17, 2020 at 16:27
• so if your B filed was .00005 Tesla strong and sent at a giga hertz. the induced voltage in one loop would be 300,000V? (.00005*2π *10^9) Nov 17, 2020 at 16:28
• I guess assuming the loop was 1 meter squared in size Nov 17, 2020 at 16:29
• @ChrisBolig, what this is telling you is that it's not easy to make a 50 uT field oscillating at 1 GHz. Nov 17, 2020 at 16:29
• @Andyaka I thought it wasn't as deep to be considered an answer, I just undeleted it.
– S.s.
Nov 17, 2020 at 17:20

The induced voltage is based on the rate of change of the B field, higher frequency = higher rate of change, so yeah, higher voltage.

• To add to this answer: to generate the same mag field at higher frequencies becomes increasingly harder to do. Like, how do you generate a field - with a coil (dummy! LOL) and the reactance of the coil rises with frequency and so you keep pushing the frequency up and reducing the dimension of the coil and by the time you get to hundreds of MHz you just cannot make anything like the original field size without massive applied voltages then you are, in effect tending towards a 1:1 transformer with the receive coil. Big volts in is approximately big volts out and all the capacitive losses etc... Nov 17, 2020 at 17:24
• can you attach a capacitor to get it at resonance, and "eliminate the reactance"? Nov 17, 2020 at 17:40
• @ChrisBolig no matter what you do with a parallel capacitor across the inductor to tune it to resonance, you still need to apply a massive voltage to get the sort of flux levels you are talking about at the very high frequencies you are implying. Nov 17, 2020 at 17:58
• ok thank you, is there an equation to explain this, i just know the basic equations Nov 17, 2020 at 18:01
• Also remember that coils have parasitic capacitance between each turn, at higher frequencies this capacitance plays a higher role and the coil becomes effectively a capacitor.
– S.s.
Nov 17, 2020 at 20:34

Once had to assist in curing the murder of MOSFET GateDriver ICs on a 10,000 horsepower speed controller PCB.

The older version had no failures at all. This new PCB had numerous failures, in the field. Were 6+ ICs, but only a few locations had failures.

I "What was changed?" of the programmer/consultant who was asked by management "Can you also make the PCB smaller." (Notice management was involved with engineering.)

One change was to move the PCB somewhat closer to the power buss, that handled 2,000 amps to a number of IGBTs, that switched the 2,000 amps in a PulseWidthModulation behavior.

OK So 2,000 amps is being switched. They would not tell me the switching speed.

But I knew SLOW SWITCHING would linger in the VDD/2 * IDD/2 = 5,000volts/2 * 2,000 amps/2 = 2,500 * 1,000 = 2,500,000 watts dissipation, and no semiconductor device can handle that much power for very long.

So I decided "1 uS switching time" and the customer said "About right."

I already knew there was a problem, because the dI/dT would be huge. So I introduced the concept of

Vinduce (between a long straight wire, and a co-planar wire loop) modeled as

• Vinduce = [MUo * MUr * Area / (2 * pi * Distance)] * dI/dT

and with Area = 10cm * 10cm (4" square region on the PCB ground traces), Distance = 2cm (between the power buss and the PCB), and 2,000 amps switching in 1 uSec === 2 Billion amps/second, we have

• Vinduce = 2e-7 * (0.1 * 0.1)/0.02 * 2Billion

• Vinduce = 2e-7 * 0.01/0.02 * 2e+9

• Vinduce = 1e-7 * 2e+9 = 200 volts induced into the Ground.

I suggested they interpose a metal_plate shield, with complaints immediately rising "That is too close to the 5,000 volt power buss" and I offered "That is why gutta percha is useful."

========================================

What does a 10,000 HP PCB look like?

This was the controller for the IGBT gate drives, which has FiberOptic_isolation to various MOSFETs that drove the IGBTs. The GateDrive ICs for the MOSFET gate were failing because Ground_here was not the same as Ground_there.

The PCB was 6" by 12" (or 15cm by 30cm, if in France).

==============================================

Notice the weakness of the 3_D (2_D?) model: the current thru a wide copper bus is modeled as a single wire.

• What does a 10kHP PCB look like? Nov 18, 2020 at 6:15