U48 circuit and resistor drop math

[trans4funks1]

I've been looking at the U48 schematic because I am interested in figure 8 patterns. I have a few questions regarding the polarization voltage. Some of the questions relate to the U47 topology too. I think they are pretty basic questions but I realize I don't fully understand what is going on.

For the polarization of the back plate there is 105vDC on one side of R9.

R9 (3Mohm) and R10 (3Mohm) act as a voltage divider and get the voltage down to 52.5vDC.

Question 1: What happens at or across R1 (100Mohm)?

For the polarization of the rear capsule there is 105vDC on one side of R5 (150Mohm) and there is another resistor, R2 (150Mohm) in series as you trace the circuit to the rear capsule.

Question 2: What is happening at R5 and R2? I assume that the voltage at the rear capsule should be approximately twice that of the back plate. If the voltage at the rear capsule is 105vDC(?) how can I quantify the voltage drop across the resistors?

Is the fact that current draw is very small a factor?

Thanks for any explanation you can offer.

 

[trans4funks1]

As a reference, the U47 has a slightly different backplate polarization. The voltage divider is not symmetrical and the backplate is at 63vDC. I've looked at this several times and never got around to thinking of the math behind the resistor values, it wasn't until I looked at the U48 that I realized I had taken stuff for granted:

 

[MagnetoSound]

You are correct, there is no significant current across the polarising resistors that would cause a voltage drop. The most important thing about the high value resistors in these positions is that the capacitors to ground (C2 and C3) form filters with the polarising resistors that keep the supply to the capsule nice and clean, as well as providing a low impedance AC ground path for the signal.

 

[trans4funks1]

Thanks for pointing me in the right direction and making the point about the capacitors, their relationship to the resistors and the way they form filters.

I think I just realized that I was trying to use Ohms law to figure this out but I was mistakenly using the plate current (0.1mA, which I have at top of mind because I've been thinking about power supplies) in my calculations when I should have been thinking about current, if any, at the capsule.

Is there any way to quantify the current at the capsule? I guess, having looked at OHMs law again this morning, that there must be almost no, if any, current.

???

Thank You.

 

[Matador]

Is there any way to quantify the current at the capsule? I guess, having looked at OHMs law again this morning, that there must be almost no, if any, current.

You can use i = C * dv/dt to ballpark a figure.  This is a constant charge system, which means dv/dt is essentially a constant.  When sound hits the diaphragm, it actually modulating the capacitance of the capsule, so you can rewrite this equation as i ~= dC/dt * V.  If the sound is changing the capacitance by 1pF at a 1kHz rate, this means that (1e-12/1e-3)*60 ~= 60nA.

 

[trans4funks1]

Thank you Matador,

That is very helpful.

I had tried Ohms law: 105vDC / 150,000,000ohm = 0.0000007Amps

I couldn't make sense of an amperage value of 700nanoAmps, but now I can see the how/why/where a bit better.

Thank You.

 

[MagnetoSound]

That is what I meant by 'no significant current' ...

I never imagined anyone would actually calculate it!  8)