Frequency and Parallel RC circuits

[lowell]

Can someone explain how to calculate this? I've looked all over the web and in my texts but cannot find an explanation. I get high-pass and low-pass series arrangements with cutoff frequency: Fc=1/(2piRC)

Say we have: 1khz into a 1M grid-leak of a 12ax7, and in series w/ the next stage's grid and after the grid-leak we have 100k ll 47n.

I realize that miller capacitance will play a part here but can we ignore it for the sake of my understanding?

[R.G.]

Can someone explain how to calculate this? I've looked all over the web and in my texts but cannot find an explanation. I get high-pass and low-pass series arrangements with cutoff frequency: Fc=1/(2piRC)

Say we have: 1khz into a 1M grid-leak of a 12ax7, and in series w/ the next stage's grid and after the grid-leak we have 100k ll 47n.

I realize that miller capacitance will play a part here but can we ignore it for the sake of my understanding?

The characteristic frequency of a single R and a single C is always F = 1/(2*pi*R*C), no matter how they're arranged. In the case of an isolated parallel RC driven by a voltage source, what changes is the current, not the voltage. Imagine a perfect signal generator, internal resistance of 0.0000000000000000000 ohms. This drives a parallel RC. The voltage always remains the same, because the generator is perfect and can supply an infinite amount of current. But the current changes as frequency rises.

For frequencies where Xc is much greater than R, the current is composed only of the current through R. As frequency rises, the Xc = 1/(2*pi*F*C) decreases until |Xc|=R, and at that point, both R and C have equal currents, and the generator is supplying twice the current it did at very low frequencies. As frequency continues to rise, the capacitor current increases as Xc falls.

In doing the calculations, you must always take into account the impedance of the voltage source, though. This causes both a voltage division with the resistor at low frequencies and a division with both R and C at high frequencies.

In your case of 100K||47n, F = 33.9Hz (if I pressed the right keys), and this will roll off everything above that.

However, there are no perfect voltage generators. Whatever is driving that 100K||47n gets into the game too. If you're driving it from the plate of a 12AX7 with rp=67K and a 100K plate resistor, then the driving impedance is about 67K||100K, and this drives the 100K||47n. The 67K||100K is about 40K, and this drives the 100K to ground paralleled with 47n. So the driving impedance is the Thevenin equivalent of 40K and 100K, or 28.6K driving a 47nF, so highs above F = 1/(2*pi*40K*47n) = 118Hz will be rolled off.

[lowell]

Thanks RG. I think in your explanation your 100k ll 47n is to ground, not in series w/ the signal right? If it were in series then will it act the opposite: everything below 118Hz will be rolled off?

[Dave H]

Thanks RG. I think in your explanation your 100k ll 47n is to ground, not in series w/ the signal right? If it were in series then will it act the opposite: everything below 118Hz will be rolled off?

Not necessarily, it will depend on the impedance that's being driven. If it's driving a very high (infinite) impedance then there will be no current through the 100k // 47k and therefore no voltage drop at any frequency. If the driven impedance is 100k then it will roll off below 118Hz at -6dB /octave.

Dave H.

[R.G.]

Thanks RG. I think in your explanation your 100k ll 47n is to ground, not in series w/ the signal right?

That's correct.

If it were in series then will it act the opposite: everything below 118Hz will be rolled off?

If it was in series, things would be different. Exactly how it would be different depends on the rest of the circuit. You have to look at what drives the network, and what senses the network.

A network of a 100K paralleled with a 47nF in series with a signal line lets current through depending on how much current the source can provide and how much current the load can take.

If you put this between the plate of a 12AX7 and the grid of a following 12AX7 (and you used a BFC to block the DC from the plate of the first 12AX7) You'd have the signal voltage at the plate of the driving 12AX7 appearing to be driven through the rp of the 12AX7 drive in parallel with the plate resistor, as in my first note; about 40K. This signal then goes through the BFC, which we'll ignore as being a magic capacitor to just block all the DC, and then through the parallel combination of the 100K/47nF, and into the grid of the next stage, with a grid leak of 1M to ground.

So the signal goes through a series 40K, through the parallel combination of the 100K and 47nF, and into the parallel combination of the following grid and the 1M grid leak.

The grid is many, many times the impedance of the 1M grid leak capacitor, so we'll ignore it and just leave the 1M.

So the signal voltage goes into a voltage divider of 40K in series with the 100K||47nF, and into the 1M. The voltage that appears across the 1M is the signal which the following stage sees. At frequencies way below the rolloff of 100K and 47nF, the cap does not matter, so we have a divider of 140K and 1M, an attenuation of 1M/1140K, or about 0.877; the signal is 12% down by the voltage division.

At frequencies way above the 100K/47nF frequency, the capacitor looks like a short circuit, so the signal is divided by 1M/1040K = 0.96, losing only 4%.

So putting this network in series with the incoming signal to a grid which is driven by another plate does ... not much. It's a little attenuation, which gets to less attenuation above the cap frequency. That's because the grid cannot accept much current, another way of saying it's a very high impedance.

If that was not a grid, but instead a low impedance input, then the change between about 100K and a short circuit in the 100K/47nF would make a big change in what signal got transferred.

You have to remember that no network exists in isolation. You have to consider what drives it and what loads it. These are part of the circuit until you can prove they're negligible.

[lowell]

Awesome, I'll be absorbing this over the next couple days/weeks. Thanks for your insight. Oh and hey if you get a minute I could really use help w/ this thread: http://music-electronics-forum.com/t20631/

[Koreth]

The behavior in parallel RC networks is how the bright cap in a Fender, or the treble boost caps in a Marshall work. If you have SPICE, you can draw up RC networks, feed them with voltage and current sources and do an AC sweep to see how they behave at various frequencies. Of course, SPICE can be used for so much more than that, but having the computer do the math and draw pretty pictures can help compliment the explanations of circuit behaviors to be found here and elsewhere on the Internets.

[lowell]

This makes me ask a question: Can phase of current and voltage be different? If so what does this mean? If voltage and current are in phase does it sound or act different than if they're say 180deg out-of-phase?

[defaced]

Can phase of current and voltage be different?

Yes, this is a big issue in industrial applications where there is a high inductive load (motors, welders, etc). Basically, you lose power because P(t) = V(t)*I(t), so if they're out of phase, you are not getting peaks at the same time, thus you don't get full power. To correct this, you need to balance the C and L parts of the circuit so the phase shifts for V and I are the same.

Phase Relationships in AC Circuits

I do not know the significance in an audio circuit.

[R.G.]

This makes me ask a question: Can phase of current and voltage be different? If so what does this mean? If voltage and current are in phase does it sound or act different than if they're say 180deg out-of-phase?

Yes, they can be different. In fact, they ARE in general different for everything except resistors. In inductors, the voltage leads the current, and in capacitors the current leads the voltage.

If you think about this, it makes sense: inductors store energy in a magnetic field, driven by the current which flows in them. The current can't change instantly in an inductor, but the voltage across it can. The reverse is true for a capacitor. You have to pump current into them for a while to change their voltage.

We learned the mnemonic "ELI the ICEman", to remind us that E (voltage) leads I (current) in Ls (inductors) while I leads E in Capacitors.

For non-sinusoidal signals, the time response is just that. For sinusoidal signals, the voltage leads current in an inductor by 90 degrees, while the current in a cap leads the voltage by 90 degrees.

Adding in resistances to these and combining Ls and Cs lead to non-90 degree shifts and changing shifts with frequency.

[Bruce / Mission Amps]

100K || 47nF.

Simple RC, I get 3.386Hz.... not 33.9Hz. Unless you meant 4.7nF.

[R.G.]

100K || 47nF.
Simple RC, I get 3.386Hz.... not 33.9Hz. Unless you meant 4.7nF.

I'm ALWAYS subject to pressing the wrong keys!

Let me try again:

F = 1/(2*pi*R*C)
= 1/ (6.28318*10^5*47*10^-9)
= 1/(6.28318*47*10^5*10^-9)
= 1/(6.28318*47*10^-4)
= 1 /(295.309*10^-4)
= 10^4/(295.309)
= 33.8627....

I think it's 33.8... Hz. Of course, I'm also subject to making the same mistake consistently.

[Enzo]

We could compromise and just agree on 1710Hz.

[Dave H]

I was too lazy to press the wrong keys. I just reposted the (wrong) value from the post above