Where does AC signal go that is blocked by coupling capacitor value?

[tubedood]

Happy Holidays and I hope the weekend is treating everyone well.


I have another question for you all!


In an amplifier that has a first gain stage with a .02uf coupling capacitor followed by a second gain stage with a .1uf coupling capacitor... how is the second stage getting a reference to the lower frequency signal that was blocked by the .02uf of the first stage?

The plates are decoupled with our filter caps so not there I don't think. I am thinking via the cathode through the grounding?

As always thanks for your explanations and time!

Dale

[g1]

Not sure I understand the question, could you clarify? You are wondering how there can be LF generated after being blocked, or are you asking 'why' they use a .1uF there?

[Helmholtz]

Happy Holidays and I hope the weekend is treating everyone well.


I have another question for you all!


In an amplifier that has a first gain stage with a .02uf coupling capacitor followed by a second gain stage with a .1uf coupling capacitor... how is the second stage getting a reference to the lower frequency signal that was blocked by the .02uf of the first stage?

The plates are decoupled with our filter caps so not there I don't think. I am thinking via the cathode through the grounding?

As always thanks for your explanations and time!

Dale

Why do you think that the second stage needs a reference to the "lower frequency signal"? The RC filter attenuations expressed in dB of the two stages just add at each frequency. In other words their frequency responses multiply.
For more detailed info please post a rough drawing of the schematic.

[tubedood]

Here's a schematic similar what I was referring to...

If the first gain stage has a .0047uf coupling capacitor blocking the lower frequencies from passing to the next stage, where is the second gain stage getting those lower frequencies to use that big ole .1uf after those frequencies were blocked by the little .0047uf?

[Helmholtz]

Here's a schematic similar what I was referring to...

If the first gain stage has a .0047uf coupling capacitor blocking the lower frequencies from passing to the next stage, where is the second gain stage getting those lower frequencies to use that big ole .1uf after those frequencies were blocked by the little .0047uf?




[ATTACH=CONFIG]51630[/ATTACH]

The .0047µ cap does not block lower frequencies, it just attenuates them. It forms a high pass RC filter with a corner frequency of 68Hz. This means that 68Hz will be 3dB or about 30% down. Lower frequencies get attenuated stronger, 6dB more each lower octave. The second coupling cap increases low frequency attenuation but there will still be low frequency content below 68Hz.

[tubedood]

Thanks Helmholtz!

This is one of those little baby steps that starts to make things clearer!
So the lower frequencies are still presented to the second gain stage they are just reduced (attenuated) a bit from the first stage!

As always I appreciate this forum and all its helpful members!

[Helmholtz]

So the lower frequencies are still presented to the second gain stage they are just reduced (attenuated) a bit from the first stage!

You got it. An attenuation of 6db/octave means that e.g. a 25Hz signal will be 50% lower/smaller than a 50Hz signal.

Thanks Helmholtz!

You're welcome. Glad to help.

[mozz]

You can also have generated intermod distortion so end up with frequencies lower.

[J M Fahey]

Or you can see it this way:
Capacitors 2 and 3, those .1 connected to PI pins 2 and 7, must "do nothing" to Guitar signal , just freely let it through, so they are chosen "larger than needed" with turnover frequencies below the Guitar signal band.
470k and .1 mean a crossover frequency of 3.4 Hz.

Also shows that "capacitors don´t really matter" since they are practically transparent at all Guitar frequencies, dropping minuscule voltage across their terminals and passing minuscule currents.
So dielectric and losses become irrelevant, ESR disappears compared to the half megohm impedance involved, etc.

But suppose designer found speaker in an open back cabinet flaps too much, there is no Audio down there, amp muddies when overdriven, etc; then maybe he prefers to cut LF somewhat.

Then he lowers *one* coupling cap value, to get some usual cutoff.

In this case .0047 and some 300k mean about 110Hz turnover.

Why "about 300k"?

Both 500k pots are in parallel, which would mean around 250k combined load, but tone pot has series capacitors at both ends, which will increase impedance.

Not doing all tye Math or simulation, I find some 300k , maybe 350k combined impedance a reasonable assumption for this "back of envelope" problem.

[Helmholtz]

470k and .1 mean a crossover frequency of 3.4 Hz.

I avoided to calculate the second crossover frequency as there is some bootstrapping involved which increases input impedance and makes calculation complicated.

In this case .0047 and some 300k mean about 110HJz turnover.

Why "about 300k"?

Both 500k pots are in parallel, which would mean around 250k combined load, but tone pot has series capacitors at both ends, which will i9ncrease imoedance.

The actual crossover frequency depends on the settings of both pots. At low volume its close to 68Hz. The 500pF cap can be ignored for low frequencies.

[Chuck H]

Interestingly... With the attenuation of the coupling cap value AND the implementation of the "tone" circuit, with both pots at full the circuits acts as both a high pass filter AND a low pass filter (with frequencies passing through the .01uf cap at half the 0V load of the rest!). Admittedly not much low pass filtration, but it's there and it's real. So this is a mid heavy circuit. For better or worse.

To address the question presented by the thread title "Where does AC signal go that is blocked by coupling capacitor value? it's actually easier than many think. Below the knee frequency (determined by cap value AND circuit impedance) the cap cuts at 6dB per octave. This attenuation is in parallel with the anode load and ultimately determines the load @ frequency that is "seen" by the amplifying tube. That's the simple answer if no other circumstances are involved. And it's much more important for power tubes than it is for signal voltage amplifiers (preamp tubes).

[Helmholtz]

To address the question presented by the thread title "Where does AC signal go that is blocked by coupling capacitor value? it's actually easier than many think. Below the knee frequency (determined by cap value AND circuit impedance) the cap cuts at 6dB per octave. This attenuation is in parallel with the anode load and ultimately determines the load @ frequency that is "seen" by the amplifying tube.

Let's look at the basic RC highpass circuit consisting of the coupling cap and a resistor wired to ground. This presents a frequency dependent AC voltage divider circuit. At high frequencies the voltage across the cap tends to zero and the following stage gets the full signal. With decreasing frequency the impedance of the cap increases and so does the AC voltage across the cap. As more signal voltage hangs across the cap and thus diminishes the signal seen by the resistor, the current through the resistor decreases and so does the voltage developed across the resistor, which is the output signal of the filter. So one might say that the missing signal voltage at low frequencies is the voltage drop across the cap. (To be precise: The input voltage of the filter is the vector sum of cap and resistor voltages).

Please ask, if the above explanation is not clear enough. I sometimes fail to precisely "transform" my german thoughts into eglish text.

Regarding the interaction between the filter and the triode feeding it:
The RC filter presents a load impedance to the triode which is always greater the the R part.
The plate output impedance of a 12AX7 input stage is typically below 60k. So a load of several hundred k will not produce much interaction effect.

[tubedood]

You all are so helpful thank you!
I am having so much fun with this 5F2-a project and I like that I am learning the "how and why" these things work!

I had always thought of the coupling cap as a "window" or "aperture" that let a certain size and smaller signal through but blocked a larger waveform... now I have a much better understanding of the way it works. Learning feels great!
Helmholtz I hail from Germany myself but haven't been back since I was a little boy ( wow 45 years ago!). I was born in Frankfurt... mom was from Berlin. The wife and I are planning a trip back maybe next year or the year after.

Happy Holidays to all and thank you for a great holiday gift to me hahaa. The gift of understanding!
Dale

[Helmholtz]

I avoided to calculate the second crossover frequency as there is some bootstrapping involved which increases input impedance and makes calculation complicated.

OK, I am not ducking out on this:

Calculation and measurement show that the signal voltage at the cathodes of a LTP is about 50% of the input voltage. The bias resistor (820 Ohm) doesn't change much. This means that the AC voltage across the 470k grid resistor is only half of what it would be if it were connected to ground. With the result that the input current also is halved. Half the input current means doubling of the input impedance. This is the bootstrap principle.

In other words, the effective input impedance of a LTP is about twice the input grid resistance. For the example circuit shown, this means a resistive input impedance of about 1M and a (second) corner frequency of 1.7Hz.

[Old Tele man]

In simple RC circuit, the AC signals are attenuated across the XC-reactance and R (in voltage divider ratio), scaled by their frequency content.