-3dB frequency calculation

[dominicbeesley]

Originally Posted by kalee20

Hi Dom,
...
Unfortunately, in practice, it's going to be somewhat lower than 659kHz. The 220nF capacitor may have significant stray capacitance to 0V, probably another 10pF. Because this is at the left-hand end of the 6.8kΩ resistor, you'll now have a 2-pole roll-off and it gets complicated! The moral is, don't press the 220nF closely against the metal chassis, so keeping the strays negligible.

Aha I hadn't thought of that! Anyway I thought I'd have a play around to see what everyone's getting heated about I used LTSpice and the circuit attached. Results with "0 impedance" I set R1 to 1R and R2 to 1M, with "0 strays" I removed C3.

Lower 3db Points:

                0 impedance     27k||45k
C1=100n            3.3Hz           3.2Hz
C1=220n            1.52Hz          1.48Hz

Not a huge amount of difference though I'm a bit surprised that the 0 impedance case gives a higher 3db point! Though its not by much. To be honest for the OP's circuit I'd go for the 100n as I'd be surprised if his power supply can actually make the HT rail anything like a virtual earth at 1-3Hz! And even more surprised if the o/p tranny and speakers can do anything with it...suppose it may matter for NFB stability...though I'd have thought a bigger margin would be left there.

Anyway I tested out the upper 3db points and here the valve impedance and strays do matter!

               0 impedance     27k||45k
Without strays      2.38Mhz      659kHz
With strays         2.34Mhz      435kHz

I have to say here the impedance of the preceding stage has made a big difference and have made the effect of strays far more pronounced.

Altogether an informative thread!

Disclaimer: the figures aren't super accurate - measured off a graph with the mouse!

Dom

[G8HQP Dave]

Originally Posted by MichaelR

Dave,

You are saying this equivalent circuit is wrong then ?

What has Dom got to do with it , It is Jimmc who says it is in series.

Mike

Your equivalent circuit is correct, but you then need to combine the voltage generator+RL+Ra into a new voltage generator with output voltage RL/(Ra+RL) and output resistance RL||Ra. I think this is the step you are missing.

My reference to Dom was about the HF point. For this the total impedance at the anode is in parallel with 470K, not in series with it, then gets 6K8 added in series.

(PS sorry about getting the sign of j wrong)

[jimmc101]

Sorry to have confused people.

I agree that if you talking about the ratio of 84 grid voltage to 83 anode voltage then the 27k is irrelevant, but then so is the value of the 83 anode voltage, it can do what it likes with frequency it will not affect this ratio;only the 220n and 470k matter.

However I took the original post as asking how to choose the value of the coupling capacitor C, and this ratio does not tell the whole story.

Dom & Dave have covered it pretty well but I thought I had better give my reasoning.

Looking at the first figure below I have represented the 83 by a current source ia, shunted by a resistor ra. This feeds across the 27k resistor RL which in turn is connected to the 220n C and the 470K RG.
The first step is to replace the 27k and ra with a single resistor R = 27k//ra.
Now convert the current source and shunt resistor to the equivalent voltage generator ia.R and series resistance R.
Since the same current must flow through R and RG (and C) they are effectively in series.
In order for this current i to be reduced by 3dB from its high frequency value (when Xc ~0) the reactance of the capacitor C must be equal to the total resistance of R + RG.

I have said before that in this case the effect is small but let me illustrate a practical case where this effect is dominant..

Looking at the second figure, consider an oscilloscope with AC coupling selected and its input connected to a low impedance sig gen.

The input current will be 3dB down (on its high frequency value) when the reactance of the 100n capacitor is equal to the 1M resistor ie at about 1.6Hz.

Now consider the effect of placing a x10 probe between the input and the generator.
At these low frequencies the probe may be treated as a simple 9M resistor.
In this case the reactance of the capacitor must be equal to 10M to reduce the input current by 3dB, this occurs at 0.16Hz.

So the LF cut-off at the probe tip is 0.16Hz even though the cut-off frequency of the 'scope (at its input terminals) is still 1.6Hz.

Jim

[kalee20]

Originally Posted by dominicbeesley

I'm a bit surprised that the 0 impedance case gives a higher 3db point! Though its not by much.

Well, if you take into account the output resistance, you have 488kΩ or whatever acting with the 0.1µF.

Ignoring the output resistance of the ECC83 stage (your 0 impedance case), you only have 470kΩ of resistance acting with the 0.1µF - a slightly smaller resistance, so yes, the 3db point will be at a slightly higher frequency.

Of course, the gain will be higher, by a small amount, for the latter case.

[Steve_Bell]

I think I'll just use 100n coupling caps!
Steve.