Hey guys,
I am trying to determine the input impedance of my amp. There is a 100K resistor at the RCA input jack from signal to ground (I am assuming this is a bleeder for the source output coupling cap?), followed by a 250K volume pot, and finally a grid grounding resistor (300K connected from grid input to ground).
What I am unsure of is: Does the 100K resistor set the input impedance? Or do I also need to take into account the 250K pot's resistive track + the 300K grid ground?
TIA
I am trying to determine the input impedance of my amp. There is a 100K resistor at the RCA input jack from signal to ground (I am assuming this is a bleeder for the source output coupling cap?), followed by a 250K volume pot, and finally a grid grounding resistor (300K connected from grid input to ground).
What I am unsure of is: Does the 100K resistor set the input impedance? Or do I also need to take into account the 250K pot's resistive track + the 300K grid ground?
TIA
Are there any input caps ??? or is it direct coupled ??
Sounds like (100K // 250K // 300K) ~ 57.7K mid band Z ....
Need more info or schematic ....
Any feedback applied to first stage ??
Sounds like (100K // 250K // 300K) ~ 57.7K mid band Z ....
Need more info or schematic ....
Any feedback applied to first stage ??
put a small signal from a function generator on the input, analyze the output voltage, take a potentiometer, put it in series with the input, adjust the potentiometer until until you see half the original voltage on the output, measure the resistance of the potentiometer, that's you input resistance
properties of voltage division allow you to do this
we do this in my linear circuit design course at my university to find input resistances
properties of voltage division allow you to do this
we do this in my linear circuit design course at my university to find input resistances
Zin changes depending on the position of the pot.
Minimum position, Zin = 100k//250k
Maximum Position, Zin = 100k//250k//300k
Minimum position, Zin = 100k//250k
Maximum Position, Zin = 100k//250k//300k
The 100k resistor does nothing useful, as it is in parallel with the pot. However, unless you have a particularly poorly-engineered source, it does no harm either.
For real benefit add a coupling cap between the pot slider and the grid. This will lengthen the useful life of the pot by removing grid current from it.
For even more benefit, add failsafe resistors to the bias pots. As things stand a failed bias pot will cook an output valve. Why do designers cut costs like this?
For real benefit add a coupling cap between the pot slider and the grid. This will lengthen the useful life of the pot by removing grid current from it.
For even more benefit, add failsafe resistors to the bias pots. As things stand a failed bias pot will cook an output valve. Why do designers cut costs like this?
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The lowest input impedance is usually with the pot wiper midway through the resistance range. The series part is added to the parallel combination of the input resistor and the shunt part of the wiper.
So that would be 100k||125+(125k||300k) in this case, or approximately 77k.
So that would be 100k||125+(125k||300k) in this case, or approximately 77k.
No, that gives the lowest output impedance of a pot. The lowest input impedance, as ballpencil said in post 4, is at maximum slider position.sangram said:
I think they had access to 250K pots....
I would get rid of the 300k and try just a 100K pot or just the 250K and see what one you like better...
I would get rid of the 300k and try just a 100K pot or just the 250K and see what one you like better...
Hmm now that's an idea. If I get rid of everything and just install a 100K pot between input and grid, (correct me if I'm wrong) I would have a constant 100K input impedance regardless of volume setting.
better solution i think would be to replace the 300k with 1Meg and use 100k pot.. that way, it's either 100k or 100k//1Meg, which is close to 100k as well (90.91k), and there is a permanent path to ground for the control grid, even though the pot wiper goes flaky.. i wouldn't trust a potentiometer to provide a grid leak current path.
why the obsession with constant input impedance?
why the obsession with constant input impedance?
Thanks for the suggestion. I am trying to select an output coupling cap for the source and want to keep the capacitance to a minimum due to size and cost. I am not too concerned with the fact that the input impedance may vary but I do want to keep it high enough to avoid any low frequency rolloff.
… then in that case, do nothing, and use the min-value as determined by the triple parallel equation.
Rparallel = 1/(1/R1 + 1/R2 + 1/R3 … …)
Its only 'problem' is that it blows up when any resistor is zero Ω. Dividing by zero and all that. But in the real world, even straight wire isn't zero Ω, so substituting a (x + 0.001) works out quite well.
57.7 kΩ … if you want –1 dB at 20 Hz, using nothing more than
Z = 1/(2 π F C) … or
C = 1/(2 π F Z) … and substituting
C = 1/(2 × 3.1415 × 20 × 57,700)
C = 0.14 μF for –3 dB
And then the rule of thumb is to use a capacitor exp(1) = 2.727 times larger in order for the actual attenuation be below –1 dB at the lowest pass frequency. Better phase coherence, as well.
In this case the answer (my opinion) is the closest modestly priced well regarded capacitor between 0.33 μF and 0.47 μF, those being standard values.
With… 930C4P47K-F (Cornell Dubilier, 0.47 μF, 275 V) being one of the most cost effective film caps out there. $3.33 at Mouser, quantity 1.
There. Simple, easy, done.
GoatGuy
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