I would suggest that you redraw the circuit with the valve's anode as a series resistor from zero impedance generator and the "top" of the inductor as connected to ground. You then have a series R (the valve's anode resistance), a shunt L, a series C, and a shunt R. So it's a simpler three-terminal circuit.
All good fortune,
Chris
All good fortune,
Chris
That is a 2 Pole high pass filter.
It is not just a single 1 Pole high pass filter.
Typically, R1 is many times higher resistance than the plate impedance, rp. That is to keep R1 from heavily loading the plate.
First pole:
Plate impedance, rp
L1 inductive reactance, Xl
At the frequency where rp Ohms = Xl Ohms, that pole is -3dB at that frequency.
Xl = 2 x pi x frequency x L
Lets pick 20Hz for -3dB frequency response, and 30H choke
2 x pi x 20Hz x 30H = 3,770 Ohms
If the plate resistance is 3,770 Ohms, the single pole is -3dB @ 20 Hz
Lets figure for a -1 dB response (you will see why in a minute)
For a 1 pole high pass filter, the -1dB frequency response requires 2 x more inductance required for -3dB.
To get -1 dB response at 20Hz, we need 60H inductance
Second pole:
Lets let R1 be at least 10 times the plate impedance, rp (so we do not significantly load rp).
Let R1 = 37, 700 Ohms.
At the frequency where R1 Ohms = Xc Ohms, that is the frequency where the response is -3dB.
Xc = 1 / (2 x pi x f x C).
C = 1 / (Xc x 2 x pi x f)
Let Xc = 10 x rp = 37,700 Ohms; and let f = 20Hz
Then C = 0.21uF
The -1dB frequency response requires 0.42uF.
Now, at 20Hz, we have the first pole at -1dB with 60H inductor; and the second pole is -1dB with 0.42uF capacitor.
That is a total of -2dB at 20Hz.
Why do that . . .
Because if our output transformer is also - 1dB at 20Hz, then we have 3 X -1dB = -3dB at 20Hz for the complete amplifier.
I am sorry for the long dissertation, I hope it makes sense to you.
It is not just a single 1 Pole high pass filter.
Typically, R1 is many times higher resistance than the plate impedance, rp. That is to keep R1 from heavily loading the plate.
First pole:
Plate impedance, rp
L1 inductive reactance, Xl
At the frequency where rp Ohms = Xl Ohms, that pole is -3dB at that frequency.
Xl = 2 x pi x frequency x L
Lets pick 20Hz for -3dB frequency response, and 30H choke
2 x pi x 20Hz x 30H = 3,770 Ohms
If the plate resistance is 3,770 Ohms, the single pole is -3dB @ 20 Hz
Lets figure for a -1 dB response (you will see why in a minute)
For a 1 pole high pass filter, the -1dB frequency response requires 2 x more inductance required for -3dB.
To get -1 dB response at 20Hz, we need 60H inductance
Second pole:
Lets let R1 be at least 10 times the plate impedance, rp (so we do not significantly load rp).
Let R1 = 37, 700 Ohms.
At the frequency where R1 Ohms = Xc Ohms, that is the frequency where the response is -3dB.
Xc = 1 / (2 x pi x f x C).
C = 1 / (Xc x 2 x pi x f)
Let Xc = 10 x rp = 37,700 Ohms; and let f = 20Hz
Then C = 0.21uF
The -1dB frequency response requires 0.42uF.
Now, at 20Hz, we have the first pole at -1dB with 60H inductor; and the second pole is -1dB with 0.42uF capacitor.
That is a total of -2dB at 20Hz.
Why do that . . .
Because if our output transformer is also - 1dB at 20Hz, then we have 3 X -1dB = -3dB at 20Hz for the complete amplifier.
I am sorry for the long dissertation, I hope it makes sense to you.
Assuming that the triode stage has an output resistance Ri (internal resIstance) and everything else is ideal, it becomes a second-order high-pass filter with a natural frequency of fn = 1/(2 π √(LC (1 + R/Ri))) and a quality factor Q = 1/(2 π fn (L/Ri + RC)).
You have lots of equations now from the previous post and this one, but are they the equations you were looking for?
You have lots of equations now from the previous post and this one, but are they the equations you were looking for?
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niclaspa,
What kind of circuit is your schematic?
Choke loaded stage driving RC coupling to the next stages grid?
Parafeed output stage?
Something else?
Knowing that can make for a more accurately aimed answer to your question.
Chris Hornbeck is right.
I was re- editing, so I have not even closely looked at MarceldvG's answer.
Sorry, I took that circuit to be a choke loaded plate, that is driving a high impedance RC coupling to the high impedance grid of the next stage.
As usually practiced, that is essentially two separate 1 pole filters that have very little influence on each other (>/= 1:10 driving:load impedance).
Instead, Suppose R1 is purposefully loading the plate, rp.
An example of that is a Parafeed amplifier output stage.
If R1 represents the resistive load of an output transformer at mid frequencies, then . . .
L1 and C1 do form a practical 2 pole high pass filter (and if the output is lightly loaded, or unloaded, LC looks like a series resonant circuit.
But it becomes more complex, the output transformer is only resistive at mid frequencies, but the resistive load the primary presents, has inductance in parallel at low frequencies, and capacitance in parallel at high frequencies.
For purely resistive R1, where rp:R1 are 3:1, 1:3, or even lower ratio, resonance begins to occur.
I have to remind myself not to jump to conclusions; instead I need to ask question(s) to the original poster, in order to further clarify the original post's question.
What kind of circuit is your schematic?
Choke loaded stage driving RC coupling to the next stages grid?
Parafeed output stage?
Something else?
Knowing that can make for a more accurately aimed answer to your question.
Chris Hornbeck is right.
I was re- editing, so I have not even closely looked at MarceldvG's answer.
Sorry, I took that circuit to be a choke loaded plate, that is driving a high impedance RC coupling to the high impedance grid of the next stage.
As usually practiced, that is essentially two separate 1 pole filters that have very little influence on each other (>/= 1:10 driving:load impedance).
Instead, Suppose R1 is purposefully loading the plate, rp.
An example of that is a Parafeed amplifier output stage.
If R1 represents the resistive load of an output transformer at mid frequencies, then . . .
L1 and C1 do form a practical 2 pole high pass filter (and if the output is lightly loaded, or unloaded, LC looks like a series resonant circuit.
But it becomes more complex, the output transformer is only resistive at mid frequencies, but the resistive load the primary presents, has inductance in parallel at low frequencies, and capacitance in parallel at high frequencies.
For purely resistive R1, where rp:R1 are 3:1, 1:3, or even lower ratio, resonance begins to occur.
I have to remind myself not to jump to conclusions; instead I need to ask question(s) to the original poster, in order to further clarify the original post's question.
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Thanks for your answers! 🙂
I posted a simplistic question because I thought that there was a simple answer. I understand now that it may not be so. I am sorry if I have wasted your time. I will be more specific now.
I have built a headphone amplifier with inverted triodes, i.e., I use the grid as anode and the anode as grid. I got the idea after reading Stevie Bench's article: http://www.jacmusic.com/techcorner/SBENCH-PAGES/sbench102/inverted.html My amplifier measures great and sounds great. However, it made me aware that I don't fully understand how an LC circuit works. In schematics that I have seen where LC stages drive another stage and therefore, don't drive any heavy load, the chokes have fairly big inductance - tens or even hundreds of H. Stevie Bench uses a 1 H inductor:
He measures a frequency response from 3 Hz to 50 kHz.
In my amplifier, I use an output capacitor because I don't think that headphones will be happy with a small amount of DC. I have tried using a cheap 0.5 H inductor and an expensive 90 H Lundahl inductor. I can neither find any difference in the frequency response nor in the sound quality. Since my RMS voltage meter seems to be inaccurate below 30 Hz, I may have missed changes in this spectrum.
Is it the heavy load that allows the use of a small inductor without adversely affecting the frequency response?
I posted a simplistic question because I thought that there was a simple answer. I understand now that it may not be so. I am sorry if I have wasted your time. I will be more specific now.
I have built a headphone amplifier with inverted triodes, i.e., I use the grid as anode and the anode as grid. I got the idea after reading Stevie Bench's article: http://www.jacmusic.com/techcorner/SBENCH-PAGES/sbench102/inverted.html My amplifier measures great and sounds great. However, it made me aware that I don't fully understand how an LC circuit works. In schematics that I have seen where LC stages drive another stage and therefore, don't drive any heavy load, the chokes have fairly big inductance - tens or even hundreds of H. Stevie Bench uses a 1 H inductor:
He measures a frequency response from 3 Hz to 50 kHz.
In my amplifier, I use an output capacitor because I don't think that headphones will be happy with a small amount of DC. I have tried using a cheap 0.5 H inductor and an expensive 90 H Lundahl inductor. I can neither find any difference in the frequency response nor in the sound quality. Since my RMS voltage meter seems to be inaccurate below 30 Hz, I may have missed changes in this spectrum.
Is it the heavy load that allows the use of a small inductor without adversely affecting the frequency response?
Could be. If the coupling capacitor has a large value and the headphone has a low impedance, then you need less inductance than with a high-impedance load.
For example, suppose the coupling capacitor is so large that it can be treated as infinite and that the internal resistance of the valve is much larger than the impedance of the headphone. You then get first-order roll-off below f = 1/(2π L/R), where R means the headphone impedance. With 0.5 H and 32 ohm, that's about 10 Hz.
The capacitor would have to be much greater than 500 uF for this approximation to hold. Is it?
For example, suppose the coupling capacitor is so large that it can be treated as infinite and that the internal resistance of the valve is much larger than the impedance of the headphone. You then get first-order roll-off below f = 1/(2π L/R), where R means the headphone impedance. With 0.5 H and 32 ohm, that's about 10 Hz.
The capacitor would have to be much greater than 500 uF for this approximation to hold. Is it?
By the way, if the coupling capacitance is not that high but the internal resistance is high enough to be neglected, the equations from post #4 can be simplified to a natural frequency of fn = 1/(2 π √(LC)) and a quality factor Q = 1/(2 π fn RC). I haven't a clue if this is applicable when you drive a triode inside-out, in fact I'm surprised the grid survives.
niclaspa,
Your schematic in Post # 6 has lots of errors.
True, it is trying to represent an inverted operation mode, where the grid is used as a plate, and the plate is used as a grid.
Why does anybody do that?
Why does anybody climb Mt. Everest?
Because both can be done, but only if they are properly implemented and properly carried out.
The grids will cook if they are connected to a 7.7V 700mA capable power supply, at the same time that the cathode is connected to a -13V 700mA capable power supply.
Glow! Meltdown!
True, the impedance of a 5687 grid is low, compared to it's plate. But low impedance is one thing, power handling is another thing.
Even the 5687 parallel plates can not handle 700mA.
Start over with a complete, accurate, and working schematic.
Then we can discuss things like chokes, high pass filters, etc.
Your schematic in Post # 6 has lots of errors.
True, it is trying to represent an inverted operation mode, where the grid is used as a plate, and the plate is used as a grid.
Why does anybody do that?
Why does anybody climb Mt. Everest?
Because both can be done, but only if they are properly implemented and properly carried out.
The grids will cook if they are connected to a 7.7V 700mA capable power supply, at the same time that the cathode is connected to a -13V 700mA capable power supply.
Glow! Meltdown!
True, the impedance of a 5687 grid is low, compared to it's plate. But low impedance is one thing, power handling is another thing.
Even the 5687 parallel plates can not handle 700mA.
Start over with a complete, accurate, and working schematic.
Then we can discuss things like chokes, high pass filters, etc.
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Post #6: "I have built a headphone amplifier with inverted triodes, i.e., I use the grid as anode and the anode as grid."
That is not my schematic. It is Stevie Bench's. In the link that I provided, he describes the amplifier. I haven't built that amplifier. So, I can't testify that it works. But, I can give the schematic of my amplifier. It has been running for at least 50 hours without problem:
niclaspa,
Your schematic:
At 20Hz, the inductive reactance of 0.5 Henry is Only 63 Ohms.
That is probably much less than the impedance of two 6N6P grids in parallel.
That would cause a lot of low frequency rolloff, regardless of the other circuits frequency response.
The input tube's cathode resistor is un-bypassed.
Is it 1.8k Ohm?
The 6N6P's u = 20.
The plate impedance With the cathode bypassed, is 1800 Ohms
That un-bypassed cathode increases the plate resistance of the input tube, rp, by another 36k (1800 x 20 = 36k), the total rp = 37,800 Ohms
The input tube's high plate impedance 37,800 Ohms, has to drive 270H at low frequency, and also at high frequency to drive the distributed capacitance of all those windings)
270H at 20Hz = 33.9k Ohms.
So, that is about -3dB at 20Hz at the input tube plate circuit.
Your schematic:
At 20Hz, the inductive reactance of 0.5 Henry is Only 63 Ohms.
That is probably much less than the impedance of two 6N6P grids in parallel.
That would cause a lot of low frequency rolloff, regardless of the other circuits frequency response.
The input tube's cathode resistor is un-bypassed.
Is it 1.8k Ohm?
The 6N6P's u = 20.
The plate impedance With the cathode bypassed, is 1800 Ohms
That un-bypassed cathode increases the plate resistance of the input tube, rp, by another 36k (1800 x 20 = 36k), the total rp = 37,800 Ohms
The input tube's high plate impedance 37,800 Ohms, has to drive 270H at low frequency, and also at high frequency to drive the distributed capacitance of all those windings)
270H at 20Hz = 33.9k Ohms.
So, that is about -3dB at 20Hz at the input tube plate circuit.
We still don't know the impedance of the headphone.
At 20 Hz, C7 in parallel with R11 has a higher impedance than the inductor, so the assumption that the inductor is connected straight to a low-impedance supply is wrong. Hence, my calculations don't apply. I wonder if you couldn't get rid of L2 and C7 altogether.
At 20 Hz, C7 in parallel with R11 has a higher impedance than the inductor, so the assumption that the inductor is connected straight to a low-impedance supply is wrong. Hence, my calculations don't apply. I wonder if you couldn't get rid of L2 and C7 altogether.
MarcelvdG,
As you noted, the B+ impedance at 20Hz is far higher than 0.5H.
C7, 47uF is 169 Ohms at 20Hz.
That is far less than R11's 2.5k Ohms.
Get rid of L2 and C7?
Well then what do you use as a load for the output tube grids (acting as plates)?
With R11 of 2.5k, you will need a lot more than 17V B+ to get lots of current.
I suspect this amplifier works as well as it does, because the 30 Ohm phones only require a few milliwatts to be damaging to ears.
At 14 Posts already, this thread may never get to a conclusion.
As you noted, the B+ impedance at 20Hz is far higher than 0.5H.
C7, 47uF is 169 Ohms at 20Hz.
That is far less than R11's 2.5k Ohms.
Get rid of L2 and C7?
Well then what do you use as a load for the output tube grids (acting as plates)?
With R11 of 2.5k, you will need a lot more than 17V B+ to get lots of current.
I suspect this amplifier works as well as it does, because the 30 Ohm phones only require a few milliwatts to be damaging to ears.
At 14 Posts already, this thread may never get to a conclusion.
R11 is already in the path from the supply to the grid, and its impedance is >> 30 ohm, so there is need why it couldn't bias the grid by itself, without L2 and C7.
I agree completely. The large value of R11 is probably also the reason why the grids survive.
My headphones output 100 dB when fed with 1 mW. The amplifier outputs 160 mW. So, plenty of headroom.
I don't see how it could deliver more than 750 uW into a 30 ohm load with R11 in the path from the supply to the headphone.