So, I'm reading that Navy NEETS thing and I'm on the cathode capacitor section. Here's what I'm troubled with:
So, is this what happens?
When the positive part of the signal hits the grid, current across the tube and Rk is increased. The voltage drop across Rk increases as electrons leave for the plate, which is what tries to drive the cathode up to +20 V. For the positive side of the capacitor to reach this point, electrons must leave it. These electrons wind up at the cathode. This becomes the source of electrons for the cathode, which lowers it's voltage and reduces current being drawn across Rk, which causes the voltage drop to decrease (and thus keep the bias of the tube).
I know none of this ever happens, which is the point, but are these the correct forces in play here?
Thanks
An externally hosted image should be here but it was not working when we last tested it.
So, is this what happens?
When the positive part of the signal hits the grid, current across the tube and Rk is increased. The voltage drop across Rk increases as electrons leave for the plate, which is what tries to drive the cathode up to +20 V. For the positive side of the capacitor to reach this point, electrons must leave it. These electrons wind up at the cathode. This becomes the source of electrons for the cathode, which lowers it's voltage and reduces current being drawn across Rk, which causes the voltage drop to decrease (and thus keep the bias of the tube).
I know none of this ever happens, which is the point, but are these the correct forces in play here?
Thanks
I am going to go on a bit of rant here. Maybe I am missing the point, but if I am someone will tell me I am sure.
This is the sort of thing that makes me want to punch someone in the nose. These hand-waving, lets explain it all with pictures explanations drive me right up the wall. I know these neets courses are held in some regard but holey moley that explanation is complete nonsense as far as I am concerned. And it is all done that way so they can try to explain it without math. I would understand that if the math involved was super duper PHD study for 40 years on a mountanin top in Tibet stuff, but just a bit of algebra and a teeny bit of calculus is all you need. If your willing to limit yourself to steady state questions then you dont need the calculus, just dont be afraid of complex numbers.
You cant get to the root of this stuff without doing the math. Youve got to be able to say V = IR,
V = 1/C int( i dt)
i = C dV/dT
know what all that means
and put it all together.
Otherwise youre at the mercy of these do it with pictures explanations.
What is all that stuff about the voltage on the resistor is 20, but the capacitor, connected to the exact same spot is 10 ?? "capacitors oppose a change in voltage " ??? HUH, then how do you charge them up ? How did it get to be 10 volts, and if it can get to be 10, why cant it get to be 20 ?
I think what they mean to say is that the current into a capacitor is proportional to the rate of change of the voltage across it, and so to change a capacitor voltage rapidly requires more current than to change it slowly. But thats a far cry from saying that one lead of the capacitor can magically be at a different potential from the point it is connected to. You have to say i = C dV/dT.
"Ck acts as the source of current for the cathode". What ?? Ck is only one source of current. There is current in the amount of 20/Rk amps coming from the resistor. That may be more than the current from Ck depending on the values of Rk, Ck, the tube, the signal voltage.
I could go on and on, but I will skip to the end of your question "forces in play here"
and give you my "handwaving" explanation
The forces in play are 1) the current through the tube is a function of grid to cathode voltage, but the voltage across the resistor is a function of that same current, so increasing the grid cathode voltage increases the current, which decreases the grid cathode voltage, tending to put you back where you started.
2) the current through a capacitor is proportional to the rate of change of volatage across it. Hence a rapidly changing voltage will cause more current to flow than a slowly changing voltage of the same magnitude. Or if you prefer, a rapidly changing current will produce a smaller voltage across the capacitor than a slowly changing current.
The net result of 1 and 2 is that the current can change in responses to the signal considerably without the voltage on the resistor changing much. R keeps the dc current and voltage more or less constant, C allows the ac current to change without changing the voltage very much.
Now my handwaving expanation is not much better, if at all, than the neets handwaving explanation. But thats my point. You have to do the math.
This is the sort of thing that makes me want to punch someone in the nose. These hand-waving, lets explain it all with pictures explanations drive me right up the wall. I know these neets courses are held in some regard but holey moley that explanation is complete nonsense as far as I am concerned. And it is all done that way so they can try to explain it without math. I would understand that if the math involved was super duper PHD study for 40 years on a mountanin top in Tibet stuff, but just a bit of algebra and a teeny bit of calculus is all you need. If your willing to limit yourself to steady state questions then you dont need the calculus, just dont be afraid of complex numbers.
You cant get to the root of this stuff without doing the math. Youve got to be able to say V = IR,
V = 1/C int( i dt)
i = C dV/dT
know what all that means
and put it all together.
Otherwise youre at the mercy of these do it with pictures explanations.
What is all that stuff about the voltage on the resistor is 20, but the capacitor, connected to the exact same spot is 10 ?? "capacitors oppose a change in voltage " ??? HUH, then how do you charge them up ? How did it get to be 10 volts, and if it can get to be 10, why cant it get to be 20 ?
I think what they mean to say is that the current into a capacitor is proportional to the rate of change of the voltage across it, and so to change a capacitor voltage rapidly requires more current than to change it slowly. But thats a far cry from saying that one lead of the capacitor can magically be at a different potential from the point it is connected to. You have to say i = C dV/dT.
"Ck acts as the source of current for the cathode". What ?? Ck is only one source of current. There is current in the amount of 20/Rk amps coming from the resistor. That may be more than the current from Ck depending on the values of Rk, Ck, the tube, the signal voltage.
I could go on and on, but I will skip to the end of your question "forces in play here"
and give you my "handwaving" explanation
The forces in play are 1) the current through the tube is a function of grid to cathode voltage, but the voltage across the resistor is a function of that same current, so increasing the grid cathode voltage increases the current, which decreases the grid cathode voltage, tending to put you back where you started.
2) the current through a capacitor is proportional to the rate of change of volatage across it. Hence a rapidly changing voltage will cause more current to flow than a slowly changing voltage of the same magnitude. Or if you prefer, a rapidly changing current will produce a smaller voltage across the capacitor than a slowly changing current.
The net result of 1 and 2 is that the current can change in responses to the signal considerably without the voltage on the resistor changing much. R keeps the dc current and voltage more or less constant, C allows the ac current to change without changing the voltage very much.
Now my handwaving expanation is not much better, if at all, than the neets handwaving explanation. But thats my point. You have to do the math.
I agree completely with Robert. As somebody who has to teach
electronics to college kids, these hand-wavy arguments which
blatantly violate Kirkhoff's laws only serve to confuse people
who are genuinely trying to understand electronics.
A much more accurate hand-wavy argument is that the resistor
is only there to set the DC operating conditions of the tube, and
since a capacitor looks like an open circuit to DC only the resistor
counts here. Any changes from that DC operating point (AC signal)
will cause AC current to be driven only through the capacitor since
this should be a much lower impedance path than the resistor for
audio frequencies. How true this is depends on the values of R and C.
In summary, the DC current goes through the resistor, while any
changes from that (AC current) is supplied by the capacitor.
The net result is higher gain for AC signals than you would get
from the resistor alone, although it is not obvious from this discussion
why that should be the case.
electronics to college kids, these hand-wavy arguments which
blatantly violate Kirkhoff's laws only serve to confuse people
who are genuinely trying to understand electronics.
A much more accurate hand-wavy argument is that the resistor
is only there to set the DC operating conditions of the tube, and
since a capacitor looks like an open circuit to DC only the resistor
counts here. Any changes from that DC operating point (AC signal)
will cause AC current to be driven only through the capacitor since
this should be a much lower impedance path than the resistor for
audio frequencies. How true this is depends on the values of R and C.
In summary, the DC current goes through the resistor, while any
changes from that (AC current) is supplied by the capacitor.
The net result is higher gain for AC signals than you would get
from the resistor alone, although it is not obvious from this discussion
why that should be the case.
Hi,
Thanks for the replies guys.
Torrence –
Why is that the case?
Without the capacitor, this is my understanding:
On the positive swing of the input voltage, more current flows through the tube, and the voltage drop across R increases, as does the voltage drop across the plate resistor. As a result of the cathode being more positive and the plate being more negative, there is less conduction through the tube (and thus less gain). However wouldn't less current through the tube decrease the voltage drops across the cathode and plate resistors again? The larger voltage difference between the plate and cathode would then make the tube start conducting more again? Help! It's a paradoxical circuit!
I never even got to what happens on the negative swing of the input voltage, hah.
Thanks!
Thanks for the replies guys.
Torrence –
Why is that the case?
Without the capacitor, this is my understanding:
On the positive swing of the input voltage, more current flows through the tube, and the voltage drop across R increases, as does the voltage drop across the plate resistor. As a result of the cathode being more positive and the plate being more negative, there is less conduction through the tube (and thus less gain). However wouldn't less current through the tube decrease the voltage drops across the cathode and plate resistors again? The larger voltage difference between the plate and cathode would then make the tube start conducting more again? Help! It's a paradoxical circuit!
I never even got to what happens on the negative swing of the input voltage, hah.
Thanks!
OK, I didn't want to get into this, and it is getting late, but...
[lecture]
A tube amplifies grid voltage into current, making it intrinsically a trans-conductance
device. Voltage gain is then trans-conductance times plate load g_m R_L. When you
add a cathode resistor, there is a sneaky negative feedback (as Horowitz and Hill
calls it) that reduces the intrinsic g_m for the reasons you say. As the
grid voltage goes up, the current increases, but the cathode voltage raises partly
to compensate this. Similarly if the grid voltage goes down, the current goes down,
but not quite as much as if the cathode resistor wasn't there as the dropping voltage
across the cathode resistor makes the grid-cathode voltage a bit larger. This is negative
feedback and hence reduces g_m.
A simple description of the voltage gain is now R_L/(1/g_m + R_K) where R_K is the cathode resistor seen "in series" with the tube. If this is large compared to 1/g_m it will dominate and your gain is just R_L/R_K. If you have g_m ~ 5 mS, this will be comparable to a 1/g_m = 200 ohm cathode resistor, and any value for R_K larger will dominate the gain. I believe the KT88 has g_m > 10 mS. To avoid this loss in high gain stages, you want to reduce the impedance seen by the AC signal through the cathode, hence an capacitor is used so the AC gain is purely g_m R_L.
[/lecture]
[lecture]
A tube amplifies grid voltage into current, making it intrinsically a trans-conductance
device. Voltage gain is then trans-conductance times plate load g_m R_L. When you
add a cathode resistor, there is a sneaky negative feedback (as Horowitz and Hill
calls it) that reduces the intrinsic g_m for the reasons you say. As the
grid voltage goes up, the current increases, but the cathode voltage raises partly
to compensate this. Similarly if the grid voltage goes down, the current goes down,
but not quite as much as if the cathode resistor wasn't there as the dropping voltage
across the cathode resistor makes the grid-cathode voltage a bit larger. This is negative
feedback and hence reduces g_m.
A simple description of the voltage gain is now R_L/(1/g_m + R_K) where R_K is the cathode resistor seen "in series" with the tube. If this is large compared to 1/g_m it will dominate and your gain is just R_L/R_K. If you have g_m ~ 5 mS, this will be comparable to a 1/g_m = 200 ohm cathode resistor, and any value for R_K larger will dominate the gain. I believe the KT88 has g_m > 10 mS. To avoid this loss in high gain stages, you want to reduce the impedance seen by the AC signal through the cathode, hence an capacitor is used so the AC gain is purely g_m R_L.
[/lecture]
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