I got my Morgan Jones book yesterday, and between that and the Tubecad site (and trying to remember Thevenin's theorem from my college classes), I'm pretty confused. The equation is here: http://www.tubecad.com/articles_2003/Grounded_Cathode_Amplifier/page5.html. It calculates a cathode bias resistor, and it says:
How do we get to that expression? Vb seems to be a constant, so the voltage at the cathode, Iq.Rk = Vb - (drop inside the tube), which is Iq.rp (assuming rp is for AC as well as DC resistance, which he says it's sometimes used interchangably for). This gives me Rk = Vb/Iq - rp, with no mu + 1. I've seen mu + 1 come in when the grid is grounded (either for real or my shorting the voltage source driving the grid while "Thevenising") - so if we're considering AC/dynamic conditions, for a change /\Vk (delta Vk), the grid-cathode difference also goes up by /\Vk, so the plate voltage changes by (mu + 1)./\Vk. But Vb is a constant. And that's where I get lost and don't know how to proceed.
I thought you picked a cathode bias resistor by looking at your operating point. You want current Iq, to get that you need the cathode to be a specific voltage Vg-k above the (grounded) grid, so Rk = Vg-k/Iq.
Any help would be most appreciated.
Thanks,
Saurav
The formula for determining the cathode resistor's value (in the absence of a plate resistor) is a simple one:
where, Rk equals the cathode resistor; Vb, the B+ voltage; Iq, the desired idle current; rp, the plate resistance; and mu, the triode's amplification factor.
How do we get to that expression? Vb seems to be a constant, so the voltage at the cathode, Iq.Rk = Vb - (drop inside the tube), which is Iq.rp (assuming rp is for AC as well as DC resistance, which he says it's sometimes used interchangably for). This gives me Rk = Vb/Iq - rp, with no mu + 1. I've seen mu + 1 come in when the grid is grounded (either for real or my shorting the voltage source driving the grid while "Thevenising") - so if we're considering AC/dynamic conditions, for a change /\Vk (delta Vk), the grid-cathode difference also goes up by /\Vk, so the plate voltage changes by (mu + 1)./\Vk. But Vb is a constant. And that's where I get lost and don't know how to proceed.
I thought you picked a cathode bias resistor by looking at your operating point. You want current Iq, to get that you need the cathode to be a specific voltage Vg-k above the (grounded) grid, so Rk = Vg-k/Iq.
Any help would be most appreciated.
Thanks,
Saurav
I'll try to help
but you must realize that I am a total newbie.
Pick your operating point, say 250v at 40 ma. That point will be somewhere on your loadline at some intersection with a grid bias, say -10v.
If you put 250v on the plate (between the plate and cathode) it will conduct 40 ma, for this hypothetical case. The grid is biased at -10v using a series resistor and, in this case, would be 250 ohm resistor (10v/.o40=250 ohm). This simply raises the cathode above the grid 10v (i.e. grid is -10v). As Joel puts it, "getting it?"
Don't know if this is what you were looking for, I don't want to do a derivation...I'll let others do that. Joel, John, Brett, Frank and others here can do better at explanation.
Rick
but you must realize that I am a total newbie.
Pick your operating point, say 250v at 40 ma. That point will be somewhere on your loadline at some intersection with a grid bias, say -10v.
If you put 250v on the plate (between the plate and cathode) it will conduct 40 ma, for this hypothetical case. The grid is biased at -10v using a series resistor and, in this case, would be 250 ohm resistor (10v/.o40=250 ohm). This simply raises the cathode above the grid 10v (i.e. grid is -10v). As Joel puts it, "getting it?"
Don't know if this is what you were looking for, I don't want to do a derivation...I'll let others do that. Joel, John, Brett, Frank and others here can do better at explanation.
Rick
The equation assumes a theretically perfect triode, and considers DC conditions only. Real valves are different, and plotting a loadline on reliable curves gives the best answer. It's worth remembering that access to full data sheets was quite limited during the valve heyday, and single-line data was common, so theoretical approximations were actually quite useful. You have the benefit of detailed data, so you can do it properly...
Your penultimate paragraph is spot-on.
Your penultimate paragraph is spot-on.
Even if the formula is correct, you still need to go back to the characteristic curves to see what rp and gm are at your actual desired current, unless you pick the one current for which the tube manual gives you those values. Bah, it's easier to calculate the resistor from the curves directly or determine the value empirically.
DO I LOVE CURVES, OR WHAT?
Hi,
Oh yes...which is why I plot them for every tube that goes into my own circuits.
A time consuming business, I tell you.
Maybe I should put a Sophia on my list for Santa.
They invariably differ from the characteristic curves as presented by the manufacturer, often not for the better.
Which also goes a long way to explain as to why people do hear differences between identical type numbers from different manufacturers, batches etc.
Cheers,
Hi,
Oh yes...which is why I plot them for every tube that goes into my own circuits.
A time consuming business, I tell you.
Maybe I should put a Sophia on my list for Santa.
They invariably differ from the characteristic curves as presented by the manufacturer, often not for the better.
Which also goes a long way to explain as to why people do hear differences between identical type numbers from different manufacturers, batches etc.
Cheers,
OK, assuming a theoretically perfect triode with constant rp and gm, could one of you explain how that formula works for DC conditions? I thought it would be DC, I didn't see why I would need to consider AC conditions to find the value of Rk. I thought mu is the ratio of the change in plate voltage to the change in grid voltage, both required to produce the same change in plate current. How does one then bring mu back into DC analysis? Using mu = gm/rp, I guess, and assuming rp is valid for DC. Even then, I don't understand how the (mu + 1) divisor comes around, even for a perfect triode.
OK, thanks, that's reassuring
And I think I know how to do that. rp is the slope of the plate curve at the operating point, gm is... ok, change in plate current by change in grid voltage for constant plate voltage, so a vertical line intersecting plate curves, measure the current change against the Y axis grid and the grid voltage change by the values on the curves I cross. Does that sound correct?
It's embarassing (and frustrating) when a book treats something as so basic that it doesn't need explanation, and I spend most of the evening trying to figure it out. Maybe I should have started with Rozenblit's book
OK, thanks, that's reassuring
And I think I know how to do that. rp is the slope of the plate curve at the operating point, gm is... ok, change in plate current by change in grid voltage for constant plate voltage, so a vertical line intersecting plate curves, measure the current change against the Y axis grid and the grid voltage change by the values on the curves I cross. Does that sound correct?
It's embarassing (and frustrating) when a book treats something as so basic that it doesn't need explanation, and I spend most of the evening trying to figure it out. Maybe I should have started with Rozenblit's book
All (valve or transistor) amplifiers are a mix of DC and AC conditions. AC conditions vary with DC conditions. Thus, DC conditions must be found first.
Early theorists tried to predict DC conditions using grossly over-simplified physical models. Hence the equation you presented. They aren't terribly good, but they are better than nothing. You have curves available, so you can ignore these poor models.
Yes, that is how to predict ra and gm. However, it's usually better to predict mu and use it to predict gm (by calculation with ra), than attempt to predict gm directly. This is because gm is very non-linear with Ia.
Early theorists tried to predict DC conditions using grossly over-simplified physical models. Hence the equation you presented. They aren't terribly good, but they are better than nothing. You have curves available, so you can ignore these poor models.
Yes, that is how to predict ra and gm. However, it's usually better to predict mu and use it to predict gm (by calculation with ra), than attempt to predict gm directly. This is because gm is very non-linear with Ia.
rp = dVp/dIp at fixed Vg
gm = dIp/dVg at fixed Vp
But, as I pointed out, if you're going to go to the trouble of determining these derivatives from the graphs, you may as well determine the operating point directly. A nice simplifying assumption is that Vk = 0, which is fairly true for most practical plate and cathode voltages.
I'm out drinking in your town tonight. If you hear some really loud, obnoxious people, that's probably us.
gm = dIp/dVg at fixed Vp
But, as I pointed out, if you're going to go to the trouble of determining these derivatives from the graphs, you may as well determine the operating point directly. A nice simplifying assumption is that Vk = 0, which is fairly true for most practical plate and cathode voltages.
I'm out drinking in your town tonight. If you hear some really loud, obnoxious people, that's probably us.
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