Ok. You were using the infinite tail impedance and infinite mu approximations, but only mentioning the former. "Small signal" does not mean no change in anode-cathode voltage and hence no change in grid-cathode voltage; it means linear change in both, in the ratio of mu.
I know what you meant by "roughly equal" in post 18, but someone new to the CF might puzzle why it was not 'exactly equal'.
I know what you meant by "roughly equal" in post 18, but someone new to the CF might puzzle why it was not 'exactly equal'.
It means that the change in Vak is negligible so it can be regarded as essentially constant.
I know CF's gain is less than unity and that high gm tubes have lower output impedance in a CF, though I'm not sure how these translate to the output level. I still think I'm missing the differences between DC bias and AC signal.
I want to say high gm leads to more current through the tube for a given (positive) signal and so more volts across a given load; thus the cathode voltage more closely follows the grid voltage. What about mu, though? (edit: Soda, come on. Mu = Gm x Rp, duh.)
SY said:
I'm going to try and wrap my head around that. Probably part of my problem is thinking about this like "If A, then B" when it would be more accurate to think about it "When A, also B."
Something about gm pulling up on the current and Vg-k pushing it back down simultaneously, but I feel like I'm still missing something. The tube is trying to convert signal voltage into modulated current, while the CCS is seeing a modulated voltage and sinking a constant current. Output would be taken at the junction of cathode and CCS? But then I don't immediately see how that is different than a resistive load other than that the CCS is keeping current constant (and has higher AC impedance?).
Thanks for the brain teasers, SY and DF96.
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Yes, the CCS is just the limit of high cathode resistance. Assume Vak constant (the small signal limit) and very high mu- the CCS sets the tube's current at a fixed value. Now if Vg changes with respect to common, the current will only be constant if the cathode changes voltage the exact same amount so that Vgk is constant.
Something about (ideally) infinite AC impedance in a CCS clicked there. So any change in AC voltage at the cathode produces no change in current because it is working into a theoretically infinite impedance. Output in the cathode follows the signal at the grid (and current is constant). I get the name now.
A cathode resistor with a value "infinity - 1" would also present a high AC impedance and result in very little AC current change [voltage/(infinity-1)]. But this would present high DC resistance and no current (ok, very little) would flow through the tube. This would be close to unity gain though, if it worked?
Ok, how is this for a re-write summation:
The difference in voltage between the grid and the cathode is what lets a tube control the flow of electrons (which in turn determines the voltage across a load). Changes in voltage on the grid (holding the cathode constant) have a reinforcing effect on this flow, but changes in voltage at the cathode (holding the grid constant) have an opposing effect. In the end, the output voltage (taken across the cathode resistor) follows the grid voltage up and down by about the same amount that the grid itself moves up and down because that’s all it can do. The cathode follower can’t make the voltage changes at the cathode any bigger than the changes at the grid because it would otherwise affect the difference between them.
The difference in voltage between the grid and the cathode is what lets a tube control the flow of electrons (which in turn determines the voltage across a load). Changes in voltage on the grid (holding the cathode constant) have a reinforcing effect on this flow, but changes in voltage at the cathode (holding the grid constant) have an opposing effect. In the end, the output voltage (taken across the cathode resistor) follows the grid voltage up and down by about the same amount that the grid itself moves up and down because that’s all it can do. The cathode follower can’t make the voltage changes at the cathode any bigger than the changes at the grid because it would otherwise affect the difference between them.
Sodacose … perhaps this will help.
First, abstract “a tube” as a voltage controlled current source (which is almost nearly the same abstraction that one uses for FETs and MOSFETs!). The math can be complicated by lots of details, or you can just use the linear form of it:
IKathode = gm • (Vgrid - Vbias)
Understand: this is a huge simplification. But it is “good enough” for looking at the cathode-follower response at an arbitrary operating point.
Second … note that 'K' is used for the kathode (from the German) conventionally. Its a little odd, but 'C' was taken by capacitance, so we all have gotten used to it.
Third, choose some parameters, and see how the math works out: its not really hard. For example… lets use a tube that offers a gm (transconductance) of 10 ma/v at a grid bias of –2 volts. Yes, numbers pulled out of a hat. Doesn't matter. We'll use a cathode resistance of RK = 1,000 Ω as well.
gm = 10 ma/V = 0.010 A/V
RK = 1000 Ω
Vbias = 2.0 V
Doing a bit of algebra:
IK = gm • (Vbias - VK + VG) … and
VK = IK • RK … with substitution yields
IK = gm (Vbias + VG) / (1 + gm RK)
Then putting in the values (above) yields (with VG = 0.0 V at quiescent)
IK = 0.010 × 2.0 / (1 + 0.010 × 1000 )
IK = 0.0018 A (≡ 1.8 ma) with
VK = IK • RK = 1.818 volts.
OK! now we have 1.818 volts on the cathode, for no input. Let's try +0.5 volts input.
IK = 0.010 ( 0.5 + 2.0 ) / ( 1 + 0.010 × 1000 )
IK = 2.272 V
What's the difference? 2.272 - 1.818 = 0.455 volts. The cathode has risen 0.45 volts, thus more-or-less following the input. NOT really well, but kind-of-sort-of.
And when we drop input to –0.5 volts? … sub in the numbers and get 1.364 volts on the cathode, or a drop of –0.454 volts. Again, more or less tracking the input.
WHAT HAPPENS when the tube has a higher transconductance? Say 25 ma/V? Simple: the higher amplification goes to cause the cathode to more closely track the input voltage. Both directions.
I'm fairly certain that the above will not have materially helped you (or anyone) understand how cathode-followers "work"; yet, if one can manipulate just first-year High School algebra, and one has confidence to throw around the symbols … it is pretty easy to see how the thing numerically works.
Just ciphering along…
GoatGuy
First, abstract “a tube” as a voltage controlled current source (which is almost nearly the same abstraction that one uses for FETs and MOSFETs!). The math can be complicated by lots of details, or you can just use the linear form of it:
IKathode = gm • (Vgrid - Vbias)
Understand: this is a huge simplification. But it is “good enough” for looking at the cathode-follower response at an arbitrary operating point.
Second … note that 'K' is used for the kathode (from the German) conventionally. Its a little odd, but 'C' was taken by capacitance, so we all have gotten used to it.
Third, choose some parameters, and see how the math works out: its not really hard. For example… lets use a tube that offers a gm (transconductance) of 10 ma/v at a grid bias of –2 volts. Yes, numbers pulled out of a hat. Doesn't matter. We'll use a cathode resistance of RK = 1,000 Ω as well.
gm = 10 ma/V = 0.010 A/V
RK = 1000 Ω
Vbias = 2.0 V
Doing a bit of algebra:
IK = gm • (Vbias - VK + VG) … and
VK = IK • RK … with substitution yields
IK = gm (Vbias + VG) / (1 + gm RK)
Then putting in the values (above) yields (with VG = 0.0 V at quiescent)
IK = 0.010 × 2.0 / (1 + 0.010 × 1000 )
IK = 0.0018 A (≡ 1.8 ma) with
VK = IK • RK = 1.818 volts.
OK! now we have 1.818 volts on the cathode, for no input. Let's try +0.5 volts input.
IK = 0.010 ( 0.5 + 2.0 ) / ( 1 + 0.010 × 1000 )
IK = 2.272 V
What's the difference? 2.272 - 1.818 = 0.455 volts. The cathode has risen 0.45 volts, thus more-or-less following the input. NOT really well, but kind-of-sort-of.
And when we drop input to –0.5 volts? … sub in the numbers and get 1.364 volts on the cathode, or a drop of –0.454 volts. Again, more or less tracking the input.
WHAT HAPPENS when the tube has a higher transconductance? Say 25 ma/V? Simple: the higher amplification goes to cause the cathode to more closely track the input voltage. Both directions.
I'm fairly certain that the above will not have materially helped you (or anyone) understand how cathode-followers "work"; yet, if one can manipulate just first-year High School algebra, and one has confidence to throw around the symbols … it is pretty easy to see how the thing numerically works.
Just ciphering along…
GoatGuy
Oh… darn… I forgot the “money quote” at the end.
The gain ("big G") =
G = Vin / Vout … which thru a little algebraic rearrangement is
G = RK gm ( VG + Vbias ) / ( 1 + gm RK )
G ≈ VG • ( β / (1 + β ) ) where…
β = gm • RK
β in this case (1000 Ω RK and gm = 0.010) is 10. Therefore
G ≈ VG • ( 10 / (10 + 1) )
G ≈ VG × 0.909090…
The 'closeness' to 1.000 gets better (rather clearly from the beta equation) when either the RK or the gm increases (and especially when both happen).
So, having a low output impedance to drive reduces the β, causing the cathode follower to not-follow-as-closely. I guess there is no free lunch.
GoatGuy
The gain ("big G") =
G = Vin / Vout … which thru a little algebraic rearrangement is
G = RK gm ( VG + Vbias ) / ( 1 + gm RK )
G ≈ VG • ( β / (1 + β ) ) where…
β = gm • RK
β in this case (1000 Ω RK and gm = 0.010) is 10. Therefore
G ≈ VG • ( 10 / (10 + 1) )
G ≈ VG × 0.909090…
The 'closeness' to 1.000 gets better (rather clearly from the beta equation) when either the RK or the gm increases (and especially when both happen).
So, having a low output impedance to drive reduces the β, causing the cathode follower to not-follow-as-closely. I guess there is no free lunch.
GoatGuy
Wow, GoatGuy. Thank you for typing all of that up. I can follow the algebra well enough as I took a lot of math/calc in school (but no EE courses).
AKA current output is the rate of change in current output with respect to change in voltage input (mutual conductance) multiplied by voltage input. I get that.
So voltage input is [Vbias - (Vk - Vg)] or bias less the difference between cathode and grid. This is the bit that I get hung up on when trying to understand how they actually "work" conceptually. I think this is because I keep wanting to call the voltage to the grid the input, when in reality it's more complicated than that.
GoatGuy said:
AKA current output is the rate of change in current output with respect to change in voltage input (mutual conductance) multiplied by voltage input. I get that.
So voltage input is [Vbias - (Vk - Vg)] or bias less the difference between cathode and grid. This is the bit that I get hung up on when trying to understand how they actually "work" conceptually. I think this is because I keep wanting to call the voltage to the grid the input, when in reality it's more complicated than that.
Also, if I restate the equation for cathode voltage:
Vk = gm * Rk * (Vbias + Vg) / (1 + gm * Rk)
So cathode voltage must increase along with the voltage at the grid, but it is reduced by a factor close to 1:
gm * Rk / (1 + gm * Rk)
Ok, just reading your next post now and I realize I'm saying the same or similar.
Vk = β * Vg + β * Vbias
Vk = gm * Rk * (Vbias + Vg) / (1 + gm * Rk)
So cathode voltage must increase along with the voltage at the grid, but it is reduced by a factor close to 1:
gm * Rk / (1 + gm * Rk)
Ok, just reading your next post now and I realize I'm saying the same or similar.
Vk = β * Vg + β * Vbias
Maybe this simple thought will help , if we simply apply ohm's low in the cathode follower we will see that a tube with high transconductance will have low Rp , because current flow through Rp of the tube to the load , low Rp = more current , therefore lower output impedance and then the voltage gain of the CF is closer to unity , because we have low voltage drop through Rp of the tube , so simple ! .
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No. For a CF the tube Rp is in parallel with the load, not in series with it as you assume.
The 'resistance' in series with the load is 1/gm.
The 'resistance' in series with the load is 1/gm.
No , in the following document , in figure 5-21 ( equivalent circuit of the CF ) it is clear that Rp is in series with the load http://www.radau5.ch/pdf_files/basics.pdf
Don't let these "know-it-all" types scare you away from asking questions OR saying things that are completely wrong.
This is an *amateur* DIY forum, and not everyone has spent decades studying or even being new has read enough text books.
Furthermore, not everyone is able to follow and/or understand what is written in most of the texts. The "english" writing is obtuse in most cases. There are a few exceptions - like The Art Of Electronics - that are written so that you can read the text and grasp the concepts without doing the math derivations and then thinking about what it means... indeed not everyone here has gone far enough in school to DO all the math.
So, the OP's attempt is perfectly reasonable.
Best to guide him and answer the details of his query rather than to tell him "go home kid and read some books, then come back and ask me a question". I used to really dislike professors who took this approach. Fwiw...
</flame>
_-_-
The effect of Rp is changed by the feedback, so although it may look like it simply appears in series with the load this is not what actually occurs.Dimitris AR said:
Two ways to look at it:
1. Rp appears in series with the transformed load (mu+1)Rk
2. 1/gm appears in series with Rk.
Look at the maths, not the diagram.
So did I, but that was because in that situation it often turned out that the professor had misunderstood the book and could not explain why his result/derivation was different from the book.bear said:
Most of the times I look at the maths , maths is to explain the equivalent circuit and not to " paraphrase " it , beside this , Rp is the transformed one .... Rp / 1+ mu and not Rk ! .
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