Why does output impedance go down when cathode R is bypassed?

[octal]

The subject heading more or less says it all- in a triode gain stage, why is the output impedance lower when the cathode resistor is bypassed? I thought that an unbypassed resistor introduced local NFB, reduces gain etc. NFB usually LOWERS output impedance. What's happening here?

Enlighten me!

Nathan

[J M Fahey]

1) I'm not quite sure that happens
2) In case it did, you are stating half the truth:
a) *voltage* NFB is the one that lowers output impedance, since it tries to keep voltage constant, a characteristic of low internal impedance generators, *but*:
b) *current* feedback tries to keep current constant, typical from high internal impedance generators or voltage sources.
In that case, you are trying to keep *voltage* constant across the cathode resistor (not across the tube itself) whick tries to keep current through the resistor constant.
Since cathode resistor and tube are in series, if current is constant through one of them, it will be so through the other.

[Dave H]

This should confuse you.

For a non-bypassed Rk as current i (say) is provided to the load from the anode the current through Rk is also reduced by i so the voltage across Rk (bias voltage) is reduced which turns the valve on more changing the anode voltage more than would have been the case had the cathode resistor been bypassed (Rk=0). The anode voltage changes more for the same load current when Rk is not bypassed therefore the output impedance is higher.

Dave H.

[octal]

Thanks, guys. I think I was thinking about it strictly as a voltage thing which was confusing me- it's making much more sense now.

Nathan

[Robert M. Martinelli]

In a nutshell:

With the cathode fully bypassed, the output impedance can be calculated as the parallel of the valve's internal anode resistance and the anode load resistor (i.e. you can neglect the cathode resistor)---> Zout = ALr||Air, where ALr=Anode load resistor, Air=Anode internal resistance).

When the cathode is not bypassed, you have to take into account the cathode resistor's value, multiplied by the amplification factor plus one. This value has to be added to the valve's internal resistance, so the above equation becomes:

Zout=ALr||(Air+Cr*(u+1)), where ALr=Anode load resistor, Air=Anode internal resistance, Cr=Cathode resistor and u=amplification factor.

This is why the output impedance becomes larger

Hope this helps

Best regards

Bob

[octal]

I knew the formula from Merlin's site, etc, I just couldn't understand WHY! I think I sort of get it now.

[kg]

simple answer:

an unbypassed Rk DOES introduce negative feedback.

however, it is not negative VOLTAGE feedback, but rather negative CURRENT feedback.

negative current feedback has the opposite effect of its voltage-derived brother, in that it INCREASES output impedance. that's just what it does.

if you want a longer description, consider what happens under dynamic conditions...

a typical thought experiment for output z is to attempt to "wiggle" the output, and see what effect that has.

1) so we grab the plate of a triode, and we attempt to raise it in potential
2) this will increase the electric field "seen" at the space charge close to the cathode
3) this will increase the # of electrons which flow from k to a
4) this will tend to "make more negative" the plate, offsetting our outside influence to some degree... its resistance to wiggling is a direct measure of rp.

now, insert some impedance in the cathode circuit.. the same thing happens as above, EXCEPT:

3.1) the increased cathode current causes the cathode potential to RISE due to V=IR across the external cathode impedance
3.2) this increased cathode potential tends to INCREASE the negative bias wrt the grid
3.3) the more negative bias voltage tends to REDUCE the increase in ip that would have happened otherwise
4) the plate is still made more negative, resisting the wiggle, but LESS than it was without the external cathode impedance.

in other words, the rp has been increased.

hth
ken

[Bruce / Mission Amps]

Adding to the other comments, and just a thought:
Follow the actual path to ground from the tube's plate back through the OT and main filter caps... then follow the path to ground from the cathode and you'll see you could figuratively "fold" the tube in half at the junction of the plate and cathode with the other ends at ground... that means they are actually in parallel to ground. Now add the NFB loop which is also referenced to ground... etc
Two or more impedances in parallel would have a total of less then either one of the impedances.

[tubeswell]

I think of it as plate resistance being the tube's AC impedance. With a bypassed Rk you are removing the AC that is otherwise produced by the cathode resistor so the tube doesn't have to deal with this when it does the work of amplifying the signal. (The AC that otherwise accompanies an unbypassed cathode resistor is effectively acting as a 'brake', restraining the tube from getting to 'full' power). When you look at a thevinin circuit of a fully-bypassed stage there is no resistance in series with the tube at the cathode side, because the tube doesn't 'see' any AC there (because Rk is (fully) bypassed). JM2CW

[Steve Conner]

These are all good points. In a triode, the plate and grid are really no different. The plate's electric field also affects the flow of electrons. Mu is just a measure of how much more "griddy" the grid is than the plate.

I find it easiest to just remember that voltage feedback lowers the output impedance, current feedback raises it. This is a general principle with wider application than just a single tube stage. Pentode power amps use voltage feedback to lower the output impedance into the ballpark that makes guitar speakers sound right. Valvestates and the like use current feedback to raise it into the same ballpark.

[kg]

I find it easiest to just remember that voltage feedback lowers the output impedance, current feedback raises it. This is a general principle with wider application than just a single tube stage.

probably the best way to go, as it is the most universal.