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Old 27th August 2010, 05:35 PM   #1
Brit01 is offline Brit01  United Kingdom
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Default Calculating output impedance

6080 in common cathode mode: I'm making calculations for output impedance.

I understand the plate resistor should be ~x10 times the plate resistance.
6080: 280 ohms

so I need around 2800 ohms resistor here.

output imp:

Zout: 280 x 2800/280 + 2800
= 254 ohms

If I parallel 2 tubes this is halved?

Also I wanted to use a CCS instead of a plate resistor but then these have a very high impedance no? This would significantly increase the output impedance??
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Old 27th August 2010, 05:38 PM   #2
rongon is offline rongon  United States
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Can I take a stab at this?

- two triodes in parallel = 1/2 output impedance

- CCS in plate, output resistance = very close to tube's internal plate resistance

How'd I do?

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Old 27th August 2010, 05:46 PM   #3
kevinkr is offline kevinkr  United States
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Quote:
Originally Posted by Brit01 View Post
<snip>

I understand the plate resistor should be ~x10 times the plate resistance.
6080: 280 ohms

so I need around 2800 ohms resistor here.

output imp:

Zout: 280 x 2800/280 + 2800
= 254 ohms

If I parallel 2 tubes this is halved?

Also I wanted to use a CCS instead of a plate resistor but then these have a very high impedance no? This would significantly increase the output impedance??
More usually 3 - 5X rp, math shown above is correct. Using a very good CCS as a load the source impedance approximates as rp/n where n is the number of tubes in parallel.
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Old 27th August 2010, 05:49 PM   #4
SY is offline SY  United States
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Quote:
Originally Posted by Brit01 View Post

Also I wanted to use a CCS instead of a plate resistor but then these have a very high impedance no? This would significantly increase the output impedance??
No, not really. That high Z is still in parallel with the plate resistance.

Note two things: plate resistance is NOT a constant- look at the graphs of plate resistance versus current on the datasheet to get an accurate value. Second, the effective plate resistance is equal to the plate resistance only if the cathode is at AC ground. If it's degenerated (e.g., unbypassed cathode resistor), the effective plate resistance will be the sum of the plate resistance and mu times the resistance in the cathode circuit.
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Old 27th August 2010, 05:51 PM   #5
Brit01 is offline Brit01  United Kingdom
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Quote:
More usually 3 - 5X rp, math shown above is correct. Using a very good CCS as a load the source impedance approximates as rp/n where n is the number of tubes in parallel.
sorry you mean source imp. equals rp with a good CCS? So really no increase in output impedance?

2 x 6080's in parallel Zout would be ~140 ohms more or less with a good CCS.

Cascode DN2540?
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Old 27th August 2010, 06:04 PM   #6
rongon is offline rongon  United States
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Quote:
Originally Posted by SY View Post
...the effective plate resistance is equal to the plate resistance only if the cathode is at AC ground. If it's degenerated (e.g., unbypassed cathode resistor), the effective plate resistance will be the sum of the plate resistance and mu times the resistance in the cathode circuit.

So for a 5687 with a 200 ohm cathode resistor left unbypassed,

rp (if bypassed) = 2500
mu = 16
Rk = 200

rp + mu = 2516
2516 * 200 = 503,200

503,200 ohms!?

That can't be right. Where did I go wrong? Do you have to figure the cathode resistance as the tube's rk in parallel with the cathode resistor?

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Old 27th August 2010, 06:17 PM   #7
Brit01 is offline Brit01  United Kingdom
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same here with the 6080. The Zout ends up as a very high number is cathode resistor is unbypassed.

Quote:
33
Output impedance:
The output impedance of this stage will become the source impedance for
whatever circuit it feeds, so it is necessary to know what it is in order to select a
suitable input impedance of the following stage. This is particularly important if the
stage is expected to drive a heavy load, such as a tone stack. Fig. 1.26 shows a
Thévenin equivalent circuit, which is how the circuit in fig. 1.18 appears to AC only.
Because the power supply is connected to ground via very large smoothing
capacitors, as far as AC signals are concerned, the power supply and ground are one
and the same: they appear to be shorted together. Additionally, rather than show a
valve with unknown properties, it has been replaced by a perfect signal generator in
series with a resistance, ra, which
represents the anode resistance
of the valve. Such diagrams are
mathematical simplifications of
the real circuit and are invaluable
for solving AC circuit problems
such as input and output
impedances. If the cathode
resistor is bypassed it is
effectively short circuited as far
as AC is concerned, and so it is
missing from circuit a. There it
can be seen that the total
resistance through which current
has travelled to reach the output,
is the parallel combination of Ra
and ra, so the output impedance is:
Zout (Rk bypassed) = Ra || ra………………………….…………………………..…...VIII
It was calculated earlier that ra is in this case about 70kΩ, so the output impedance
is:
Zout = 100k || 70k
100k x 70k
Zout
100k + 70k
=
= 41kΩ
If the cathode resistor is unbypassed, which it will be at very low frequencies, then
the circuit appears as in fig. 1.26b, and Rk is now in series with ra. Because Rk is
below the cathode its value appears multiplied by μ+1. Therefore the output
impedance when the cathode is unbypassed becomes:
Zout (Rk unbypassed) = Ra || (ra + Rk(μ+1))………………………………
If Rk is 1.5kΩ and μ is 100 then:
Zout (Rk unbypassed) = 100k || (70k + 1.5k x (100 + 1))
Zout (Rk unbypassed) = 100k || 251.5k
(Rk unbypassed)
100k x 251.5k
100k + 251.5k
Zout =
= 72kΩ
This clearly indicates that an unbypassed cathode resistor considerably increases the
output impedance. If the stage is only partially bypassed it will therefore have a
higher output impedance at low frequencies than at high frequencies, and this will
somewhat increase the degree of bass attenuation caused by the output coupling
capacitor

Last edited by Brit01; 27th August 2010 at 06:19 PM.
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Old 27th August 2010, 06:29 PM   #8
SY is offline SY  United States
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Quote:
Originally Posted by rongon View Post
So for a 5687 with a 200 ohm cathode resistor left unbypassed,

rp (if bypassed) = 2500
mu = 16
Rk = 200

rp + mu = 2516
2516 * 200 = 503,200

503,200 ohms!?

That can't be right. Where did I go wrong? Do you have to figure the cathode resistance as the tube's rk in parallel with the cathode resistor?

--
Actually, small correction- it's mu + 1, not mu.

(mu + 1)Rk = 17*200 = 3400 ohms.

rp = 2500 ohms

Total effective plate resistance is then 3400 + 2500 = 5900 ohms.
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Old 27th August 2010, 06:30 PM   #9
kevinkr is offline kevinkr  United States
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Quote:
Originally Posted by rongon View Post
So for a 5687 with a 200 ohm cathode resistor left unbypassed,

rp (if bypassed) = 2500
mu = 16
Rk = 200

rp + mu = 2516
2516 * 200 = 503,200

503,200 ohms!?

That can't be right. Where did I go wrong? Do you have to figure the cathode resistance as the tube's rk in parallel with the cathode resistor?

--
Nope, closer to 5.9K.. As a rough approximation take the unbypassed cathode resistance, multiply by mu+1 and add the result to rp. This is usually close enough.

Were it truly 500K then this could make one heck of a ccs which it really doesn't..
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Last edited by kevinkr; 27th August 2010 at 06:32 PM.
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Old 27th August 2010, 06:58 PM   #10
SY is offline SY  United States
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We seem to be in synchronous mode today, Kevin.
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