Calculating output impedance
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 27th August 2010, 05:35 PM #1 Brit01   diyAudio Member   Join Date: Jul 2008 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??
 27th August 2010, 05:38 PM #2 rongon   diyAudio Member     Join Date: Jul 2009 Location: Across the river from Rip's big old tree... 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? --
kevinkr
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Join Date: Sep 2004
Location: Boston, Massachusetts
Quote:
 Originally Posted by Brit01 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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SY
On Hiatus

Join Date: Oct 2002
Location: Chicagoland
Quote:
 Originally Posted by Brit01 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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Brit01
diyAudio Member

Join Date: Jul 2008
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?

rongon
diyAudio Member

Join Date: Jul 2009
Location: Across the river from Rip's big old tree...
Quote:
 Originally Posted by SY ...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?

--

Brit01
diyAudio Member

Join Date: Jul 2008
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.

SY
On Hiatus

Join Date: Oct 2002
Location: Chicagoland
Quote:
 Originally Posted by rongon 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.
__________________
"You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."

kevinkr
diyAudio Moderator

Join Date: Sep 2004
Location: Boston, Massachusetts
Quote:
 Originally Posted by rongon 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.

 27th August 2010, 06:58 PM #10 SY   On Hiatus     Join Date: Oct 2002 Location: Chicagoland We seem to be in synchronous mode today, Kevin. __________________ "You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."

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