Calculating output impedance

[Brit01]

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??

 

[rongon]

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]

<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.

 

[SY]

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.

 

[Brit01]

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]

...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]

same here with the 6080. The Zout ends up as a very high number is cathode resistor is unbypassed.

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

 

[SY]

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.

 

[kevinkr]

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..

 

[SY]

We seem to be in synchronous mode today, Kevin.

 

[Brit01]

So in short; lowest Zout is acheived by using bypassed cathode resistor.

This weekend hopefully going to try a simple common cathode 6080 headphone amp.

Just 1 to start with a CCS (cascode DN2540, max 500mA right? Have to check the ratings).

 

[Poindexter]

You guys have it exactly, from a modelling standpoint. Zout will be (the plate resistance + (μ+1 times the cathode resistance if unbypassed)) in parallel with Rload.

The problem I have found is that the characteristics of the device are almost always optimistic, and taken at optimistically unrealistic operating points. Look in the datasheets; they're pounding the devices right at all the rating limits. Even if you take your chars off the plate curves, which gives a depressingly more realistic estimate, they're still feeding you a happy meal by 10% or so. Usually.

I've found that the best method is just to set the output voltage at a nice even number like 1Vac, then cliplead a pot across the output. Turn the pot down from max until the voltage is one half reference, in this case 0.5Vac. Measure the resistance across those (disconnected) clipleads, that's ur Zout. Revise upward a percent or so to account for the existing ground-referencing resistance on the output if you like.

Be prepared for a little disappointment, if you've calculated with the above model, though.

Aloha,

Poinz

 

[Globulator]

So in short; lowest Zout is acheived by using bypassed cathode resistor.

This weekend hopefully going to try a simple common cathode 6080 headphone amp.

Just 1 to start with a CCS (cascode DN2540, max 500mA right? Have to check the ratings).

I found out the other night that when I bypassed Rk (=1k) on my 6N1P input tube with a 470u cap that the thing sounded better. This was a surprise to me, I did it as an experiment as I wanted to remove all feedback from that tube so I'm not sure if the improvement was due to lower feedback or lower output impedance, but it was better.

This input stage was a simple 47K Ra to a 350V rail.
I can't bypass the Rk on the lower triode in my SRPP however as I'm using that leg for feedback entry (until I work out a better way!).

 

[Wavebourn]

Нou can bypass it to lower resistance feedback voltage divider.

 

[Brit01]

The HA-2 headphone amp boasts a 100ohm Zout.

I can't work out how they came to this.

input tube is a 6SN7,plate resistance of 6700 and plate resistor of 15K+4.7K (split).

Zout=4999.6

CF 6AS7 with transductance 5500 microohms.
Zout of CF = 1/transductance
= 181 ohms.

How do they work out a Zout of 100 ohms as stated in the secs?

Basically what is the Zout of a common cathode followed by a CF?

 

[Brit01]

Or is it?

Zout= Ra/u+1
= 280/2+1
93 ohms.

Is the Zout determined by the CF?

 

[Brit01]

Something just spring to mind.

if I'm thinking of putting 6080 in parallel, each stage the Zout will be halved.

Could I take a tap from each stage and use these to drive different impedance headphones?

lets say the first stage is ~300, a jack output for a set of senns for example.

Then the next parallel tube would be less Zout with another tap out.
and so on down to ~35 ohms (4 in parallel).

Bit of a crazy idea but wonder is its feasible.

This is where the fun in DIY is.

 

[Wavebourn]

In order to get cleaner sound it is better if amp's output resistance in respect to load resistance is lower. You've already discovered it recently.