Converting SF Bandmaster PI and Bias to Blackface?

[shocki]

I just got an early SF Bandmaster (68 I think) Reverb Head.

It looks like it has been blackfaced partly. Bias was changed to the Blackface specs but with two 100K Resistors and not two 220K. Also the Plate Load Resitors at the PI are 47K and not 100K and 82K.

I am not 100% sure if these things should be changed. The 47K gives me less Gain and more gain I think. I am not really sure about the two 100K resistors in the bias supply.. What effect would changing those to two 220K have?

Thanks alot.

[JoeM]

By the bias circuit being changed, you mean it's been re-wired to set bias, rather than the SF style bias balance?

There's some discussion regarding the 100k vs 220k, and in some amps the 100k is recommended to reduce grid blocking. Overall, I think the SF values (i.e. 47k PI plate resistors) give the amp a bit more clean headroom, but it may just be a matter of taste. If you have both values around, why not try them and see what you prefer?

[shocki]

By the bias circuit being changed, you mean it's been re-wired to set bias, rather than the SF style bias balance?

Yes correct. But with two 100K resistors instead of two 220K. I am not sure what effect this would have to the sound.

[JoeM]

Not much difference either way, if you like the way it sounds now just leave at as is. Some make a big deal over the BF values, but the single most important improvement is having good speakers.

[shocki]

I think I will leave it. It sounds great in my opinion. Using it with a Celestion Gold which I like alot. I don't really like how the Amp breaks up when pushed with pedals. It just doesn't really stay tight compared to my Badger 18 (which goes more into Marshall direction). But I think this is just normal with those Fender Amps.

[jpfamps]

Reducing the grid resistors to 100k from 220k was probably done to improve reliability.

Having lower grid resistors reduces the effect grid current from the power valves has on bias.

The 47k plate loads in the PI reduces gain a bit, but actually not be as much as you might think, but from what I remember do allow a bit more voltage swing, and give a lower output impedance, which will give less attenuation than the BF values when 100k grid resistors are used.

You can investigate the effects of different circuit values with this useful online calculator.

Balancing Long-Tailed-Pair Phase Inverter Gains

[JoeM]

I've thought the 82k, 100k PI plate resistors were a 'leftover' from the earlier Brown (and Tweed bassman) values when a 12AX7 was the PI tube. The 12AT7 has a lower Rp, maybe a better match for the 47K.

[shocki]

I've thought the 82k, 100k PI plate resistors were a 'leftover' from the earlier Brown (and Tweed bassman) values when a 12AX7 was the PI tube. The 12AT7 has a lower Rp, maybe a better match for the 47K.

I agree this makes sense. So the blackface conversion is technically not reall correct. Only if you would like to use AX7. Soundwise it still might be good.

[Alan0354]

From theory, the 82K/100K serves to unbalance the signal to the push pull tube so it give you more harmonics. The 47K is more balance. The open loop gain of the silver face power amp can be increased a little by changing the 68K from the negative grid biasing to 220K and higher. You can always try changing one of the 47K to 51K or 56K to simulate the imbalance of 82K/100K and see whether you like it better. I decided not to change the 47K and pickup the gain by other means. My reason is I want a lower impedance drive to the power amp. 47K able to drive a little more current when the grid of the power tubes are driven +ve. Because of the rp is relatively low in the PI tube, you really don't pick up that much more gain by increase the plate resistance of the PI.

Another thing, the overall gain of the power amp depends more on the negative feedback, not the gain of the PI stage.

[jpfamps]

I've thought the 82k, 100k PI plate resistors were a 'leftover' from the earlier Brown (and Tweed bassman) values when a 12AX7 was the PI tube. The 12AT7 has a lower Rp, maybe a better match for the 47K.

I agree.

I think the 82k and 100k values were calculated to give best balance in the Tweed circuits using a 12AX7 and a smaller tail resistor, and nobody bothered to redo the calculations.

In the BF circuit with a 12AT7 and larger tail resistor 91k would be the nearest standard value to balance the PI. Two 100k resistors actually give better balance than 82k/100k.

The 82k/100k pair crops up everywhere; not just in Fender amps.

Fender now use 91k and 100k in the HRD series.

In the BF circuit I can't here any difference between 100k/100k and 82k/100k.

[JoeM]

From theory, the 82K/100K serves to unbalance the signal to the push pull tube so it give you more harmonics. the plate resistance of the PI.

I don't think that's correct, there's some imbalance in the PI, hence the difference in the PI resistors to obtain better balance. And then it's in the feedback loop anyway.

[Alan0354]

I don't think that's correct, there's some imbalance in the PI, hence the difference in the PI resistors to obtain better balance. And then it's in the feedback loop anyway.

There should be no inherent imbalance on the long tail pair. Why do you think there is an imbalance in that?

[JoeM]

There should be no inherent imbalance on the long tail pair. Why do you think there is an imbalance in that?

Read this:

The Valve Wizard -Long Tail Pair

Quote from the article:

Balance: Because this circuit does not use a constant-current sink in the tail, the two outputs will not be perfectly balanced if both triodes have the same value anode resistors (although balance will be pretty good all the same). The difference in gain between the two outputs is given by:
A1/A2 = 1 + [(Ra+ra)/(Rk(mu+1))]
Where Rk is the total tail resistance Rb+Rt.
In this case we can see from the characteristics graph that ra = 30k, mu=45:
A1/A2 = 1 + [(82k+30k)/(33.82k(45+1))]
= 1.07
So the inverting output will be 7% higher than the non-inverting output. This could be corrected by making Ra1 7% smaller in value, but in practice it is not necessary as an unbalanced phase inverter is often quite benificial to guitar tone, due to the additional 2nd harmonic it introduces Nevertheless, this is why the 'traditional' version of the circuit also uses mis-matched anode resistors.

[jpfamps]

There should be no inherent imbalance on the long tail pair. Why do you think there is an imbalance in that?

The long tail pair is unbalanced.

There are two ways of looking at the LTP that I find helpful in understanding it's action.

Firstly, if there was constant current drawn through the LTP, then if the current through one side was reduced then current through the other side must increase by the same amount and thus it would be perfectly balanced.

The tail resistor act as current source (the larger the tail resistor the better the balance, but the lower the headroom), and is some way from ideal, so the LTP is unbalanced.

Substituting an active current source to the tail of the LTP would improve balance. I'm only aware of one guitar amp that uses this approach though.

The other way I look at the LTP is consider it as two gain stages, with the second driven from the cathode of the first. Although the cathode output impedance is low, it isn't zero. The output forms a voltage divider with the cathode resistor, you lose some signal, and the second stage needs to have more gain than the first to try to maintain balance (hence the larger plate load resistor on the second stage).

There are plenty of books that describe this in more detail including Morgan Jones' Valve Amplifiers, and Richard Kuehnel's book on Power Amplifiers. There is also a good calculator at Richard Kuehnel's website which allows to to see the effects of varying the component around the LTP.

I'm sure it's also covered in the RDH4, but I haven't read that part of it.