Determining necessary bias voltage from tube data sheets

[Michael Allen]

I'm trying to put this amp together with what I have, and the OT i want to use is a 10W, with 8k, 4k, 2k, 1k primaries and 8 and 4 ohm secondaries. The only speaker cab I have right now is a single 8 ohm Private Jack. So the highest impedance I could get would be 8 ohm cab into 4 ohm secondary to give 16k reflected load. That's short of the suggested 22k. But 22k is well short of the mathematical result of 36k.

Just what determines the acceptable reflected primary load? If I use this 16k load, then with 130V on the plate, the current would be about 8mA. So I would need to bias colder to, say, -2.5V?

[Old Tele man]

Just what determines the acceptable reflected primary load?

...one hyphenated-word answer: turns-ratio (TR), ie: TR = SQRT[Np/Ns]

...remember, a transformer does NOT itself set ANY resistance / impedance value, it merely TRANSFORMS up or down (proportional to TR) the value that's presented to the output or secondary winding, ie: speaker(s) in our case. Connect a 1.6K ohm load to the primary of a 20:1 Turns-Ratio OT and the secondary will 'see' a 4 ohm load. Likewise, connecting a 4-ohm load to the secondary will "reflect" a 1.6K ohm load back to the primary winding! The 4X thing is a function of Class-AB push-pull (PP) operation, hence:

Zpp ~ rp/6-to-rp/8
RL = Zo*(TR)^2 ...Zo = speaker impedance
Zpp = 4*RL ...due to Class-AB push-pull operation.
Zpp = 4*Zo*(TR)^2

...we want the "reflected" resistance / impedance that's presented at the PRIMARY windings (where the tubes are) to "match" (roughly) the 1/6-1/8th ratio of the tubes' working plate resistance value (rp).

...so, with some "creative" selection of the available OT output taps and the total speaker impedance value(s), you might be able to come up with a combination that gets you pretty close to the desired OT primary value.

Zi = Zo*(TR)^2 or, RL = Z.spkr*(TR)^2 and thus: Zpp = 4*Z.spkr*(TR)^2

...for example, a 4-ohm speaker load connected to an OT with TR=20:1, will reflect an effective RL = 1.6K and a Zpp = 5.6K, and would sometimes be designed as: 5.6Kpp:4.

[Michael Allen]

Awesome, thanks so much for all the help guys.

So if I plug my 8ohm speaker into the 8ohm secondary and use the 8k primary, i'll get Zpp = 4*8*1000 = 32K. Should be perfectly suitable.

Thanks so much again!

[raiken]

Awesome, thanks so much for all the help guys.

So if I plug my 8ohm speaker into the 8ohm secondary and use the 8k primary, i'll get Zpp = 4*8*1000 = 32K. Should be perfectly suitable.

Thanks so much again!

No, if you plug your 8 ohm speaker into the 8 ohm secondary and use the 8k primary, you'll get 8k.

If you plug your 8 ohm speaker into the 4 ohm secondary and use the 8k primary, you'll get 16k.

RA

[Michael Allen]

Right, but then 8k * 4 since it's AB push pull for a Zpp of 32k?

[raiken]

Right, but then 8k * 4 since it's AB push pull for a Zpp of 32k?

No, push-pull output transformers are spec'd for plate-to-plate impedance. You'll get 16K plate-to-plate if you connect an 8 ohm speaker to the 4 ohm tap and use the 8K primary.

If you connect a 16 ohm speaker to the 4 ohm tap and use the 8K primary you'll get 32K.

Note, however, that the low frequency cutoff point will change because the transformer was designed to run into the matched impedances set on the secondary. The formula for the lower -3dB point is f = Z/(2*pi*L), where L is the primary inductance. Since the primary inductance is constant and fixed by the number of turns and the core characteristics, the lower -3dB point will go up by a factor of 4 when you change the reflected impedance from 8k to 32k, so your low end will decrease.

For more info:

http://www.aikenamps.com/OutputTransformers.html

RA