02-28-2019, 03:12 AM

I've got a Bogen CHB-35A with fixed bias 7868 power tubes.

I'd like to reconfigure the PA to cathode bias. First, I'll drop the plate voltage down to 450V. Then I'll need to deal with the output impedance. According to the 7868 datasheet, a cathode biased pair of 7868s at 450v have a load resistance of 10,000.

https://frank.pocnet.net/sheets/168/7/7868.pdf

Here is what I have calculated for the Bogen's output transformer.

The output transformer measures as follows:

With 7.16vac applied across the primary, I get 0.184vac at the 4ohm tap.

7.16 / 0.184 = 38.8 turns. 38.8 * 38.8 * 4 = 6,039 primary impedance.

If I connect an 8 ohm speaker to the 4 ohm tap, the primary impedance would then be 38.8 * 38.8 * 8 = 12,128

To get a 10,000 impedance with this transformer, I would need a speaker load of 6.64 ohms. (38.8 * 38.8 * 6.64 = 10k)

And if all that is both correct and safe, can I then place a 39R 10W resistor in parallel with the 8 ohm speaker to get the 6.64 ohm load and the calculated 10,000 impedance?

I'm also calculating about 17% of the output being dropped across the paralleled 39R, less than 5 Watts

Is this math correct and will this work without frying anything?

I'd like to reconfigure the PA to cathode bias. First, I'll drop the plate voltage down to 450V. Then I'll need to deal with the output impedance. According to the 7868 datasheet, a cathode biased pair of 7868s at 450v have a load resistance of 10,000.

https://frank.pocnet.net/sheets/168/7/7868.pdf

Here is what I have calculated for the Bogen's output transformer.

The output transformer measures as follows:

With 7.16vac applied across the primary, I get 0.184vac at the 4ohm tap.

7.16 / 0.184 = 38.8 turns. 38.8 * 38.8 * 4 = 6,039 primary impedance.

If I connect an 8 ohm speaker to the 4 ohm tap, the primary impedance would then be 38.8 * 38.8 * 8 = 12,128

To get a 10,000 impedance with this transformer, I would need a speaker load of 6.64 ohms. (38.8 * 38.8 * 6.64 = 10k)

And if all that is both correct and safe, can I then place a 39R 10W resistor in parallel with the 8 ohm speaker to get the 6.64 ohm load and the calculated 10,000 impedance?

I'm also calculating about 17% of the output being dropped across the paralleled 39R, less than 5 Watts

Is this math correct and will this work without frying anything?