O/P tran impedance

[alastair_in_oz]

Can anyone tell me how to work out the primary and secondary impedance of my O/P trany? It is an old OZ brand and I can't find any exact info on it. Its called A&R model 2783, made in Melbourne, Australia.
The amp was rated as 30w with 6L6 tubes. So I am assuming its should have a primary around 6k.
Any idea folks?

[tubeswell]

You need to determine the OT impedance (which is resistance to AC).

But 1st - with your Rmeter check that the secondary(s) aren't open and that the primary isn't open, and that the primary isn't shorting through to the secondary. If you get some resistance measurement in ohms. albeit however small, you can assume the windings aren't open. If there is no resistance reading between primary and secondary you can pretty much proceed with step 2, but you need to be careful

2nd - if that is all okay, hook up a 1VAC source to the secondary whilst measuring the primary with your Vmeter (on the 1000V setting). be careful not to electrocute yourself (the primary and secondary could still be dodgy in a way you couldn't determine from the resistance measuremnt in 1 above)

3rd - The reading you get off the primary VAC will be in proportion to the 1VAC on the secondary, because of the ratio of turns on the primary winding in relation to the number of turns on the secondary winding. e.g. if you get 30VAC on the primary to 1VAC on the secondary, the turns ratio is 30:1. The impedance ratio is the square of the turns ratio, so the impedance ratio would be (30x30):1 or 900:1. Therefore in this case if we hooked up an 8R speaker to the secondary, there would be a reflected load of 8R x 900 = 7200R (7k2) on the primary.

[R.G.]

Can anyone tell me how to work out the primary and secondary impedance of my O/P trany? It is an old OZ brand and I can't find any exact info on it. Its called A&R model 2783, made in Melbourne, Australia.
The amp was rated as 30w with 6L6 tubes. So I am assuming its should have a primary around 6k.
Any idea folks?

Output tranformers don't have impedances. They have impedance RATIOS.

Let's take an example. I have here in my hypothetical hand a hypothetical transformer intended for 6L6s, and its hypothetical box is marked "6.6K plate to plate; 8 and 16 ohm output." What that really means is "If you put an 8 ohm resistive load on the marked 8 ohm secondary, the transformer will present that to the plates as 6600 ohms. If you put a 16 ohm resistor on the 16 ohms secondary, the plates will still see 6600 ohms."

So far, so good. But what happens when I put a 5 ohm resistor on the "8 ohm" secondary?

The transformer as stated converts 8 ohms to 6600, a ratio of one to 825 on the 8 ohm secondary. If you put a 5 ohm resistor, it does not change the ratio. The plates will then see 5ohms times 825 or 4125 ohms plate to plate.

The impedance ratios are fixed by the the number of windings on the core in each part (i.e. primary, secondary, etc.) of the winding. Fortunately, we can find out the transformation ratio by measuring the voltage ratio. The math works out that the impedance ratio is the square of the voltage ratio. For the hypothetical 8:6600, the voltage ratio is 28.7228 to one. You must cause 27.7228 volts of signal to appear across the two plates to get one volt at the 8 ohm winding. And this presents a way to measure things: put in an AC voltage signal, then measure the other winding, and find the ratio. The impedance ratio is just that ratio squared.

In your case, you have one hint that makes it possible - that it put out about 30W. We can compute the AC voltage needed to give 30W into 4, 8 and 16 ohms (since you haven't said you know the speaker impedance) and pick the most likely voltage.

For instance, since P = V*V/R, we can compute V = SQRT(P*R). This gives
for 4 ohms, V = SQRT(30*4) = 10.95Vac (it's in RMS); for 8 ohms, V = 15.49Vac, and for 16, V = 21.91Vac. If we put, say, 12Vac into the secondary and measure the open primary plate to plate voltage, we could then compute the voltage ratio the transformer gives, and could then compute the plate to plate swing. (Hang on, we're getting there...)

Now, square the voltage ratios and multiply each by the ohms assumed, 4 8, and 16. It is highly likely that one of these will be in the range of 4K to 6.6K, which is the range that most 6L6 amps use. That's the one to use.

Did I confuse you?

[alastair_in_oz]

Confuzzed, fuzzed, fused???

Just kidding,
Thanks guys for your replies.
Perhaps I should elaborate just a little.
I picked up an old amp a couple of years ago and have been rebuilding and modding it, searching for that elusive Fender tone. I have a C12K speaker.
The amp was made in Sydney, Australia and called a Fi-Sonic. I was able to get some schematics (photo copies of hand drawn). The schematic showed the various versions 30-50 watts. I worked out mine was the 30w version because it uses cathode resistor biasing, other versions did not.

The power transformer has an un-tapped 170v secondary, and uses a voltage doubler & SS rectifier. There is no choke. I think this design probably adds to the dullness of the tone

The output transformer has 2 taps on the secondary with 1 tap connected to the speaker jack and the other in unconnected (And not labled). So I assumed that the connected tap was 8 ohms, well, best guess.

The mission: I want to upgrade the amp to a deluxe reverb style with 6V6GT tubes and a tube rectifier, with minimal outlay
I have a power transformer that might just work: 365 - 0 - 365 secondary.
I have some GZ34 valves.
So the reason I need to work out the primary impedance of the O/P trany is to see if it could be compatible with 6V6 tubes.

Any help would be appreciated

Alastair

[tubeswell]

What R/G said about using a 6k6 load with 2 x 6L6 in push pull, you can also run 2 x 6V6 in PP into a 6k6 load - or really into anything between about a 5k load and a 10k load. So odds are your OT will be fine for your 2 6V6s (assuming R.G.'s guesstimates were on the ball - which I bet they are).

[alastair_in_oz]

Progress so far

OK, I have done the calculation on the O/P trany.

For the unused tap:
Voltage Ratio= 17.86v\360v = 20.16:1 Sq = 406.4

406.4 x 4 = 1625
406.4 x 8 = 3251
406.4 x 16 = 6502

For the used tap:
Voltage Ratio= 17.47v\482v = 27.6:1 Sq = 761.76

761.76 x 4 = 3047
761.76 x 8 = 6094
761.76 x 8 = 12188

So it looks like the currently used tap is 8 ohms with a primary impedance of 6100 ohms, and the spare tap is 16 ohms.

Tubeswell;

So 6000k primary is OK for 6V6 ?
From the place I buy parts from here in OZ, the deluxe reverb O/P trany has a primary of 8500k.
Will the biasing take care of the current flow through the valves?
I am just worried that I would be driving the valves to hard.


Thanks very much.

[tubeswell]

6k will be fine for 2 x 6V6s (FWIW the Deluxe Reverb uses a 6k6 OT Zed). It will sound more compressed than an 8k. As I understand it - rule of thumb is that the steeper the load line, the more assymetrical the transfer characeristic is (which means you get a more assymetrical voltage swing on the plate compared to what you put into the grid, and therefore more harmonic distortion), so long as the load line doesn't go above the 'knee' of the Vg0 curve on the tube characteristics chart. If it worries you, build it with a small choke at the screen supply (instead of a resistor) to lift the screen voltages a bit, and then think about adding 3W-5W 470R-1k5 screen grid resistors if the screens come up more than 3 or 4V above the plate voltage. Screens should sit somewhere between 2V-50V below the plate voltage usually (depending on whether you use a choke or a resistor respectively).

[alastair_in_oz]

Thanks for that.
Should I start a new thread about the power trany question?
(Will 365 - 0 - 365 be too much for the 6v6 valves?)

[tubeswell]

If your PT has a 5V recto winding and you use a 5Y3GT you will get 400VDC, and the B+ goes up from there - 440V with a 5U4G, 470V with a GZ34. A 5Y3GT will make it sag the most. I understand JJ 6V6s can run at up to 500V, but I never tried it. I wouldn't want to try other 6V6s at that voltage. FWIW the Deluxe Reverb, which is fixed bias, runs at about 420V B+ (although it uses a 330-0-330VAC HT winding and a more powerful rectifier that supplies more current)

If it doesn't have a 5V secondary you will get over 510V with diodes. What is the heater (and recto winding if applicable) max current rating? If your heater is beefy enough you could run more powerful output tubes.

[alastair_in_oz]

Ok, the tran has the 5v for the GZ34, and its 360 - 0 - 360 (not 365), 2 x 3A CT 6.4v heaters, and 2 other 6.4v heaters with no rating. So a 5U4G would give closer to the 420v. Could I use a large resistor, 10w or so right off the B+ to drop to 420v? Or will that have some other effect?

[tubeswell]

If you want to run a 5U4G or 5U4GB you need at least a 3A rating on the 5V winding.

Using a resistor to drop the B+ Voltage uses up current, so it will induce more sag.

A better alternative, since your HT winding has a centre tap, is to insert one (or more in series) reverse-biased zener diodes between the HT Center Tap and the chassis ground, (with the banded/anode end of the diode(s) pointing to ground) to drop the HT voltage. that way you don't lose current. Make sure the rating of the diode(s) is sufficient to dissipate the heat/energy from the voltage drop.

[alastair_in_oz]

I had better stick with the GZ34 as I don't know what the 5v winding is rated at. Great idea with the diodes, I was doing the maths with the resistor idea while at work, I think it would have to be a 50w resistor! Not a good idea! Thanks for your help

Alastair

[alastair_in_oz]

The idea with the diode got me thinking about something I read about: back biasing. So this will kill 2 birds with 1 stone, I need to loose about 35v off the B+, and I will need -35v for the biasing. Great.
I did a little search and found this: http://www.aikenamps.com/BackBiasing.html
So it looks like its going to work then.
Thanks

Alastair

[black_labb]

If you want to run a 5U4G or 5U4GB you need at least a 3A rating on the 5V winding.

Using a resistor to drop the B+ Voltage uses up current, so it will induce more sag.

A better alternative, since your HT winding has a centre tap, is to insert one (or more in series) reverse-biased zener diodes between the HT Center Tap and the chassis ground, (with the banded/anode end of the diode(s) pointing to ground) to drop the HT voltage. that way you don't lose current. Make sure the rating of the diode(s) is sufficient to dissipate the heat/energy from the voltage drop.

using the resistor on the b+ doesnt draw current itself, it just knocks down the voltage. it does create sag though when the tubes draw more current (V=IR). a good way to get some sag and get a cleaner power supply, that is if you want some sag.

[tubeswell]

using the resistor on the b+ doesnt draw current itself, it just knocks down the voltage.

OK I never said it drew current, I said it uses up current. By the very fact of dropping the voltage, the resistor uses up current. A quantum of electrical energy is converted into heat in the resistor, ispo facto there is less current