Metropoulos Forum

Tell what impedance load an amp head needs to have, when it isn't marked?

Mars Hall wrote:Tell what impedance load an amp head needs to have, when it isn't marked?

For Class A Single Ended-

B+^2 / Max Plate Dissipation Rating = Zplate

For Class AB Push Pull -

B+^2 / (Max Plate Dissipation Rating x Number of Tubes) = Zplate-plate

"^2" means "squared", or multiplied by itself.

So if you have a Class AB Push Pull amp with 2 x EL34s, max dissipation rating for an EL34 is 25 watts. Let's say you're running a B+ of 450V -

450V^2 / (25 x 2) = 4,050 ohms...4K Zp-p would be close enough.

If it were a 100 watt at the same voltage -

450V^2/ (25 x 4) = 2,025...2K would be close enough, which is 1/2 the 50 watt required Zp-p, which allows the OT primary to draw double the current at the same voltage, which means double the power and the reason why two more tubes are needed (creates a 2nd path for that extra current to flow since one tube can only handle so much).

Of course, the power transformer would have to be able to source double the current, but you get the idea.

flemingmras wrote:

Mars Hall wrote:Tell what impedance load an amp head needs to have, when it isn't marked?

For Class A Single Ended-

B+^2 / Max Plate Dissipation Rating = Zplate

For Class AB Push Pull -

B+^2 / (Max Plate Dissipation Rating x Number of Tubes) = Zplate-plate

"^2" means "squared", or multiplied by itself.

So if you have a Class AB Push Pull amp with 2 x EL34s, max dissipation rating for an EL34 is 25 watts. Let's say you're running a B+ of 450V -

450V^2 / (25 x 2) = 4,050 ohms...4K Zp-p would be close enough.

If it were a 100 watt at the same voltage -

450V^2/ (25 x 4) = 2,025...2K would be close enough, which is 1/2 the 50 watt required Zp-p, which allows the OT primary to draw double the current at the same voltage, which means double the power and the reason why two more tubes are needed (creates a 2nd path for that extra current to flow since one tube can only handle so much).

Of course, the power transformer would have to be able to source double the current, but you get the idea.

LOL! I think the OP was asking how to figure out what load should be connected, if it is not silkscreened on the chassis of an existing amp. You gave him an answer, but in a round about way. The OP would need to disconnect the OT and figure out reflected impedance on the primary, by means of the turns ratio. with a small AC input signal on the seconday, a 1Vpk-pk signal @ 800hz, should be fine, you should measuer a much larger one on the primary. He would then use your method above to find the most Ideal load for the valves, and voltage used in the amp.

joey wrote:

flemingmras wrote:

Mars Hall wrote:Tell what impedance load an amp head needs to have, when it isn't marked?

For Class A Single Ended-

B+^2 / Max Plate Dissipation Rating = Zplate

For Class AB Push Pull -

B+^2 / (Max Plate Dissipation Rating x Number of Tubes) = Zplate-plate

"^2" means "squared", or multiplied by itself.

So if you have a Class AB Push Pull amp with 2 x EL34s, max dissipation rating for an EL34 is 25 watts. Let's say you're running a B+ of 450V -

450V^2 / (25 x 2) = 4,050 ohms...4K Zp-p would be close enough.

If it were a 100 watt at the same voltage -

450V^2/ (25 x 4) = 2,025...2K would be close enough, which is 1/2 the 50 watt required Zp-p, which allows the OT primary to draw double the current at the same voltage, which means double the power and the reason why two more tubes are needed (creates a 2nd path for that extra current to flow since one tube can only handle so much).

Of course, the power transformer would have to be able to source double the current, but you get the idea.

LOL! I think the OP was asking how to figure out what load should be connected, if it is not silkscreened on the chassis of an existing amp. You gave him an answer, but in a round about way. The OP would need to disconnect the OT and figure out reflected impedance on the primary, by means of the turns ratio. with a small AC input signal on the seconday, a 1Vpk-pk signal @ 800hz, should be fine, you should measuer a much larger one on the primary. He would then use your method above to find the most Ideal load for the valves, and voltage used in the amp.

Oops...lol.

What's funny is that I searched everywhere for those damn equations...didn't find them, and ended up having to figure them out for myself via drawing up load lines and all that engineering shit you've gotten me into.

flemingmras wrote:
What's funny is that I searched everywhere for those damn equations...didn't find them, and ended up having to figure them out for myself via drawing up load lines and all that engineering shit you've gotten me into.

In either case I don't think he liked the answers too much (mine just doubled the work load), as it is not as simple as just setting the meter to the ohms setting. and putting the probes between two points

joey wrote:

flemingmras wrote:
What's funny is that I searched everywhere for those damn equations...didn't find them, and ended up having to figure them out for myself via drawing up load lines and all that engineering shit you've gotten me into.

In either case I don't think he liked the answers too much (mine just doubled the work load), as it is not as simple as just setting the meter to the ohms setting. and putting the probes between two points

Not everything on an amp can be done "paint by numbers" style.

Thanks guys for the responses. I'll be figuring it out hopefully this weekend. What can I use as a signal generator?

Mars Hall wrote:Thanks guys for the responses. I'll be figuring it out hopefully this weekend. What can I use as a signal generator?

Disconnect the OT wires from the amp. Plug the primary of the OT into a wall socket (i.e. the wires that came off of pin 3 of the output tubes) and measure the voltage at the secondary.

Then perform these equations -

120V / Measured secondary voltage = Turns Ratio

Turns Ratio x Turns Ratio = Impedance Ratio

Then multiply the impedance ratio by 4, 8 and 16 and this will tell you what plate load it will reflect to the output tubes with those 3 impedances.

Example, lets say you plug it in and you measure 12 volts at the secondary -

120 / 12 = Turns Ratio = 10

10 x 10 = Impedance Ratio = 100

On a 16 ohm load this will reflect a load of -

100 x 16 = 1600 ohms, or 1.6K

And yes the OT primary can more than handle wall voltage since it sees lots more voltage under normal operation than what comes out of the wall. It won't need a load hooked to it either...it just becomes a simple power transformer at this point and since there are no tubes hooked up to it that can short due to an open load condition, a load on the output isn't needed. With no load hooked up to it there will be very little primary current to speak of.

Ok know it's either an 8 or 16 ohm, Using Vac setting on my meter, I got 3.6 volts on the secondary. 120v / 3.6 = 33.33 turns ratio. 33.33 squared = 1110.89 x 8 = 8887.12 ohms or 8.8k. With a 16 ohm load 1110.89 x 16 = 17774.24 or 17.7k. Which is more ideal for 6V6 power tubes?

Mars Hall wrote:Ok know it's either an 8 or 16 ohm, Using Vac setting on my meter, I got 3.6 volts on the secondary. 120v / 3.6 = 33.33 turns ratio. 33.33 squared = 1110.89 x 8 = 8887.12 ohms or 8.8k. With a 16 ohm load 1110.89 x 16 = 17774.24 or 17.7k. Which is more ideal for 6V6 power tubes?

In consulting with the datasheets, in push-pull mode they call for a 10K load plate-plate at 250V.

What kind of amp is this in? What's the plate voltage? Single ended? Push-pull? Class A? Class AB?

More than likely it's set up for an 8 ohm load, but would like to know the other info I'm requesting before I say for certain. DEFINITELY not a 16 ohm, as a plate load of 17.7K would be WAY high for even 6V6s.

I was told it was patterned after a Deluxe. BF or tweed, I'm not certain. The original speakers were 2 16ohm Celestians, so I figured they could only be wired to 8 or 16 ohms. The owner removed the speakers and doesn't remember how they were wired. That is why I took it upon myself to help him. As I have guessed everything is pointing to 8 ohms. I can check the plate voltage if you would like. The PT is very small.

Mars Hall wrote:I was told it was patterned after a Deluxe. BF or tweed, I'm not certain. The original speakers were 2 16ohm Celestians, so I figured they could only be wired to 8 or 16 ohms. The owner removed the speakers and doesn't remember how they were wired. That is why I took it upon myself to help him. As I have guessed everything is pointing to 8 ohms. I can check the plate voltage if you would like. The PT is very small.

With two 16 ohm speakers, the only usable impedance you can wire them to would be 8 ohms by paralleling them. If you series wired them, it would be a 32 ohm load.

Knowing this bit of info I'm thinking it's safe to say that the OT you have is set up for an 8 ohm load with 6V6s.

flemingmras wrote:With two 16 ohm speakers, the only usable impedance you can wire them to would be 8 ohms by paralleling them. If you series wired them, it would be a 32 ohm load.

Knowing this bit of info I'm thinking it's safe to say that the OT you have is set up for an 8 ohm load with 6V6s.

The plate voltage is 430v does that sound about right for a Deluxe type circuit?

Mars Hall wrote:

flemingmras wrote:With two 16 ohm speakers, the only usable impedance you can wire them to would be 8 ohms by paralleling them. If you series wired them, it would be a 32 ohm load.

Knowing this bit of info I'm thinking it's safe to say that the OT you have is set up for an 8 ohm load with 6V6s.

The plate voltage is 430v does that sound about right for a Deluxe type circuit?

Yep that sounds about right. With an 8.8K load at that voltage that should get you roughly around 30 watts.

flemingmras wrote:Yep that sounds about right. With an 8.8K load at that voltage that should get you roughly around 30 watts.

Thanks for all your help.