Fender Twin Reverb 135W Power Transformer specs

[nickb]

Anyone have a definitive specification for the max RMS current rating on the HV secondary for the 135W Twin Reverb?

Please don't go and just look up existing replacements as they are all over the shop ( 450mA to 1A) and I can do that myself anyway.


Ta!

[SoulFetish]

Anyone have a definitive specification for the max RMS current rating on the HV secondary for the 135W Twin Reverb?

Please don't go and just look up existing replacements as they are all over the shop ( 450mA to 1A) and I can do that myself anyway.


Ta!

What’s the Fender part #? Ill check and sse if I have one. If I do, I can take some measurements and run a simulation of the power supply

[nickb]

It's 013830. I can't think of anyway to easily measure or sim the required max current parameter. You'd have to load it up and bury a thermocouple inside then guess at the the max temp. It's something you either have the spec or you don't.

[SoulFetish]

It's 013830. I can't think of anyway to easily measure or sim the required max current parameter. You'd have to load it up and bury a thermocouple inside then guess at the the max temp. It's something you either have or spec or you don't.

Okak, I misunderstood the end game. Yeah, even then you’d have to know the temperature rating of the insulation and derate it somewhat for some breakdown over time, correct? Maybe try and contact Schumacher?

[nickb]

Exactly.

Well here's my best guess.

The 100W version uses a center-tapped winding rated at 450mA. That give 630mA RMS after rectification ( 630 = 450 * 1.4 since it's RMS). So we could estimate the 135W version as 135/100 * 630 = 708mA.

The Hammond replacement is rated at 1A.

Any other ideas gratefully received.

[Helmholtz]

Exactly.

Well here's my best guess.

The 100W version uses a center-tapped winding rated at 450mA. That give 630mA RMS after rectification ( 630 = 450 * 1.4 since it's RMS). So we could estimate the 135W version as 135/100 * 630 = 708mA.

The Hammond replacement is rated at 1A.

Any other ideas gratefully received.

I am not sure and I am tired. But I don't understand the 630mA after rectification. I would think that the 450 mA are already RMS. (The DC current must be lower than the AC RMS current as the DC voltage is higher than the RMS voltage and DC wattage cannot be higher than AC wattage.)

[nickb]

Does this help you HH?

The 450mA from two out of phase windings will give 630ma after the diodes and before the filter cap. Why? because we're adding RMS currents so the total is ( I1^2 + I2^2)^0.5 = (450^2+450^2)^0.5 = 630mA.

[Helmholtz]

Does this help you HH?

The 450mA from two out of phase windings will give 630ma after the diodes and before the filter cap. Why? because we're adding RMS currents so the total is ( I1^2 + I2^2)^0.5 = (450^2+450^2)^0.5 = 630mA.

Need to think about it tomorrow.
It would be much easier, if I knew the AC voltages of the two transformers.

[nickb]

Need to think about it tomorrow.
It would be much easier, if I knew the AC voltages of the two transformers.

Ooo! Good point! I had completely forgotten that the 100W version is 340V and the 135W is 363V.

So for equal power 708 *340/363 we get to 663mA as the best guess.

[SoulFetish]

I would only compare wattages. In a bridge rectifier circuit the RMS current will be twice as high compared to a two-way center tapped circuit (as here the two legs only carry current half the time, right? Not quite sure about the RMS values.) for the same DC voltage.

I don't know if this is relevant to your point, but my understanding was that each phase carries 1/sqrt2 the total RMS ripple current during each half cycle in a typical CT rectifier.

[Helmholtz]

Does this help you HH?

The 450mA from two out of phase windings will give 630ma after the diodes and before the filter cap. Why? because we're adding RMS currents so the total is ( I1^2 + I2^2)^0.5 = (450^2+450^2)^0.5 = 630mA.

Thanks, I could verify this relation for resistive loads.
But according to the link below, things seem to change for capacitor input, requiring a higher RMS current for the bridge rectifier for a given DC current.

https://www.tube-town.net/cms/userfi...mond-5c007.pdf

[Helmholtz]

Thanks, I could verify this relation for resistive loads.
But according to the link below, things seem to change for capacitor input, requiring a higher RMS current for the bridge rectifier for a given DC current.

https://www.tube-town.net/cms/userfi...mond-5c007.pdf

After consulting some serious standard literature, I came to the conclusion that the Hammomd info must be wrong regarding capacitor load.

Schröder ("Elektrische Nachrichtentechnik") shows that (also) for capacitor load the sec. RMS current of the bridge circuit is higher by a factor of 1.41 compared to the CT circuit for the same DC current.
He gives the following relations:
CT: Irms = (1.36...1.11)*Idc
Bridge: Irms = (1.93...1.57)*Idc
The variation takes care of different conduction angles between 60° and 90°.

[nickb]

I was going to say that pdf has some obvious errors but I see you found out for yourself.

We probably don't care much about the relationship between the average output current and the RMS current after the rectifier since the conduction angle is likely to be about the same in either case.

So, unless someone can produce some actual specs I'm going with 663mA as the best guess.

[Helmholtz]

I was going to say that pdf has some obvious errors but I see you found out for yourself.

We probably don't care much about the relationship between the average output current and the RMS current after the rectifier since the conduction angle is likely to be about the same in either case.

So, unless someone can produce some actual specs I'm going with 663mA as the best guess.

Of course this means assuming that wire type (cross section and insulation class) as well as hot spot temperature are the same.
But I don't have a better proposal