Power Transformers - I'm the Worst at Math

[industrialarts]

I have a Fender Performer 1000 on my bench with a bad power transformer. It seems the transformer is no longer available (PN# 027668) but it's a simple 120vac single primary CT secondary, 30-0-30vac. So I could spec a replacement but I do not know how to calculate the current / VA of the transformer.

The internal fuse is 4A so let's go with that as primary draw (which should be overkill) Since Vs/Vp=Ns/Np=Ip/s then Vs/Vp=Ip/Is and if we take Ip as 4 then does 60/120=4/x solving for x = 8A? I think I did the math right but that intuitively seems wrong.

I know there are smart people here so any help appreciated. I did google the part number and found nothing available after about 15-20 minutes of searching

 

[Khron]

Why not, at least at first, look for ready-made transformers? 60v center tapped, or 2x30v, or something close; 2x24v may be more common though.

 

[scott2000]

it's a simple 120vac single primary CT secondary, 30-0-30vac.

The Hammond 290BAX has this....
Says Fender chorus replacement.... Not sure what model that is but there is mention of 125watt Chorus?

I'm bad at math too....
VA/120=4 is where I was heading but that doesn't seem right to me either...

 

[JohnRoberts]

Most transformer specs I've seen are basically secondary winding voltages @ current.

JR

 

[industrialarts]

Looking closer at the schematic it rates the amp at 94watts into 4 ohms, so using ohm's law the maximum load current would be around 5A DC. If you look at this sheet from Hammond the full wave capacitive load current is I DC=.62*I AC of the secondary, so again my bad math says around 8 amps. This is of course maximum pull but the previous transformer blew up. I have seen this transformer in many Fender solid state amps and have had to replace several, but the last time I did it, the part was available.

 

[abbey road d enfer]

You're overcomplicating things. the amp is rated at 94W. The efficiency is about 50%, which computes at 190W at the primary. You can add 10% losses so a xfmr rated at about 220VA should be right. that would translate as ca. 2 A continuous current, which is coherent with a 4A fuse.

 

[industrialarts]

That makes sense to me except I don't understand the 50% efficiency. Where does that come from? Would I be correct in thinking that since it has 94W out and the turns ratio is approx 2:1 then reflected back to the primary is 190W (198W?) The rating on the back of the amp is 300VA

Thanks!

 

[trobbins]

The 50% efficiency relates to the amp circuit parts and configuration used to transfer a DC voltage rail to a signal output that then forces 50W in to your speaker load. The latest trick of class D amplifiers improves that 'efficiency', and going back to using valves degrades it.

There is also a form of 'efficiency' in how the AC mains power is transferred to DC power for the amp voltage rail(s), which some simply insert a 50% factor as well when ball-parking the PT VA rating.

 

[abbey road d enfer]

The rating on the back of the amp is 300VA

Why didn't you start with that? Would have cut short on any speculation.

 

[MisterCMRR]

If you're sure of the output power rating of 94 W, and we assume that it's measured at the customary few percent THD or onset of clipping, here's how the math would go, working backwards. To simplify the math, let's assume the output power is 100 W. Since P=E^2/R, E = sq rt ( RP) = sq rt (100 x 4) = 20 Vrms, which equals 20 x (2 x sq rt 2) = 20 x 2.828 = 56 V pk-pk or ±28 V pk-pk. If we assume the output stage clips at voltages 2 V less than the DC rails, then the DC supply would need to be ±30 VDC. This is consistent with a raw power supply using 30 VAC on either side of the power transformer center tap. Now, how much DC current is required for the output stage? Class AB or class B output stages have a theoretical maximum efficiency of 78.5%, but real-world circuits run around 65% (that other 35% is what makes the heatsinks hot!). So, to get 100 W of AC output power, the output stage needs 100 W/0.65 = 154 W of DC power. Since there's other small-signal circuitry, indicator lights, etc. that may consume 6 W, total DC load on the power supply would likely be 160 W at full output. With total rail voltage at 60 VDC, and since P = E x I, I = P/E = 160 W/60 V = 2.7 A. Allowing for a typical 90% efficiency of the rectifiers and typical DC resistances in the power transformer secondary, that would translate to roughly a 3 A rating for the power transformer secondary. Just for added information here, if we assume the power transformer has losses of 5% to 10% (what makes it get warm), then the actual total input power to it is only 5% or 10% higher than the 160 W DC output power. At 170 W input AC power, the line current at 120 VAC would only be about 1.4 A. The "rated" power of "300 VA" is related to a worst-case consumption, usually with some sort of fault conditions, for fire safety considerations. Likewise, the fuse rating is likely 4 A because UL considered it adequate to prevent a fire under worst-case fault conditions ... and Fender likely wanted the rating high to prevent turn-on inrush currents from popping the fuse. Usually "slo-blo" time-lag fuses are specified for that purpose, but Fender likely also knows that, in a pinch, people will install any type of 4 A fuse, including the faster blowing standard type ... so 4 A will save a lot of annoying phone calls. But neither the fuse or VA rating gives an accurate picture of actual circuit operation. I hope this helps!

 

[industrialarts]

Hey Mister CMRR,

This is exactly the kind of math I was looking for - the kind that scrambles my brains. Probably why I never finished my engineering degree Thanks so much!

It all makes sense to me, I kind of figured that the 4amp fuse was overkill and that 300VA was worst case. I should be able to source something at this point now that I have numbers to work with.

I have just copied and saved your reply for future reference. I'm sure I'll be doing this again sometime in the future.

Thanks to all who gave advice, I'll post a part number when I find something in case anyone else stumbles across this thread. I think I mentioned this part is used in several Fender amps from this era, I was surprised that a replacement isn't readily available.