Simple question I hope

[g1]

It is puzzling why Peavey didn't go with a 20mA part. Verify that R51 is really 1.2K as the schematic indicates. Maybe it is a typo or they changed the value in production.

Agree. Seems odd that Peavey would use a non-generic part. The description in the BOM seems very standard. I can't see that you would need to order a peavey specific LED or it burns up.

[Oldmactech]

Waiting on LED to arrive from Mouser. Going to replace R 51 but the resistors I have have a much smaller footprint? Is this a problem?

Would it make sense to keep the R 51 and add a second resistor connected to it to increase the resistance? Does that even work?

thanks

[Helmholtz]

Would it make sense to keep the R 51 and add a second resistor connected to it to increase the resistance? Does that even work?

Yes, you can wire a 1k/0.5W resistor in series with R51.

[Pixel]

Waiting on LED to arrive from Mouser. Going to replace R 51 but the resistors I have have a much smaller footprint? Is this a problem?

Would it make sense to keep the R 51 and add a second resistor connected to it to increase the resistance? Does that even work?

thanks

The resistor you show in the picture has a smaller footprint because it is lower wattage, too low for this application. Whatever current goes through the resistor is going to heat it up, if the wattage rating is too low the resistor won't be able to shed the heat fast enough and will burn up. Typically when you want a resistor to shed heat you mount it a little bit off the board so air can flow around it better.

I hadn't really thought about it until now, but there is power being wasted here. (Yeah I know it is a tube amp...but still...) Why didn't they grab power a little further along where it is -15V?

[g1]

If you lift one end of the existing resistor and add one in series, you can get away with a lower power (smaller footprint) resistor. Like Helmholtz mentioned above, adding a 1K in series would pretty much split the power between the two. And there will be less current through them which further helps the issue.

[Helmholtz]

I hadn't really thought about it until now, but there is power being wasted here. (Yeah I know it is a tube amp...but still...) Why didn't they grab power a little further along where it is -15V?

No matter how/where you connect the LED, the 36V supply will always have to deliver the LED current.
So with an LED current of 20mA, the power supply always has to supply 36V x 20mA = 0.72W for LED operation, no matter where this power is dissipated.

Apart from that, the -15V supply is high impedance and connecting the LED here would completely load down the voltage.

[Pixel]

No matter how/where you connect the LED, the 36V supply will always have to deliver the LED current.
So with an LED current of 20mA, the power supply always has to supply 36V x 20mA = 0.72W for LED operation, no matter where this power is dissipated.

Interesting. No such thing as a free lunch.

[Oldmactech]

So, I will make an attempt to add a second resistor in series. If I pull one leg I need to solder the leg I loosened to one leg of the new resistor then the other end goes in the PCB? And on to the LED?

Fire hazard?

[Tom Phillips]

Another common approach is to unsolder one end of the existing resistor from the PCB, put the new resistor in that hole then stand up both resistors in an upside-down "V" shape. Then solder the resistors together at the base of the "V".

[Oldmactech]

Well, put the new LED in, turned it on, nothing. I decided to simply install the LED as a replacement to the one I pulled out without making any changes to the resistance. . (Just wanted to see it light up I guess)

Checked the resistor R51 shows -47v on both sides. Schematic shows -36v as normal. What the heck?

Thanks.

[g1]

Do you have all the tubes in? Are they lit up? I think the -36V runs the heaters, so it will be high if the heaters aren't running.
With open LED (or reversed or bad connection) you will read full voltage at both sides of that resistor.

[Tom Phillips]

...Checked the resistor and showing -47v on both sides. Schematic shows -36v as normal. What the heck?...

If you are measuring -47V on both sides of the resistor with respect to system ground, that indicates no voltage drop across the resistor which means that there is no current flowing through the resistor. I would therefore figure that the LED is installed backwards, or is faulty or there is a bad connection between the LED and the resistor.

[Oldmactech]

Thanks. Yes, no tubes installed.

I’ll pull it apart (again) tomorrow, re- check continuity (again) and possibly reverse the polarity of the LED. Good grief but thanks for the tips.

[Pixel]

Forward voltage on the LED would be about 3.2V

With the tubes in and heaters in the circuit, R51 would be dropping about 33V and letting 27.5mA through. Too much for a bog standard 20mA LED so we wanted to increase R51.

If we are talking 47V without the heaters, we have R51 dropping about 44V and letting 37mA through. Our Cree LED has an absolute max current rating of 25mA.

Anytime one of the heaters on V1, V2 or V3 goes we are going to have this happen. Maybe this is why Peavey wanted a special LED in here.

Have I got this right?

[Helmholtz]

Reverse voltage limit is only 5V .