80s ODS Half Power 50/100 Switch Diagram

[ic-racer]

I searched and could not find this, so this is my guess on how it is wired. I'm just a guitarist, so the EE's will have to help me out on this one to know if it is correct.

Here is a repost of the picture of the back of the switch (#124) and my guess at how it might work.

Why do I care? Just trying to fill some holes in the back of the chassis for 'correct' cosmetics (and a working LED also!) So, as long as I put the switch and LED in there, it seems simple just to hook it up so it works.

What do you all think?

[Tonegeek]

I searched and could not find this, so this is my guess on how it is wired. I'm just a guitarist, so the EE's will have to help me out on this one to know if it is correct.

Here is a repost of the picture of the back of the switch (#124) and my guess at how it might work.

Why do I care? Just trying to fill some holes in the back of the chassis for 'correct' cosmetics (and a working LED also!) So, as long as I put the switch and LED in there, it seems simple just to hook it up so it works.

What do you all think?

Assuming you have a positive voltage on the LED, I would think the top 2 terminals on each side of the switch in your drawing should be reversed. That way the LED is grounded when the switch is in half power, turning it on. My self-taught opinion....

[ic-racer]

Assuming you have a positive voltage on the LED, I would think the top 2 terminals on each side of the switch in your drawing should be reversed. That way the LED is grounded when the switch is in half power, turning it on. My self-taught opinion....

Sorry the diagram is not real clear but a tiny "BIAS" CURRENT (as some call it) from the unused tubes is driving the LED (no other power source). So the diagram is supposed to show the LED already grounded on one end.

I wasn't shure about the polarity of the LED but, the way I think it should work is like this. In the 50 w mode the resistors and LED will allow a tiny bias current of electrons flowing from ground that will go through the LED then through the resistors then the switch and through the unused tubes; to reach the positive cathodes. The 'electrical convention' uses 'flowing positive charge,' which is in the opposite direction. This would mean the diode arrow should point in the opposite direction, or toward the ground.

Another subtle detail in the pictures I noticed is that (unlike the schematic I drew) the RIGHT tube in each pair is hooked to the switch (not the middle two). Maybe MR D does this stuff randomly or maybe there is a reason...

A little more searching and I finally found this post by Gil and I think my diagram shows what he is explaining for the Dumble method:

If I recall correctly, Dumble did indeed put the LED in series with the tubes' cathodes, with a 10K resistor in the path too. Short the resistor and LED and you get your 100 W mode, but flip the switch and two tubes barely pass a couple of mA's, but that's enough to juice the LED.

In my amps, I chose to lift he ground of two of the tubes on one side of the switch, while tapping off the 6.3 VAC supply with a diode + 390 ohm resistor + LED. In my case, LED on = 100W mode.

Cheers,

Gil

[heisthl]

When you "lift" 2 tubes don't forget your speaker impedance changes to a smaller number . i.e. run an 8 ohm speaker on the 4 ohm tap.
Ideally your switch could incorporate the impedance change.
You barely notice the difference in power so a better solution might be to: a: lower the PI voltage at the same time you lift the tubes or
b: run 2 tubes w/cathode bias for the lower power setting

[ic-racer]

When you "lift" 2 tubes don't forget your speaker impedance changes to a smaller number . i.e. run an 8 ohm speaker on the 4 ohm tap.
Ideally your switch could incorporate the impedance change.
You barely notice the difference in power so a better solution might be to: a: lower the PI voltage at the same time you lift the tubes or
b: run 2 tubes w/cathode bias for the lower power setting

My OT is just going to have the 4 ohm tap (fender 022889 style). Funny that the back plate graphics on all the pictures of 50/100 amps I have seen do not mention this impedance change 'feature.' ....I guess it's explained in the Dumble owner's manual...

[glasman]

Your schematic looks correct. The two resistors in the circuit are 10K. When the switch is placed in the half power position, the two tubes still conduct a little current (about 2ma) and the LED will have a soft warm glow .

[rfgordon]

This is a cool design!

As it happens, I'm working on an amp for a guy that will run a pair of 6V6 and a pair of 6L6, all cathode bias. So it looks like I could adapt this design to my amp by putting a switch w/ LED on each pair's cathode line. So either or both pairs could be played by switching the cathode from the 250 ohm resistor to a 10k/LED line to ground, right?

[ic-racer]

Your schematic looks correct. The two resistors in the circuit are 10K. When the switch is placed in the half power position, the two tubes still conduct a little current (about 2ma) and the LED will have a soft warm glow .

Thanks for the info on the resistors; I was imagining frying a lot of LEDS by trial and error to find a correct resistance

[tele_player]

Seems you could also just disconnect input signal from the grids of a pair of tubes, leaving them idling. Then the impedance doesn't change.

[ic-racer]

Seems you could also just disconnect input signal from the grids of a pair of tubes, leaving them idling. Then the impedance doesn't change.

I thought about that but I think the impedance will still mismatch in that situation. Like if you were doing mechanical work with 4 human arms on a lever. If two of the arms were still holding on the lever but not 'helping' any more, you would need to change your mechanical advantage on the lever to do the same work.

Personally I'll probably just ignore the mismatch on my amp (and use my 8 ohm cab all the time). I 'mismatch' for 'tonal' considerations on my other amps with no problems in 20 years. I don't see my self flogging this Dumble output stage into distortion like I do with some of my small class A amps.

[heisthl]

I could be wrong but its' possible HAD didn't care either, aren't there "real" combo style amps out there with 4 ohm iron and an 8 ohm speaker?

[tele_player]

Seems you could also just disconnect input signal from the grids of a pair of tubes, leaving them idling. Then the impedance doesn't change.

I thought about that but I think the impedance will still mismatch in that situation. Like if you were doing mechanical work with 4 human arms on a lever. If two of the arms were still holding on the lever but not 'helping' any more, you would need to change your mechanical advantage on the lever to do the same work.

Personally I'll probably just ignore the mismatch on my amp (and use my 8 ohm cab all the time). I 'mismatch' for 'tonal' considerations on my other amps with no problems in 20 years. I don't see my self flogging this Dumble output stage into distortion like I do with some of my small class A amps.

No, the impedance doesn't change when you stop driving a pair of tubes.

[heisthl]

On a fixed bias amp the bias voltage will rise somewhat when 2 tubes are disconnected.

[groovtubin]

I could be wrong but its' possible HAD didn't care either, aren't there "real" combo style amps out there with 4 ohm iron and an 8 ohm speaker?

YES MINE IS! LOL!

jim

[tele_player]

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