Fender Super Champ (Rivera era) - 1N5234B/6.2v Zener ?

[captntasty]

Hello all... I am working on a Rivera era Super Champ... one of the issues was the 1N5234B/6.2v Zener diode needed to be replaced... the diode keeps a negative reference (-6.2VDC) voltage for the opto switching/ reverb, etc... the only thing I have on hand is a 1N5230B/4.7v Zener which keeps the ref voltage at ~4.95Vdc. I subbed it in temporarily and the amp works fine but I do not have the pedal to test footswitching... in this function is there any difference between a 6.2v reference or 4.9v reference?

The Free Information Society - Fender Super Champ Electronic Circuit Schematic

[Enzo]

Look at the schematic. The only thing the -6.2v does as far as I can see if provide current for the LED in the opto. If your -4.95 is sufficient for good operation of this, then it is fine.

[captntasty]

So if the Lead is switching in and out from the push/pull pot... then it's good? Now, what about good operation (more for educational purposes)... the opto is a VTL5C1 - data sheet info = LED Reverse Breakdown Voltage: 3.0V, LED Current: 40 mA, LED Forward Voltage Drop @ 20 mA: 2.0V (1.65V Typ.). What, out of these specs, determine good operation with respect to how it is deployed in this amp? -6.2v vs -4.9v? Should I have any concerns about differences in current using -4.9v vs. -6.2v? When/If the footswitch is used, which introduces 2 more LED indicators into the circuit (supplied by the same -6.2v(-4.9v)), should it be a cause for the long-term health of the opto?
Thanks Enzo

[Enzo]

Don't over-think it. The VTL is nothng more than an LED next to a photoresistor inside. You run current through the LED and it lights up, shines on the photocell, and causes the photocell resistance to go down. It's just an LED.

Is it a problem? It depends on the results. The LED controls the photocell. The photocell is used simply as a switch - on or off. And considering its placement in the circuit, it either grounds that 470k ohm resistor or it does not. So the question is: does the reduced current still light the LED enough to turn the photocell off and on? Well, if the circuit works, then I guess it must.

Imagine you have a 6.2v battery, an LED and a 150 ohm resistor. all in a series loop. (See the resistor between the opto and the reverb pedal plain jack? If you ground that jack terminal, then you have a 150 ohm resistor and an LED in series across 6.2v.) The LED lights up from the current in that circuit. We can even do some math. If the LED has 2v drop across it, then that leaves 4.2v across the 150 ohm resistor. Ohm's Law tells us that the resistor must have 4.2/150 = 28ma flowing. Imagine your experimental circuit has 28ma flowing too. Now what happens if we drop the 6.2v down to 4.9v? It is like a weaker battery. The LED still has the 2v drop, so then that leaves 2.9v dropped across the 150 ohm resistor. Ohm's Law tells us then that the resistor has 2.9/150 = 19ma flowing. Current is the same throughout a series circuit, so whatever current flows through the resistor is also flowing through the LED. So what happens when we reduce the4 current through the LED from 28ma to 19ma? The LED doesn;t glow as brightly. WIll that hurt the LED? I don't see how, if anything it is easier on the LED. In neither case do we come very close to that 40ma maximum. And I used 2v drop for the LED, but yo0u can do the same math for 1.65v or you can measure the actual drop across the LED in the thing and use THAT number. The actual amounts will be a little different, but regardless, at 4.9v it will conduct less current than at 6.2v by roughly the same amount.

If your footswitch adds an LED in series, then there will be another 2 less volts (or whatever) across the resistor.

And just to be thorough, the plain footswitch jack grounds the opto LED through that 150 ohm resistor. But the red footswitch jack just grounds the -6.2v supply completely. Shorts right across the zener. Well, if the zener has a short circuit across it, then it doesn't care WHAT voltage might be applied. No matter what, there will be zero volts across the diode. And no matter what zener you install, when the red jack is grounded, the result is the two 390 ohm 2W resistors across -55v to ground. Those resistors total 4w dissipation. With a 6.2v zener, they have 48.8v across them for about 3.1w total dissipation. with a 4.9v zener, then they drop 50.1v for about 3.1w total dissipation for the pair. I don;t think the extra tenth of a watt will bother them.

Isn't Ohm's Law amazing, it just shows up everywhere?

If it works it works.

[g1]

The only possible glitch I can think of is if you are playing somewhere with bad AC power. If the wall voltage is lower than normal (120V for US) then the switching may stop working sooner than it would stop working with the higher voltage zener. Not very likely but something to remember if you ever get somewhere with poor AC and the switching stops working.

[52 Bill]

...But the red footswitch jack just grounds the -6.2v supply completely. Shorts right across the zener....

The footswitch for this amp has leds and dropping resistors in it. The -6.2 volt supply is used to light the leds and is never shorted out.

[guitician]

What ever happened to using the little loops for jumped lines in schematics? It sure makes reading them easier.

[J M Fahey]

Agree.
This schematic drawing is junk, pure and simple.
They should at least have used the dots which mean wires *are* connected. They didn't.

[Enzo]

Thanks, Bill, I never looked up the pedal wiring, just looking at potentials for trouble within the amp. Your drawing eliminates that extra potential.

[J M Fahey]

Since the "oficial" pedal adds Leds in series, now you are "losing" almost 4V, so actual voltage to push current through those resistors (which is the exact same current which will drive the optos) *is* quite affected by Zener choice.
A 6.2V one means you have useful 6.2-4=2.2V
A 4.9V Zener means you end up with only 4.9-4=0.9V ... a *lot* less.
Using the footswitch may be marginal or even not work at all.
So go get your 6.2V Zener and use it.

[Enzo]

Juan, are you looking at a different "official" pedal than the one in post #6? I see no series LEDs in that. I see -6.2 rail applied to either LED, and a switch to ground to light each LED.

[J M Fahey]

You are right, of course, the Leds are not in series but on their own separate power rail (- 6.2V).
The mechanical implementation of which I find horrible, because now this amp *needs* to have its special dedicated pedal, plus 2 stereo cables to work.
I BET most users by now do not use the original system and just use a generic single button pedal , mono cable and plug into the red jack, to enable the Lead mode . Of course, that shorts the 6.2V line and the fancy leds stop working
I didn't dedicate more than a few seconds to that schematic because that kind of drawing gives me a headache
Plus , "you see what you want to see" , I always switch DC , never remotely switch audio because of the cable capacitance , so for me a natural is to add a Led in series and call it a day
Thanks for catching my mistake

[Leo_Gnardo]

If it turns out you MUST have 6.2V, you could wire two diodes forward-biased in series with your 4.9V Zener, for instance the very common 1n400X where x is 1 thru 7. Whatever you have on hand will do if it is rated for 1 amp or more. The 0.6V forward drop x 2 will add 1.2V to your 4.9V and get you within a gnat's hair of the 6.2V you're looking for. ALSO pay attention to the dropping resistor, usually 47 ohms for this series of amps. Fender put half watt carbon resistors in, and these almost always overheated and went off-spec, sometimes to a lower resistance. If it already hasn't been done, replace that resistor now with a higher watt rated part.