amp help please, my brain's frazzled

[makaze808]

Hi, I am looking to mod my small 70's marshall hybrid amp but I have gone into fuzzy brain world.

Schematic here  http://www.amparchives.com/Amp%20Archives/Marshall/Schematics%20&%20Layouts/Marshall%20Amp%20Schematics/Miscellaneous%20Valve%20Combos/2060%20Mercury/2060.gif

What I initially wanted to do was cut out the tremolo circuit.
But what's troubling me is how the hell is the 275v (X on the schematic) being dropped to 26v by the time it arrives at the collector of the transistors.

Will cutting out the tremolo circuit (point A on the schematic) leave the voltage on the other trannies at 26v? I am worried that the 26v is dependent upon the tremolo circuit being intact but can't fathom out what's going on to ease my worries.

Many thanks for your help.

[pinkjimiphoton]

why not just leave the trem in? put a switch on it or whatever, someday you may regret removing it. taking it out probably isn't gonna help anything...

jmo!

[J0K3RX]

I don't know if I would depend 100% on that schematic? Get your meter out and get voltage readings at different points maybe? Myself, I wouldn't disturb that amp unless it was broken and I was going to restore it to original working order... It's a vintage so I would have to agree with pinkjimiphoton.

[thelonious]

What I initially wanted to do was cut out the tremolo circuit.
But what's troubling me is how the hell is the 275v (X on the schematic) being dropped to 26v by the time it arrives at the collector of the transistors.

+1 on what pinkjimiphoton and J0K3RX said. Do you have a particular reason for wanting to get rid of the tremolo?

The 1K dropping resistor (R17), 27k dropping resistor (R16), and the 220k collector resistor of each transistor will each drop some voltage. If you cut out the trem, it would mean one less transistor drawing current... which would mean less voltage dropped across the resistors... which means you end up with more than 26V at the other trannys, as you said. You'd have to change that 27k to something larger to compensate.

[pinkjimiphoton]

if it has a footswitch jack (probable) just plug a plug in that's got hot and ground shorted (or open, as the case may be) and be done with it.

then the marshall will retain the value. believe me...give it time, it will be worth keeping stock. some of the unwanted marshalls are real sleepers until someone on the net raves about them (like the mosfet JCM800's of the late 80's...don't ask how i know this... but i quadrupled what i paid for mine after raving about 'em on the net for a couple years)

[makaze808]

Thanks for the replies.

Should have made it clear, I am wanting to build a clone of the amp without the tremolo (which is terrible).

So a 1k  27k and 220k series of resistors can drop 225v? 

The schematic (not the figures given) is accurate apart from the capacitors from collector to base labelled with a ? which are not present.

[thelonious]

Should have made it clear, I am wanting to build a clone of the amp without the tremolo (which is terrible).

Ah, I get it. Yes, you could build a clone without the tremolo, you'd just probably want to change the value of one or both of dropping resistors R17 and R16.

So a 1k  27k and 220k series of resistors can drop 225v?  

Yep. Even a 1 Ohm resistor can drop 225v if there's enough current draw through it and it's rated for enough power. You can rearrange Ohm's Law to read V=I*R, which states that the voltage dropped = current * resistance. I don't know how much current those transistors are drawing, but it wouldn't take much current to drop a lot of voltage across a 220K resistor. Let's say they're drawing 0.75mA each. In that case, one of those 220k collector resistors would drop
V=0.00075*220000=165 volts
by itself. Check out the calculators at http://www.the12volt.com/ohm/ohmslawcalculators.asp

The schematic (not the figures given) is accurate apart from the capacitors from collector to base labelled with a ? which are not present.

The real world voltages in your amp will be different. Measuring the real voltages would be a good next step, but BE CAREFUL. I know people say it on this and every forum all the time, but the voltages in these things can stop your heart.

[PRR]

> how the hell is the 275v (X on the schematic) being dropped to 26v

By sucking on a resistor.

275V to C18, tube OT. 1K drop to tube G2, say 265V at C17.

Now 27K R11 from 265V to ???V.

> dropped to 26v

Where do you see 26V?? At the _collectors_, not the low-volt rail.

T1 T2 are noted as 8V across 10K emitter resistor, 0.8mA. The collector resistors are 220K. 0.8mA in 220K is 176V. The low-volt rail is clearly at least 8V + 176V + some drop in transistor.

*Assuming* the 26V number is right, then low-volt B+ rail is 26V+176V= 202V.

To get 202V from 265V through 27K means 2.3mA in the 27K R16.

Check: T1 T2 are noted as 0.8mA. T3 has a similar collector resistor and is likely pulling similar current. Three 0.8mA loads is 2.4mA, very close to the 2.33mA calculated the other way.

Alternatively, raise R11 by a factor 3/2. 39K, 47K, sumthin like that.

> Will cutting out the tremolo ....leave the voltage on the other trannies at 26v?

Per figuring above, three transistors drops 265V to 202V, so two live transistors will drop 265V to  223V. Really, 202V or 223V is insignificant. We should check the voltage ACROSS the transistors, because there are limits. Stock is apparently 26V-8V= 18V. With two live suckers, probably 10% higher or 20V. It is unlikely the transistors are 19V break-down. It is not a disaster if they are, because voltage breakdown doesn't kill transistors, they die when break-down passes infinite current and heat. Here the _worst_ current is 265V/220K 1.2mA; the worst power happens at half of 265V in 220K or 0.08 Watts.