Is there a better method to power the LDRs for a Mesa MarkIIC+ project?

[capehead]

Hi again guys; I'm in the process of trying to build an accurate as possible MarkIIC+ clone from scratch and the one thing that is giving me some mental grief (aside from reproducing the PCB layout of the preamp, lol!) is the way that Mesa designed the LDR switching circuit. It's not that I don't at all understand it, rather I don't get the logic of it, particularly the LDR power supply. I mean what sense is it to tap off the 60V bias supply with a 680R resistor feeding a 15v zener diode? From all indication that resistor gets very hot and is the source of failure. Second is how the voltage supply is split between a 470R and 3.3k resistor, what is the point of this? Wouldn't putting a larger series resistance in between the diode and zener help alleviate some of the wasted power situation? And what if I were to somehow regulate one power supply rail to accommodate the rating of the LEDs and use a regular grounded/ungrounded line to switch the LDRs on and off like a relay? Sorry if I'm sounding like a noob here, it is just that my experience with LDRs thus far has been very limited as I usually work with relays. I am not necessarily stuck on LDRs for authenticity's sake, rather I already bought a Fender style 100 watt power transformer without any 'extra' taps for switching/solidstate stuff. For quick reference, I'll post the well known IIC+ and MarkIII (more clear) schematics here: Mesa_Boogie_MarkIIC+_original.pdfMesa_Boogie_MarkIII_original.pdf

[Pedro Vecino]

Regulation is well made. First, an approach to 15 volts across the zener (the 680R resistor / 1 watt will probably require a little more power for safety, though I've never seen problems with it) and thereafter through a 470R resistor, a second reference to ground via a 6 volt zener. What is obtained with 470R/3300R voltage divider is a 1.6V potential difference in the optocouplers line. Bridging this line 0V is obtained to deactivate the optocouplers.

[capehead]

Regulation is well made. First, an approach to 15 volts across the zener (the 680R resistor / 1 watt will probably require a little more power for safety, though I've never seen problems with it) and thereafter through a 470R resistor, a second reference to ground via a 6 volt zener. What is obtained with 470R/3300R voltage divider is a 1.6V potential difference in the optocouplers line. Bridging this line 0V is obtained to deactivate the optocouplers.

Hi Pedro, thanks for the reply. You are right in that the way Mesa designed this circuit is generally very reliable; what I was referring to is what I had read on the Boogie Board forum about how if the 470uF dies, it usually takes the 680R resistor with it; although it was acknowledged that resistor normally runs warm/hot. What you mentioned about how the channel switching is activated by bridging the 2 lines to deactivate the LDRs doesn't make sense to me either as from every gut shot I've seen the 'pull lead' grounds out the 3.3k side of the LDR supply, activating the LDRs; also the footswitch does this as well. I'm wondering if this is a mistake in the schematic or if it was shown like that to mislead (as the official Mesa schematics are known to do).

What I am brainstorming as a possible alternative to this stock arrangement is to use a LM317 adjustable voltage regulator set to around 1.5-2.0V and simply ground out the LDRs to activate them, does this sound like a better and simpler alternative? What I would do is use a 10k/2w resistor after the first diode, have it feed the 15v/5w zener to allow it to work with the regulator and then bias the regulator to the proper output voltage. Am I totally off base here in thinking of this problem as only a voltage issue? Is there a current component to all this? I'm not really an LED expert or anything, but I far as I know LED's consume very little current, right?

[Merlinb]

I mean what sense is it to tap off the 60V bias supply with a 680R resistor feeding a 15v zener diode?

Not much sense. It kinda looks like that 15V zener is there simply to protect the crappy 16V 470u capacitor from overvoltage. With a higher-voltage cap, that zener might be unecessary.

Second is how the voltage supply is split between a 470R and 3.3k resistor, what is the point of this?

It appears the 470R is feeding extra current into the 6V zener, presumably to improve its regulation. Not sure why they thought this necessary. The zener is there so there is enough voltage across the footswitch to light its LED. The circuit is unecessarily elaborate, considering all it does is turn a bunch of LEDS on or off! It doesn't even do that very well, because all those LEDs in parallel will share their current randomly. I'm surprised it works at all. Even the footswitch LED is drawn backwards.

Really they should have put all the LEDs in series, fed from a constant-current source, which would be easy with that 60V supply just sitting there!

I couldn't follow your description, but a 317 can't handle 60V input so might be a dodgy choice, but a TL738 can take 125V.

[capehead]

From the man himself! Thanks for the info Merlin and at the risk of excessive flattery; I have found your work and insight to be enormously helpful which I know has been an invaluable resource to many more people than just myself! Anyways though, I absolutely agree that this switching circuit is more complicated than it should be. What I was referring to specifically with the idea of using a LM317 (which I know only has a maximum 40V input to output differential) was to first zener it down to 15v and then to the regulator. What I was wondering if I were to use a higher resistor in place of that 470R, would that negatively effect the operation of the LDRs? I was thinking about using a 10K/1w resistor there after the diode; after that, the 15v/5w zener with a 470-1000uF/35v cap and then after that, the LM317 regulator with it's associated circuitry. As for the idea about putting the LDRs in series (which I gave some thought on doing), it would work most certainly but there is the LDR that controls the treble shift (the added 750p across the 250p treble cap in the tone stack), so that one would have to be powered independently as it is activated by a separate front panel switch.

[Merlinb]

I was thinking about using a 10K/1w resistor there after the diode; after that, the 15v/5w zener with a 470-1000uF/35v cap and then after that, the LM317 regulator with it's associated circuitry.

That doesn't really sound any more convenient than what is already there, to me...

As for the idea about putting the LDRs in series, it would work most certainly but there is the LDR that controls the treble shift, so that one would have to be powered independently as it is activated by a separate front panel switch.

In the Mesa schem, that treble LED is controlled both by a separate switch, but also by the footswitch. So I would do it like this:

[capehead]

That does make alot more sense Merlin; a question though is it only the 5.6k/1W resistor that would be sufficient to drop the voltage enough or should there be a zener there?

[Merlinb]

That does make alot more sense Merlin; a question though is it only the 5.6k/1W resistor that would be sufficient to drop the voltage enough or should there be a zener there?

You seem awfully keen on zeners! When you think about it, a string of LEDs is more or less like a zener... so do you really need another zener?

When the footswitch is closed, the resistor is shorted to ground and gets the full 80V, draining 14.3mA. When the switch is open I estimate there will be 74V across it, making 13.2mA (about 1.3mA for the footswitch and the rest for the string of LEDs). In other words, a high voltage plus a resistance acts a lot like constant current source!

[loudthud]

That 1W 5.6K will be almost hot enough to melt solder. Best to use a 5W if you can make it fit.

[Merlinb]

That 1W 5.6K will be almost hot enough to melt solder. Best to use a 5W if you can make it fit.

Oops, miscalculation on my part!