Another Microsynthesizer question - the power supply

Started by Kevin Mitchell, February 27, 2019, 09:39:18 AM

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Kevin Mitchell

Long story short... over the weekend I took a trip out to an old friends studio to check things out and jam. Of course I walked out of there with equipment that needed some TLC.

I've got a 90s reissue bass microsynthesizer on the bench. Seems it only needed some new slide pots and wire touch-ups. But I'm also using the unit for confirmation on a personal (soon to be shared) project of my own.

So my question is on the power transistor. The schematic shows a 2N6110 while the unit has a B826 (2SB826?). It's tough to source these outside of the ebay realm and I'm hoping someone could school me in the use of complementary PNP transistors and it's role in this schematic so I can safely source alternatives. There has been a few clone builds with the use of an alternative transistor but I don't quite understand why they had chosen what they had

Here's the schematic for the PS if you're unfamiliar with it;


I'm also considering working up an alternative power supply circuit. But why fix what aint broken?  :P
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Kevin Mitchell

For consideration but not to confused the topic - here's a drawing of the most recent microsynth power supply. It runs off 9 volts but wondering why the old one isn't as simplified if in theory - it should work the same with a higher voltage. Perhaps the old version demanded more current and a more thought out configuration for a bipolar supply?

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bean

Here are three I found on Mouser that are at least somewhat close in spec to the 2N6110 and same pinout:

https://www.mouser.com/ProductDetail/ON-Semiconductor-Fairchild/KSB596YTU?qs=sGAEpiMZZMshyDBzk1%2FWi8oN7VHZ91OkSn8NMa56zOs%3D
https://www.mouser.com/ProductDetail/ON-Semiconductor-Fairchild/KSB1017YTU?qs=sGAEpiMZZMshyDBzk1%2FWi8oN7VHZ91OkRctAINJWIv4%3D
https://www.mouser.com/ProductDetail/ON-Semiconductor-Fairchild/BD244BTU?qs=sGAEpiMZZMshyDBzk1%2FWi4G1GLBZKHK1NaKAlaM4eWE%3D


2N6110 Bipolar Transistor
Characteristics of the 2N6110 bipolar transistor

    Type - p-n-p
    Collector-Emitter Voltage: -70 V
    Collector-Base Voltage: -80 V
    Emitter-Base Voltage: -5 V
    Collector Current: -7 A
    Collector Dissipation - 40 W
    DC Current Gain (hfe) - 30 to 150
    Transition Frequency - 10 MHz
    Operating and Storage Junction Temperature Range -65 to +150 °C
    Package - TO-220

bean

Quote from: Kevin Mitchell on February 27, 2019, 10:05:00 AM
For consideration but not to confused the topic - here's a drawing of the most recent microsynth power supply. It runs off 9 volts but wondering why the old one isn't as simplified if in theory - it should work the same with a higher voltage. Perhaps the old version demanded more current and a more thought out configuration for a bipolar supply?


Well, EHX already used a 24v supply in a lot of its vintage effects so it might have been a matter of economics at the time. But, that's pure speculation. And, again, for the modern one it's probably economics again since 9v supplies are so ubiquitous in guitar effects now.
Anyway, I've been working on a clone of the modern one and definitely prefer the simpler approach. Getting the thing to work...that's another story.

Kevin Mitchell

Quote from: bean on February 27, 2019, 10:14:46 AM
Anyway, I've been working on a clone of the modern one and definitely prefer the simpler approach. Getting the thing to work...that's another story.
Me too, Bean. Me too  ;)

I was laying out the hardware for the enclosure and started to feel morality kick in since it's near impossible to lay out the circuit without it looking like an exact clone of the new one. So I'm shooting for the old circuit in a 1590xx enclosure since I found a source for the CA3094E chips that doesn't break the bank. And of course - a substitute PCB for the 13700 version as an alternative is in the game plan.

I've got the new circuit working on breadboard after applying corrections to the schematic floating around. Shoot me a msg if you'd like the edited drawings.

And thanks for the recommended subs! Good old mouser to the rescue.
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Mark Hammer

I ran into the same thing with a friend's early-issue Deluxe Memory Man that needed repair.  Some here suggested that since the unit simply needed -15V, that I should opt for a simple 7915 regulator, which I did, with great success.

You may want to simply sidestep the transistor replacement and go with a more modern form of regulation.

Kevin Mitchell

#6
Quote from: Mark Hammer on February 27, 2019, 04:04:00 PM
I ran into the same thing with a friend's early-issue Deluxe Memory Man that needed repair.  Some here suggested that since the unit simply needed -15V, that I should opt for a simple 7915 regulator, which I did, with great success.

You may want to simply sidestep the transistor replacement and go with a more modern form of regulation.
Thanks, Mark. Your input is always appreciated. I guess this is a job for the breadboard before I get too far ahead of myself.

If you can, please elaborate how you used the regulator while omitting the power transistor? Trying to wrap my head around it since we're aiming for a +9v/GND/-10v supply.
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Mark Hammer

I just removed the components used to provide a regulated -15V and reused the pads - with any relevant cuts and jumpers - to feed -15V to the point on the board where it begins to be distributed, using a standard 3-pin regulator and basic appnote circuit.

Rob Strand

#8
Quotef you can, please elaborate how you used the regulator while omitting the power transistor? Trying to wrap my head around it since we're aiming for a +9v/GND/-10v supply.

Keep in mind the regulator is creating a dual supply out of a single supply.

The part at the bottom A1B, D21, R110 create a low lower negative rail -10V.

The part at the A1A tracks the negative rail.   Feedback keeps pin 3 at 0V, so
10V / 15k  = Vout  / 12k     =>  Vout = 10 * (12/15) = +8V
Q5 boosts the current drive on the positive rail.

The other way to look at is, *relative to the -V rail*:
- The bottom part (A1B etc) produces +10V.
- The top part (A1A + Q10) produces +18V ;
   Feedback relative to -V rail: 18V = 10V * [(15k + 12k)/15k] = 1.8 * 10V = 18V

As is, the power supply ensures that the +V rail never drops below the 0V rail.  That's probably a good thing since you don't want things connected between 0V and +V to get a negative voltage.    Also, the negative rail and positive rail rise more or less equally.  If the 0V point  rose up slow slow it could put 18V between 0V and +V. 

Another subtly is the zener is powered from the +V rail, and the negative rail depends on the zener.  So there's some magic at turn on so the power supply powers up with gettting stuck with 0V output.   Here the tick come from the fact if A1A's output is at zero it turns on Q10, then that kicks the whole thing off.

The point here is there's some subtle things going which make the circuit work.  If you do regulator mods you could open-up a can of worms by not handing all the power-up cases.  You don't want any weird stuff going on when you power-up the unit as it could fry something every 1 in 100 power-ups.

One option might be to make +V the master rail.  You create +V using an +18V regulator relative to the -V rail (kind of treat the -V rail as "0V for the regulator").   Then you replace D21 with a 5.6k resistor so now the output of A1B will be at just over one half the voltage of +V *relative to the -V rail*.    In that way it should power up more or less like the original.   Is it correct?   Well that's the problem with messing with the circuit you would need to test that new regulator configuration to make sure +V to 0V doesn't have any voltage undershoots or overshoots when you power-up.  Similarly for -V to 0V.

If you were doing a clone it might be worth modernizing the PSU but for a repair it might be best leaving it as is.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

PRR

> schematic shows a 2N6110 while the unit has a B826

It's just a transistor.

The question is: how much current does it have to pass? Normally, AND when some fool shorts the output (which is why the question comes up).

Price is no object. An over-kill transistor could be $3 while your labor to replace it *again* will be $15+.

TIP42 comes to mind. Gain is decent. Will carry 6 Amps if you can get the heat out. Comes in a large package so you can get the heat out.

In any case, depending on the fools who will use it, I would be inclined to put SOME resistance in series, so current can not try to become "infinity".

Taking a wild guess from the 100uFd main cap: it's nominal 30V input and 100mA load. A 10 ohm resistor in series with emitter would limit short current to <3 Amps. This resistor will idle at 0.1 Watts, but try to throw 100 Watts in a short (I suspect the PT won't come close). Therefore a 1 Watt resistor will not get hot normally yet burn-up cheaply when abused.
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Kevin Mitchell

#10
Now that's what I call being schooled  :icon_lol:

Rob - that was very informative and more than I could ask for. Thank you very much! I'll test things out on breadboard once I have the ICs. I might even give the modern microsynth supply design a shot with the higher voltage.

Mark - I overlooked the DMM part and know that it doesn't require a bipolar supply so apologies for the confusion. I was reviewing your thread from last month where you mentioned the old microsynth you were working on with a 5554 regulator. Thought that was relevant at first glance.

PRR - thanks for your input. The power supply sure could use an update. Using Robs information I'll tinker a bit and test the alternatives - hopefully cutting the part count & build cost a little.

Thanks, Guys! I love the feedback. You're all saints.
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Rob Strand

QuoteUsing Robs information I'll tinker a bit and test the alternatives - hopefully cutting the part count & build cost a little.
It occurred to me later if you need the -10V rail or +8V to be precise for some reason you might need a slight tweak on what I've said there;  the rest of it still holds.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Kevin Mitchell

Uhg. Wishful thinking with the repair on my bench. I've got it running (kind of) but the filter sweep doesn't seem to be doing it's thing. All of the voices seem to work fine.
A10 and A12 are buzzing at the outputs but not A11, swapping around some of the 3094s that prove to work didn't change anything. I should probably check out the start and stop detector. Not sure of the chances of A7 (RC4558, start & stop detector) going bad.

Hmm...
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Rob Strand

QuoteA10 and A12 are buzzing at the outputs but not A11, swapping around some of the 3094s that prove to work didn't change anything. I should probably check out the start and stop detector. Not sure of the chances of A7 (RC4558, start & stop detector) going bad.

Assuming I'm looking at the same circuit,   that's a bit weird.  For A12 to produce an output it has to get a signal good signal from A11.

At the start you take your best shot at what is wrong then see if those ideas fix the problem.   If you get a hit it's a quick fix.

Often with complex circuits like this you have to work though each block and convince yourself it is working.  Sometimes that means disconnecting blocks from the rest of the circuit and injecting your own AC and DC signal into the blocks.   Also watch out disconnecting something doesn't blow-up something in the surrounding blocks.  It's a slow process but you eventually get confidence in certain blocks and suspect other - it's a much slower process than the quick fix.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Kevin Mitchell

Quote from: Rob Strand on March 08, 2019, 06:12:52 PM
Assuming I'm looking at the same circuit,   that's a bit weird.  For A12 to produce an output it has to get a signal good signal from A11.
The outputs of A10 and A12 have loud buzzing, A11 is much lower but it comes back at the output of A12. Maybe it has to do with the circuit itself along with other underlining issues. I'm just poking and guessing here.

I'll hopefully figure it out this week.

Meanwhile... I couldn't help myself and got this for a steal. Should have it later in the week;


It's the internal transformer version (not the 90s reissue) - claims to be in working order just needs a new slider. Apposed to this reissue bass one on my bench that needed 5... The most difficult part of fixing a broken box is not knowing to entire story  :o At least I'll have voltages to compare.
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Rob Strand

QuoteAt least I'll have voltages to compare.
Another unit makes life a lot easier  ;D
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Kevin Mitchell

#16
Quote from: Rob Strand on March 11, 2019, 04:55:12 PM
QuoteAt least I'll have voltages to compare.
Another unit makes life a lot easier  ;D

If only it was indeed "fully functional"... Those bastards lied to me.

I'll be back later  :icon_lol:

EDIT: The problem with this one seems to be at the "attack delay" section (last part of the circuit after the filter). Signal is not passing through the last OTA (A14). I suppose it could be worse. I'll keep at it.
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Kevin Mitchell

After years of tinkering I've finally decided to try out simulations. I'm having much fun as it's more visual than the usual breadboard work.

Here's the vintage PS;


Seems to work just as Rob has noted in his analysis - with +V being +8 and not +9 as the schematic claims.
Though after a few seconds the voltages jump around but they all seem relevant to +8v, 0v & -10v. Possibly a simulation error.

I'm going to experiment with a 9v supply and charge pump to see about creating the voltages in other ways. I intend to etch my clone prototype within the next couple of weeks  8)

-KM
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Kevin Mitchell

Any reason why this configuration wouldn't be ideal?



-KM
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