MXR Dyna Comp (1982) Pedal Repair Help

[misterc57]

Hey Folks

Not sure how many here are into pedal repairs. Thought I'd give this a shot. Thanks for the help! MC

Trying to repair my 1982 Dyna Comp. I am referencing from the attached schematic which I hope is the correct version.

Pedal does not turn on. Negative wire from battery clip is flakey, sometimes high resistance, sometimes not.

Positive cable has a diode attached in series that reads open in both directions when doing a diode check. I am guessing this diode is to protect the circuit from a reversed polarity connection at the DC adapter?

I figured I can clip in a 9V battery directly for a quick test. Negative to input jack, positive to DC adapter jack. That test failed to turn on the unit, and the 9V battery got hot and drained 1.5V in about one minute.

Always learning, what did I do wrong? Does that 1N5817 need to be in circuit to test this unit?

BTW, I did remove all the rotten foam that was in there and I realize I need to put something in there to prevent shorts. I was thinking of cutting some cardboard sections.

[Kevina]

The 1N5817 is there so that when the pedal is connected to a dc supply the diode is reverse biased, and the supply will not try to charge the battery.

For such a large current drain, either the 10mF capacitor is short circuit, or the CA3080 is toast.

Remove the battery and connect your ohmmeter across where you connected the battery; it will read a very low resistance. Carefully lever the CA3080 from its socket, if the resistance has gone up, the CA3080 is the culprit; if it is still low, remove the capacitor and retest, do not forget to replace the CA3080 as it is always possible both of them are shorted.

Unfortunately, if someone has connects a dc power supply with the wrong polarity, it will have destroyed all of the semiconductors.

[Jon Snell]

Some body will have used the wrong polarity wall wart, a very common occurance.
Replace the CA3080 plus fit a new battery lead for when your wall wart is mislaid and all will be fine.

[misterc57]

The 1N5817 is there so that when the pedal is connected to a dc supply the diode is reverse biased, and the supply will not try to charge the battery.

For such a large current drain, either the 10mF capacitor is short circuit, or the CA3080 is toast.

Remove the battery and connect your ohmmeter across where you connected the battery; it will read a very low resistance. Carefully lever the CA3080 from its socket, if the resistance has gone up, the CA3080 is the culprit; if it is still low, remove the capacitor and retest, do not forget to replace the CA3080 as it is always possible both of them are shorted.

Unfortunately, if someone has connects a dc power supply with the wrong polarity, it will have destroyed all of the semiconductors.

Can I use any diode to replace the 1N5817 when installing the new battery leads and clip?

So, I had low resistance (cable needs to be inserted in the input jack). I slowly lifted the CA3080 and still have low resistance (no change).

I am trying to find that 10mF cap. I am not seeing it if I follow the positive supply trace on the board. See the orange clip attached to the board, that is where the red positive cable enters the board. There is the yellow 10mF cap on the other end of the board. But that is not the one I am trying to find per the schematic..

What are those little blue rocket shaped components?

Also I noticed that the socket is installed backwards on the board. But the chip was installed correctly (not following the socket pin 1 designation).

[Helmholtz]

1N 5817 is a Schottky diode that drops about 0.5V less than a standard rectifier diode.

The blue drops are tantalum ecaps.

[vintagekiki]

https://www.electrosmash.com/mxr-dyna-comp-analysis#schematic

http://www.generalguitargadgets.com/pdf/ggg_dnr_dc_sc.pdf

[misterc57]

OK, that electrosmash page is helping. Seems like the 10uF is missing from original designs. My board has 10 caps, and no caps in the box.

Looks like I can use a 1N 4148 (which I have) instead of the 1N 5817 Schottky.

Lifting the CA3080 did not change my earlier resistance monitoring. So, to test. I am thinking....

I could connect a1N 4148 and a 10uF in parallel in the power supply. I was testing with a 9V battery but I do have a DC power supply I could use instead and monitor the current. What would be the expected current draw and if I have a high current draw at what reading would I want to back off? I am not use to working on 9V DC supplies.

Sorry but I am one of those that has to overthink everything...

Thank you, MC

[Helmholtz]

I don't like the parallel protection diode as shown in the schematic.

When a reverse polarity power supply is connected, the diode shorts the power supply and chances are that the power supply and/or the diode are destroyed.

Best way is a series Schottky. Second best a normal series diode. As indicated it will lower supply voltage by around 0.7V, but that shouldn't be a problem.

[misterc57]

I don't like the parallel protection diode as shown in the schematic.

When a reverse polarity power supply is connected, the diode shorts the power supply and chances are that the power supply and/or the diode are destroyed.

Best way is a series Schottky. Second best a normal series diode. As indicated it will lower supply voltage by around 0.7V, but that shouldn't be a problem.

Ok, so back to my original schematic and I will add the missing 10uF cap. For testing is it correct to say I do not need either the diode or the 10uF since I would be coming in from a DC power supply?

Also, I still need to know what level of current is safe or excessive in this circuit.

Thank you

[g1]

Just a side note.
It's actually 10uF. 10mF is 10,000 uF.. With modern advances in capacitors, it matters. Farads used to be a strictly theoretical value, now they are seen in real manufacture.

[g1]

Best way is a series Schottky. Second best a normal series diode. As indicated it will lower supply voltage by around 0.7V, but that shouldn't be a problem.

I don't trust that schematic from post #1. Why would you need a diode in series there? The DC jack switch disconnects the battery line so the diode seems pointless.

[vintagekiki]

I could connect a1N 4148 and a 10uF in parallel in the power supply.

Instead of 1N4148 uses 1N4004.
1N4148 is a signal diode, as opposed from 1N4004 which is a power current diode (for 1A).

[Mick Bailey]

With the 9v battery getting hot this usually means a dead short. You already removed the IC with no change to the supply input resistance, so consider the failure mode of components and possible current draw scenarios. Other than the IC, and (absent) 10uf capacitor, there are no other likely places where a single component short could cause such a heavy current draw due to the various resistors which would limit the current. Resistors nearly always fail open anyhow. So if any of the other semiconductors failed short the current draw would not be excessive. However, looking at the schematic the transistor and 10uf capacitor combination top right make me suspect that this is a possible failure point - the transistor collector connects directly to the 9v supply. Check the transistor legs for a short to ground - the 10uf cap/transistor could be shorted.

[vintagekiki]

Compare pcb layout ... ...

https://music-electronics-forum.com/forum/effectification/guitar-effects/50505-hekp-needed-with-vintage-script-logo-mxr-dyna-comp-problem

https://music-electronics-forum.com/forum/effectification/guitar-effects/16440-mxr-dynacomp

https://music-electronics-forum.com/forum/effectification/guitar-effects/10419-mxr-dynacomp-script-1975-schematics

1)

https://reverb.com/item/7894519-vintage-mxr-dyna-comp-pcb-works-tested

2)
Voltages at the transistors and IC pins

http://www.generalguitargadgets.com/pdf/ggg_dnr_dc_instruct.pdf

[Enzo]

Or a misplaced wire or connection, or a solder short between traces.