Mark Levinson No23 repair help

Did you completely shut off the right amp ?
It is not wise to expose it to 111 Volt, that will cause more damage.

1) 22mV on R20 of the left amp is a bit low, should be 27.5mV after warming up. You know how to adjust it.

2) You could check again the shut down circuit on the right amp without power on.
Disconnect P332 and measure the cable from here the resistance to gnd.
This should be practically zero ohm.
When you interupt the line through the temp sensors, preferably WH.11, the line should go from zero ohm to infinity.

3) Did you find the User Manual with the value of the fuses ?

4) I have made a new simulation with the lamps in the lines, here visible as R14 and R22, both having 50V voltage drop.

Plse make a copy of this circuit diagram, and use it to report the voltages to me on those places where voltages are shown (14 different values) after you have repaired the Vcc Reg.
That will save you and me a lot of time.

Hans
 

Attachments

  • ML23-17.jpg
    ML23-17.jpg
    265.7 KB · Views: 173
Last edited:
Thanks again for the schematic. That is good idea to use reporting the numbers.

I disconnected the right amp when measuring the left one.

I could not find my own 23 manual but 23.5(basically same amp) fuse size is 8A so I think I have the right size.

This left channel biasing is funny because I need to increase it every time I check. Hopefully it is soon stabilized.
 
I have another question; my left amp is not connected(No Vcc unreg, Vcc reg or OL-2 connected.) but I connected the P332 cable. I cannot turn the amp on because it will immediately shut down. Is it working right?

What is turning the amp off in this case?

That's O.k. because I gave you the wrong number, P332 belongs to the right channel :D I thought your right channel was defective.
Take off P333, that's the left channel.

Hans
 
Now I'm really confused.
Taking off P332 or P333 should never result in switching off the Amp no matter right or left channel.
If it does, the circuit diagram you gave me does not match your amp.
See image below.

A current has to flow through the diode of the 4N33 opto coupler to activate the triac that energises shut off relay SW300.
No connection, means no current can flow.

So with both P332 and P333 detached from this circuit, you completely disable the protection circuit on OL-2 for both amps.
But only when you have this circuit diagram.

Hans
 

Attachments

  • ML23-18.jpg
    ML23-18.jpg
    166.2 KB · Views: 220
Yesterday I did not have the temperature sensors soldered in the main board. Now OL-2 board is connected + the P332 and P333 connected the amp turns and stays on. Left channel connected, right disconnected. So temperature sensors need to be soldered otherways the amp will not turn on. By default temperature sensors are shorted and when the heatsinks get too warm sensors disconnect the line and amp will be turned off.

Yesterday I biased the left amp to 27.5mV over R20 after fully warmed. Now when I turn the amp on it will start from 50mV and will go slowly low towards 27.5mV. I quess that is OK?
 
Yesterday I did not have the temperature sensors soldered in the main board. Now OL-2 board is connected + the P332 and P333 connected the amp turns and stays on.
The OL-2 Board needs a VCC supply to function, but the schematic doesn't tell which Vcc, probably +Vcc unreg ?
Maybe you can send a picture of this OL-2 board.
Left channel connected, right disconnected. So temperature sensors need to be soldered otherways the amp will not turn on. By default temperature sensors are shorted and when the heatsinks get too warm sensors disconnect the line and amp will be turned off.
Correct
Yesterday I biased the left amp to 27.5mV over R20 after fully warmed. Now when I turn the amp on it will start from 50mV and will go slowly low towards 27.5mV. I quess that is OK?
There is no other way so this is it.
When driving a 2 Ohm speaker at 700Watt, peak current is 26.5 Amp.
So the 0.5Amp at start up seems completely harmless to me.

You could check the ripple on both left amp's Vcc's.
Using again the formula C=I*dt/dV gives the the momentary value of the 36.000 uF capacitors C300 and C303.
After warming up I = 0.32Amp, dt = 0.01sec, so with 36,000uF, dV should be ca 90mV.

Hans