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SMABB 22nd November 2020 08:47 AM

Help needed understanding this prepreamplifier
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Dear members,

I have recently acquired a fantastic sounding prepreamplifier from way back in the eighties; Electrocompaniet MC-2. An all class A device originally marketed as having a self-sensing system for detecting optimal load on cartridge based on current flow.
The idea was said to improve headroom and dynamics. The load and feedback was set through resistors (using dip-switches).

My unit is a later one with exchangeable resistors for optimising load and gain. My problem is, I simply do not understand how it works!
Ideally I want to be able to adjust MC-2 to any chosen cartridge. But how....?

I hope the community is able to shed some light on this :)

How is the load resistor correlated to the gain resistor?
Do the cartridge only see the load resistor value or is some of the circuit part of the total load seen by cartridge?

How to calculate and adjust gain using feedback resistor? I guess the feedback resistor value is dependant on the load resistor, but otherwise I have no clue.

I found a drawn schematics on the internet which correlates well with my unit except for trim resistor setup for feedback/gain. I have only a fixed resistor in mine.
This is for the first version with load resistor being (maybe?) 10 Ohm on input.
Feedback/gain reistors made up from trim resistor in series with 160 Ohm; both in parallel with 2.7 kOhm. This should result in values between 151 Ohm and 393 Ohm.

MarcelvdG 22nd November 2020 08:58 AM

The schematic has a 10 ohm resistor to a virtual ground, so its input impedance is ideally 10 ohm - which is unusually low. Maybe they later changed to a non-inverting configuration, with the signal applied to the bases of the right two 2SB737s and the 10 ohm connected to ground?

SMABB 22nd November 2020 09:06 AM

2 Attachment(s)
Well - I am as surprised as you and wondering what is really going on here...

I found some guidance in the manual of the preamplifier which had the MC-2 build in..

Still left me wondering...

MarcelvdG 22nd November 2020 09:12 AM

OK, so you have to change the resistor to the virtual ground point depending on the cartridge.

SMABB 22nd November 2020 09:20 AM

Agree - but I simply have no clue what the cartridge see as a load...

How can I measure the total gain applied by the Circuit?

MarcelvdG 22nd November 2020 09:24 AM

Assuming the amplifier has a decent amount of loop gain, which it probably has, the cartridge sees the input resistor as a load - so the 10 ohm, 33 ohm, 43 ohm or 100 ohm to the virtual ground point also sets the input impedance.

SMABB 22nd November 2020 09:39 AM

Allright - point taken. Glad you could help in making this crystal clear :checked:

Does anyone have an idea on how to calculate the gain resistor or maybe just measure gain from input to output?

MarcelvdG 22nd November 2020 09:56 AM

The gain from the terminal voltage at the input to the output voltage is ideally -Rfb/R1, where R1 is that 10 ohm, 43 ohm or 100 ohm and Rfb the resistance of that circuit consisting of a 140 ohm resistor, a 300 ohm potmeter and a 2.7 kohm resistor.

The gain from the cartridge's open-circuit voltage (Thevenin voltage, EMF) to the output voltage is lower, ideally -Rfb/(R1 + Zcartridge), where Zcartridge is the impedance of the cartridge. You would get the same gain reduction with cartridge impedance with any amplifier having an input resistance of R1, it's just the voltage division between the amplifier's input resistance and the cartridge impedance.

SMABB 22nd November 2020 11:15 AM

Wow - thanks Marcel :)
Now I'm finally getting somewhere in understanding this circuit.

Thanks a lot

SMABB 22nd November 2020 11:22 AM

One last question for now ;)
Is the circuit inverting the input signal? :confused:

Meaning I will have to alter the wiring connector pins on the cartridge?

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