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How to set gain on solid state amp

post #1 of 8
Thread Starter 

I have a 1999 Headroom Maxed out Home headphone amp I've owned since new and love it. Internally it uses OPA627 opamps. I'm driving Audeze LCD-2 headphones and it works well. However, it's too loud. It has so much gain, the volume knob is around 7:00 to 9:00 for my listening. This volume knob position is so low, it's very sensitive and sometimes it gets into the volume knob's non-linear area where the channels aren't balanced. Unity gain position on the knob is around 12:00 to 1:00.

 

I'd like to reduce the gain. I can think of 3 ways to do it, what I think of as "best" first:

1. Change (reduce) the gain across the OPA627 opamps internally. I know they're stable at unity gain, but they might already be at unity gain.

2. Install a different volume pot to reduce the gain at any given volume position (Unity gain would move from 1:00 to 3:00 or maybe 4:00).

3. Put a ladder voltage divider on the input to reduce the input signal voltage.

 

The short question: can anyone help me with options (1) or (2) above? I have basic EE skills - designed and built my ladder stepped attenuator - and decent soldering skills but I haven't worked with opamps or power amps. For example to display my ignorance, for solution (2) for less gain do I want a potentiometer with a lower, or higher, value?

 

I've already done (3), wired a cable with 8 kOhm Dale metal film resistors so it's 16 kOhm input impedance and -6 dB of attenuation. This puts the volume knob pretty much exactly where I want it - that -6 dB made about 1/4 turn difference in knob position. And there is no audible deterioration in the sound quality.

 

However, I know (3) reduces the S/N ratio. And it's reducing the input impedance of my amp from 100 kOhm (its actual input impedance) to only 16 kOhm. This may be an issue since my preamp is a 10 kOhm ladder stepped attenuator, and source devices see the headphone amp in parallel with my 10 kOhm attenuator.

 

Normally I'd be fine with step (3) above but reducing the MOH input impedance from 100k to 16k brings the overall load impedance too low. That's 16 kOhm in parallel with 10 kOhm which is only 6.1 kOhm. My CD player (Onkyo DX-7555) has a 470 ohm output impedance, and I'd like a 20:1 ratio, which would be about 10 kOhm total. The MOH's native 100 kOhm input impedance is fine but I don't want to lower it.


Edited by MRC001 - 4/11/14 at 1:15pm
post #2 of 8

Changing the pot shouldn't do anything to the overall gain.  It's just acting as a voltage divider, so as long as the taper is the same on two pots (say 10k and 100k, both audio log taper), they will both adjust the volume the same, IIRC.

 

Do you have any pics of the inside of the amp?  I doubt the 627s are already being run at unity gain, so it may be trivial to adjust the gain, particularly if the resistors are through-hole parts.

post #3 of 8

I agree with the post above, chances are the gain is no where near 0dB. Maybe more like 14-20dB (X5-X10). It should be easy to change that. Have a look inside the amplifier, and find the voltage divider that's built around the opamp's feedback (output to input). Then its just a question of checking the values of the resistors that are used, and replacing one of them (or even adding another resistor in parallel) to lower the gain.

post #4 of 8
Thread Starter 

Problem is, I can't even get the thing open. See attached pics. To open it, either the rear or the front faceplace must be removed. The rear can't be removed without unsoldering the wires to the RCA jacks. The front can't be removed without removing the volume knob. And it doesn't want to come off. There is no set screw and when I pull straight out it just doesn't want to budge. I'm afraid pulling harder might break it.

Sounds silly, and I've taken apart lots of different equipment before, but this one is stumping me.

Before I pull harder on this knob, I wonder if there is a trick to loosen it, like putting it in the freezer or heating it.

P.S. I believe it's a Spectrol potentiometer, if that matters.

Created with GIMP

 

Created with GIMP


Edited by MRC001 - 4/12/14 at 10:59am
post #5 of 8

How long are the wires to the RCA jacks?  If you unscrew the IEC from the back panel, do the wires give you enough leeway to tilt the back panel and pull it through the chassis?

 

Could you try squrting some lubricant up in the back side of the knob?  You would probably want to use something that wouldn't hurt the pot.  Maybe some contact cleaner or something.

post #6 of 8
Changing the gain on the OPAmp could be as simply as changing a single resistor, like on a gain clone. Most standard non-inverting opamp circuits function to make the voltage equal at both inputs. If you can locate a schematic for the amp, I would be more than happy to do some calculations if I can
post #7 of 8
Thread Starter 
Quote:
Originally Posted by DutchGFX View Post

Changing the gain on the OPAmp could be as simply as changing a single resistor, like on a gain clone. Most standard non-inverting opamp circuits function to make the voltage equal at both inputs. If you can locate a schematic for the amp, I would be more than happy to do some calculations if I can


I called the folks who made it (Headroom in Montana). Unfortunately they don't have any schematics. They said it used OPA627s in class A, and from my own testing I know it's non inverting. I couldn't get the faceplate full off but I could tilt it enough to see inside; the circuit board has a toroidal transformer and the layout looks simple and clean; looks like dual mono.

 

I found a simpler solution: use my 16kOhm / -6 dB ladder on the input, and install a switch on the "tape out" of my 10kOhm attenuator, so I can switch the headphone amp out of the circuit when I'm not using it. That way the source device doesn't have to drive the attenuator in parallel with the headphone amp. And when I'm using the headphone amp, I already have a switch for the source to drive the headphone amp directly, bypassing the 10kOhm attenuators.

 

This solves the input impedance problem. However, I'd still like to learn how to reduce the gain on this headphone amp. Just curious, it would be fun and educational. I just need to find a way to get that volume knob off without breaking it.


Edited by MRC001 - 4/13/14 at 2:36pm
post #8 of 8



In theory, no current goes into either input. Since the opamp wants to make the voltage across the inputs 0, and ground is 0v, you can determine that the voltage drop from the input, accross R2 into ground must be the same as the input voltage. Using Ohms Law, you Solve Vin= R2*I. that same current then, since it doesn't go into the input, is the same current that goes through R4. Since the Voltage AFTER R4 = Vin, you can solve Vout-I*R4 = Vin.

So the Gain, Vout/Vin, = (Vin+I*R4)/Vin. Since I= Vin/R2, you can substitute to get
Gain = (Vin + Vin*R4/R2)/Vin

So Gain = 1 + R4/R2.


That's the simplest opamp circuit around, but the same principles apply when you make it more complex, but that's beyond the scope of my basic EE knowledge haha
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