Matching two different amp's input sensitivity?

JakeJ · « **on:** 20 Sep 2008, 09:09 pm »

Is there a formula that I can calculate the resistor value to match one amp's sensitivity to another. Amp 1 has a rating of .72 VRMS for full output and Amp 2 is 1.5 VMRS for full output. I just need to change one resistor and I'd rather not put a potentiometer in due to mounting issues.

So, how do I match two different amplifiers for equal sensitivity?

Thanks,
Jake

art · « **Reply #1 on:** 20 Sep 2008, 09:57 pm »

If you know the actual value of the resistor that goes to ground on the amp that you want to lower, you can stick a resistor in series with it at the input.

But...........

Most amps already have a resistor between it and the input. And a capacitor across the (load) resistor to act as a low-pass filter for the input. Raise the value of that input resistor, and you will get treble roll-off.

Yes, there is a simple formula, but the correct answer may be more complex than you had hoped for.

Pat

Albert Von Schweikert · « **Reply #2 on:** 20 Sep 2008, 10:03 pm »

You also need to know the Input Impedance of both amps in addition to the Input Sensitivity. Please pm and I'll get back to you with exact values, but check out the various formulae inherent in Ohms Law and you can easily work out the values with a pocket calculator. By the way, what speakers and amplifiers are you working with? I'm a big fan of DIY.
AVS

Davey · « **Reply #3 on:** 20 Sep 2008, 10:14 pm »

Jake,

Unless "full output" from both amplifiers is the same you can't make a determination from the specifications you quoted. You need to know the voltage gain of each amplifier.....then you can determine the proper attenuation that must be applied to one amplifier.

You can find voltage gain in some manufacturers specifications. However, I would make a measurement to be sure. Apply a sinewave input at some arbitrary value and measure the output with a voltmeter. 20*log(Vout/Vin) will yield the amplifier voltage gain in db.

Once you've determined which amplifier is "hotter" wire a potentiometer in series with the input to that amplifier and adjust it so the two amplifiers now have equal voltage gain. Remove the potentiometer (without disturbing the setting) and measure the resistance. That will be the value of resistor to add in series.

Pat makes a good point about it possibly altering an input filter......but that might not happen and you can cross that bridge if you come to it.

Make sense?

Dave.

PLMONROE · « **Reply #4 on:** 22 Sep 2008, 02:58 am »

What would be the downside of using a REALLY, REALLY good passive attenuator to reduce the sensitivity of the hotter amp? The Placette passive preamp which for a long time has been rated class A in Stereophile comes to mind as a possible candidate. Obviously this is too simple an answer so I must be missing something.

:

mjosef · « **Reply #5 on:** 22 Sep 2008, 03:43 am »

I have used a basic TVC for this same purpose, works as well as any as far as I can hear.

JakeJ · « **Reply #6 on:** 22 Sep 2008, 09:02 pm »

Sorry for such a late response, one of the foibles of a graveyard lifestyle.

While the passive volume control is a good idea it is a rather expensive solution as once the balance between amplifiers is found then the setting never changes and you can eliminate a lot of ancillary wire and mechanical connections. The Placette is very nice and 125 steps is really sweet, but in the end it's still just an expensive adjustable resistor. I am considering building a pot-in-a-box using a Goldpoint pot just for this and future mismatches, but ultimately a single resistor is the best solution.

I think Davey has the right idea. Measure gain then use a pot on the hot amp, the HK in this case. However as Albert Von Schweikert stated in the VSA owner's circle (here) the input sensitivity is the key to successful integration not the final output of the amps being used. We may only need 15 to 20 Watts of tube power on the M/T module but need 150 to 200 Watts of high current SS power on the bass modules to get excellent integration in a given space. Space changes and so do the power requirements.

In my case the HK is more sensitive (1.25 VRMS for 150 Watts out) and my Dyna Mk3 CAE updated tube amps are 1.8 VRMS for 60 Watts out. Too much bass! So I had hopes of a simple "change a resistor and your problem is solved" solution but it will be a bit more complex than that.

Thanks for all who have responded. Davey: I have a couple of high power sand block 10 ohm resistors. These should suffice to get a delta of voltage but ultimately just placing the pot on the input is the quick and dirty way, eh? Only thing is there is one of them fancy RC networks on the input of the HK that Art mentioned. Hmmm...it gets curiouser and curiouser

Jake

art · « **Reply #7 on:** 23 Sep 2008, 07:05 am »

Oh......so it is a matter of getting the levels to match up. Now you have to take into account the efficiency of the drivers...............

And you were hoping for a simple answer! Then dorks like me come along and ruin the day.

Adjust the levels on the input. Don't try fixing it on the output with a handful of "cement boat" power resistors. Stick a 10 kilohm pot in the input of the louder one. Use it to replace the series resistor that is in there now. (Probably around 1 kilohm or so.) Adjust it to where it sounds right to you. Repeat on the other channel. Put the lid back on and don't futz with it.

Unless you have treble loss.

Don't tell your audiophile friends what you did. Just enjoy it and pretend that ignorance is bliss. It very well could be.

Pat

JakeJ · « **Reply #8 on:** 24 Sep 2008, 12:29 pm »

Again I apologize for my late reply.

Quote from: art on 23 Sep 2008, 07:05 am

Oh......so it is a matter of getting the levels to match up. Now you have to take into account the efficiency of the drivers...............

Yes, but I am also driving the crossover. Maybe it would help if I had mentioned what speakers I am using (VSA VR4 G3 HSE) but I would think that it holds true for any speaker in a bi- tri- or however many channels are required, active or passive, one must try to match the amp to the speaker's efficiency. Many people have success bi-amping with less power to the mid/tweeters than to the woofers, pretty normal for all drivers.

Quote from: art on 23 Sep 2008, 07:05 am

And you were hoping for a simple answer! Then dorks like me come along and ruin the day.

Yes, sometimes there is a simple answer. My apologies if I have given you the impression that I think you are a dork. I realize that input impedance is a function of the circuit's inherent RLC components so just changing a single resistor is not going to "fix" everything. But if I don't ask I don't learn.

Quote from: art on 23 Sep 2008, 07:05 am

Adjust the levels on the input. Don't try fixing it on the output with a handful of "cement boat" power resistors. Stick a 10 kilohm pot in the input of the louder one. Use it to replace the series resistor that is in there now. (Probably around 1 kilohm or so.) Adjust it to where it sounds right to you. Repeat on the other channel. Put the lid back on and don't futz with it.

Hmmm...I thought I was clear in that I am trying to fix it at the inputs. Oh, I see, you thought that my reference to the "10 ohm sand block resistors" meant I was attempting to solve my problem with them. No, I planned to use them as a dummy load for the method Davey outlined above in his post for determining an amp's overall gain expressed in dB. I only mentioned the value as it probably doesn't need to be exactly 8 ohms (nominal speaker ratings) to get the measurement just as long as the same load is applied to both amplifiers under test. Pot-in-a-box is looking like a better solution all the time.

Quote from: art on 23 Sep 2008, 07:05 am