diyAudio

diyAudio (http://www.diyaudio.com/forums/)
-   Solid State (http://www.diyaudio.com/forums/solid-state/)
-   -   Changing Input Z on Adcom (http://www.diyaudio.com/forums/solid-state/104395-changing-input-z-adcom.html)

WithTarragon 29th June 2007 09:33 PM

Changing Input Z on Adcom
 
Hello,
Here is my situation. I am DIYing a horn loaded system that will be quite efficient (about 104 dB / watt). However the bass bin uses dual drivers so the impedance may dip as low as 3 Ohm.
I will be bi-amping the speakers after an electronic crossover (Behringer DCX). It has plenty of voltage output and an output Z of 180 Ohm. After the crossover I will going through a 4-channel pot (resistance betwen 10 - 25K). The cables (short & low capacitance) will then go to a pair of stereo amps. The amp for the bass bin might be an Adcom (GFA 535 or 545). It will have plenty of gain, however it only has an input Z of about 20KOhms. I would prefer the bridging impedance between the resistor pot (source) and the Adcom (load) to be about 10x different.

My questions are:
1. Does it make sense to increase the input Z of the Adcom amp? Or would this be a headache or ill-advised?

2. If I don't have a 10:1 bridging impedance what are the consequences. I assume the voltage transfer is poorer (although the power transfer would be better). Would this limit the bandwidth or create distortion?

3. Having the higher input Z at the Adcom (if I go this route), would this make the cable run more susceptible to picking up noise, hum and RF (remember the efficiency of the speakers)?

4. The other idea is to solder fixed resistors in parallel with the pot (volume control) and brings its resistance (the source) down to lower value. Although at some point, wouldn't a very low resistance (load) require that the Behringer (source) is providing more current than it can deliver?

BTW, since I am using very short, low capacitance cable after the pot & the pot is not that high of a resistance value, I am not worried about low pass filtering the signal ( 1 / 2pi F C). Also BTW, the idea of using multi switch resistor pad or a transformer as a volume control is not practical in this case

Thanks,
-Tom

Nelson Pass 30th June 2007 01:15 AM

Last time I looked, these amps had Mosfet inputs. If you aren't
running balanced inputs, then you can change the input resistor
to ground to any value you like.

:cool:

WithTarragon 3rd July 2007 05:30 PM

Input resistance on Adcom
 
Re: increasing input Z on an Adcom 5x5 amp so I can have a bridging Z of about 10x (for use with a passive pre-amp)

This attached schematic is for a GFA 555 amp, but I assume the resistor in question is R2 (for both the Left and Right channels).

I will be incresing this from 22k to close to 4 times that resistance.
Is this the correct resistor?

Thanks for your help,
-Tom

Nelson Pass 3rd July 2007 07:31 PM

No schematic. Is R1 from the input to the Gate, and then
R2 is from the Gate to the ground? If so, you should be
able to make that increase.

:cool:

WithTarragon 3rd July 2007 09:31 PM

Whoops, here is the attachment
 
2 Attachment(s)
Sorry about that.
Here is the attachment, I think (it does not show in my " preview")

Nelson Pass 3rd July 2007 10:56 PM

Sorry, wrong amp. These inputs are bipolar, and you can
go to a higher value for R2, but you will see some DC offset
as a result. Most likely you'll be better off leaving it alone.

Are you sure your passive preamp won't drive it?

:cool:

WithTarragon 4th July 2007 03:48 PM

Thanks, I will be an empiricist
 
Thanks for leading me through this.

I will probably take an empirical approach to this and just give it try with a 15K or a 25K pot and see if it works. There should be plenty of voltage from the Behringer crossover so I am not worried about the gain.

I assume if the resistor pot has a too low of resistance, the Behringer (source) may have problems providing enough current. I assume this would result in some sort of clipping and the consequent harmonic distortion. This should be easily seen on a scope, if the problem is severe enough.

However if the input Z on the Adcom (load) is not sufficiently higher than the resistor pot resistance I am not sure what the consequence is. Folks always mention having a 10:1 difference between the load Z and the source Z. I assume this is to get an adequate voltage transfer. But I am not sure what happens if I deviate too much from the 10:1 "recommendation". If I simply lose a bit of gain, I am not worried. However, if I lose dynamic range or get clipping then I would be worried.

The problem with these "rules of thumb" is that the consequences are never explained in physical terms.

In either case, I will give it a try. The schematic I posted was labelled as a GFA 555, for all I know this could have been from a later series (I found it on the Web).

Thanks for helping me out,
-Tom


All times are GMT. The time now is 02:44 AM.


vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2014 DragonByte Technologies Ltd.
Copyright 1999-2014 diyAudio


Content Relevant URLs by vBSEO 3.3.2