How To Set Up A Stereo Amplifier

Thread Starter

JDR04

Joined May 5, 2011
367
Here's a newbie question for you folks! I've got a stereo amplifier, the type that is used for communications at a bank counter. It has one button type microphone and one 8 ohm speaker on each side of the seperating glass.

The amp has volume controls on the front for the staff to use. It also has pots at the rear that are used to set the sensitivity of each channel/customer and staff side respectively. If the rear pots are turned up too far then there is feedback. The trick is to set each channel at 50% to get maximum effectiveness. If one channel is turned higher than the other then it will be louder than the other one.

Is there a way of injecting a signal into each microphone input at the same time and then measuring the output as one adjusts the rear pots? To complicate things, I do not have an oscilloscope but would be willing to tackle a small measuring circuit of some description. Any help out there??:confused:
 

russpatterson

Joined Feb 1, 2010
353
I've used a scope that used the soundcard on my pc, I think it's called Winscope. There's a few of them. I've also used scope apps for the iPhone/iPhad. You should make the circuit to protect your mic input on the soundcard (although I didn't). To make the test signal you need a signal generator of some kind. Preferably a sine wave generator. Look on SparkFun. They sell a signal generator kit and a scope that plugs into the usb on your PC for about $250 that's pretty good.
 

wayneh

Joined Sep 9, 2010
17,498
There are also some nice function generators for iPod or iPhone. You can set the frequency and get a tone at an adjustable volume. Another approach is to play a pre-made "song" that includes a tone sweep or white noise or whatever you want. Now that I think of it, AV receivers often generate a white noise tone so you could set your relative speaker volumes. Avoids quirks related to any individual frequency, or quirks with the hearing of the user.
 

Thread Starter

JDR04

Joined May 5, 2011
367
I've got a MP3 player with a 1Khz tone on it which I use for setting up induction loops for hearing aid users. This is easy because the amp has a Line In facility so the MP3 player connects directly in. But I'm told I can't connect directly into the Audio amp because its a MIC INPUT and not a LINE IN INPUT. How does one get around this??:confused:
 

#12

Joined Nov 30, 2010
18,224
Hold the MP3 player next to the microphone? That would be a way of injecting a tone into each microphone input. The other way is to use resistors or a variable resistor to turn the line level signal down to mic level voltage. I think 100 to one will do it.
 

Thread Starter

JDR04

Joined May 5, 2011
367
Thanks #12, I get the idea but don't I also need some sort of a capacitor to decouple the signal etc? I've heard something about this but am not sure to the construction or design of such a circuit . Do I need to insert a capacitor in both lines ie + and - and if so what sort of value would I use:confused:
 

#12

Joined Nov 30, 2010
18,224
Measure the output of the MP3 player and see if it has any DC voltage on it. If there is no DC component, you don't need a capacitor because it is already provided inside the MP3 player.

If there is DC, you need the input impedance of the mic input to calculate the capacitor in series with the signal line only. Assuming 47k input impedance and 1000Hz, Xc=1/(2piFC)
.0033uf or larger.
 

Thread Starter

JDR04

Joined May 5, 2011
367
Thanks for the post #12. I'm a newbie so I dont really understand your formula etc. I understand what you mean by input impedance of the mic but the 1000hz and the formula I'm kinda lost. It really would be appreciated if you could explain it further for me. What does 2pifC represent for example? I do so far understand that the required capacitance is derived by dividing 1 by whatever is in the brackets, I think............:confused:Thanks very much.
1000Hz, Xc=1/(2piFC)
 

Pencil

Joined Dec 8, 2009
272
Maybe this helps:

Xc=Reactance (input impedance) in OHMS (Ω)
f=Frequency in Hz
C=Capacitance in Farads (F)
∏=pi (3.14151926...)

From #12's post: Xc=1/(2∏fC)

Using algebraic rules solve for C:
........
........
........

C=1/(2∏fXc)


Plug in your numbers:

C=1/(2*∏*1000*47000)

Solution: C=.00000000339 or .00339uF
Nearest common capacitor value .0033uf or 3.3nF or 3300pF

Hope this helps.
 
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