I have NO technical training or education, so pls don't think me stupid--I'm just uneducated. 
I posted today another question about level controls, etc., but I think this one makes more sense.
I want to reduce my poweramp's Voltage gain by 15dB. I 'understand' that there's a way to connect series and shunt resistors so that the input impedance of the circuit doesn't change but the Voltage output will be 15dB lower. Can someone pls tell me what resistors to wire in series and what value to wire to ground from between the 2 series resistors?
Or do you need more info?
I posted today another question about level controls, etc., but I think this one makes more sense.
I want to reduce my poweramp's Voltage gain by 15dB. I 'understand' that there's a way to connect series and shunt resistors so that the input impedance of the circuit doesn't change but the Voltage output will be 15dB lower. Can someone pls tell me what resistors to wire in series and what value to wire to ground from between the 2 series resistors?
Or do you need more info?
If you google 'attenuator calculator' you will find many calculators for Pi, Tee and L-pad attenuators. Just plug in the cut in dB and the impedance you want to match to. You can even match dissimilar impedances to a degree.
As Nigel says though, why not just use a pot? Impedance matching is not that critical generally, or pots wouldn't be as common as they are...
As Nigel says though, why not just use a pot? Impedance matching is not that critical generally, or pots wouldn't be as common as they are...
Lost indeed. If the stated input impedance is 47K, set by a 56K resistor, what value do I 'normalize' to? As 'counter' so aply infers, these kinds of calcs are WAY beyond me.
You also have to consider the impact of the added resistors on the input network of the amp. Adding any series resistance will change the frequency response of the input network.
I would suggest a simple divider, simulating a potentiometer:
22k in series with the source to amp IN and then 4.7k from amp IN to amp GROUND.
The amp now 'sees' Rs + 4.7k // 22k at its input. Assuming Rs (source output impedance) is small this is about 3.9k, more than 10x smaller than the original input impedance of the amp. This minimizes unwanted interaction between the new divider and the original input network of the amp.
The attenuation would be approx. 4.7/26.7 = 5.68 = 15.09 dB.
The input impedance of the amp will change to ~26k, still acceptable for sources with output impedance < 2.6k.
I would suggest a simple divider, simulating a potentiometer:
22k in series with the source to amp IN and then 4.7k from amp IN to amp GROUND.
The amp now 'sees' Rs + 4.7k // 22k at its input. Assuming Rs (source output impedance) is small this is about 3.9k, more than 10x smaller than the original input impedance of the amp. This minimizes unwanted interaction between the new divider and the original input network of the amp.
The attenuation would be approx. 4.7/26.7 = 5.68 = 15.09 dB.
The input impedance of the amp will change to ~26k, still acceptable for sources with output impedance < 2.6k.
Simply get a pair of these. No need to calculate anything or do any soldering.
Ask Google/Yahoo/Yahoogle/whatever about "RCA attenuator" for a supplier in your neighbourhood.
An externally hosted image should be here but it was not working when we last tested it.
Ask Google/Yahoo/Yahoogle/whatever about "RCA attenuator" for a supplier in your neighbourhood.
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