This novice needs help with Voltage divider

jeffreybehr · 21st October 2011, 12:39 AM

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?

nigelwright7557 · 21st October 2011, 01:06 AM

Why not use a potentiometer ? That way you can adjust it at will.

counter culture · 21st October 2011, 02:09 AM

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...

454Casull · 21st October 2011, 02:16 AM

I doubt he's worried about impedance matching but rather impedance dropping.

counter culture · 21st October 2011, 02:23 AM

What's in a name?

A rose by any other name would smell as sweet.

Don't you think the difference is likely to be lost on him, given what he says in his OP?

jeffreybehr · 21st October 2011, 03:33 AM

Quote:

Originally Posted by nigelwright7557

Why not use a potentiometer ? That way you can adjust it at will.

Level pots and degrees of attentuation

jeffreybehr · 21st October 2011, 03:48 AM

Quote:

Originally Posted by counter culture

Don't you think the difference is likely to be lost on him, given what he says in his OP?

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.

jeffreybehr · 21st October 2011, 03:49 AM

Quote:

Originally Posted by counter culture

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.

TY; found one. Didn't know they existed.

sreten · 21st October 2011, 04:21 AM

Hi, why do you want to reduce voltage gain by so much ? its unusual ...., rgds, sreten.

discrete · 21st October 2011, 08:23 AM

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.

21st October 2011, 12:39 AM	#1
jeffreybehr diyAudio Member Join Date: Mar 2005 Location: Phoenix, Arizona USA	This novice needs help with Voltage divider 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?

21st October 2011, 01:06 AM	#2
nigelwright7557 diyAudio Member Join Date: Apr 2008 Location: Carlisle, England	Why not use a potentiometer ? That way you can adjust it at will. __________________ http://www.murtonpikesystems.co.uk PCBCAD40 pcb design software.

21st October 2011, 02:16 AM	#4
454Casull diyAudio Member Join Date: Jan 2004 Location: Toronto, ON, Canada	I doubt he's worried about impedance matching but rather impedance dropping. __________________ Building a 2.1 system out of a 3/4"x4'x8' sheet

21st October 2011, 04:21 AM	#9
sreten diyAudio Member Join Date: Nov 2003 Location: Brighton UK	Hi, why do you want to reduce voltage gain by so much ? its unusual ...., rgds, sreten. __________________ There is nothing so practical as a really good theory - Ludwig Boltzmann When your only tool is a hammer, every problem looks like a nail - Abraham Maslow

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21st October 2011, 02:09 AM	#3
counter culture diyAudio Member Join Date: Jun 2011	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...

21st October 2011, 02:23 AM	#5
counter culture diyAudio Member Join Date: Jun 2011	What's in a name? A rose by any other name would smell as sweet. Don't you think the difference is likely to be lost on him, given what he says in his OP?

21st October 2011, 08:23 AM	#10
discrete diyAudio Member Join Date: Sep 2010	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.

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