Voltage Divider at input to Power Amp - diyAudio
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Old 28th July 2013, 06:32 PM   #1
diyAudio Member
Join Date: Oct 2004
Default Voltage Divider at input to Power Amp

Hi, be gentle, I'm trying to make the move from being a consumer buying shiny new gear, to making my own simple designs and mods...

Would be grateful for someone more experienced to audit my work - I'm making a voltage divider at the inputs to a pair of monoblock amps. The goal is to reduce audible hiss from the preamp's mosfet output stage. I don't have nor can I obtain a schematic.

The specs are:

Speaker sensitivity ~ 107dB/w/m (front loaded horns)

Power amp output = 18W

Desired attenuation = 15dB

Power amp input Z = 41K shunted by 47pf

Preamp output Z = 130R 'corrected' (per manufacturer)

There's a parallel load on the preamp from the monoblocks and a Velodyne SMS-1 bass management processor. I've searched the documentation and web for the input impedance of the SMS-1 but come up empty. So I'm assuming it is 'high', ~100K.

Click the image to open in full size.

My math is based on these formula:

1) Vd = 1 + (R1 / R2)

2) dB = 20 * Log10 (Vd)

3) Zin = R1 + R2

4) Zout = (R1 * R2) / (R1 + R2)

Following are 6 R1 / R2 combinations for a 15dB voltage divider:

R1 R2 Z(in) Z(out)

4K99 1K06 6050 874

6K8 1K5 8300 1229

10K 2K2 12,200 1803

15K 3K3 18,300 2705

22K 4K7 26,700 3873

47K 10K 57,000 8245

If my math is correct, any of the above R1/R2 combinations will work to attenuate the signal by 15dB. My lack of experience is around matching impedances with the pre and power amps. I'm trying to design a voltage divider with adequately high load for the preamp and low impedance for the power amp.

Since engineering is a process of balance and compromise, at what point does the output Z of the voltage divider become too high for the power amp inputs? Is it a simple 10X ratio?

Power amp input Z (unmodified) = 41K -> 41K / 10 = 4100

Does this imply the max output Z of the voltage divider < 4100R?

1) Which R1 /R2 combo would be best in this application and why?


Much thanks!
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Old 28th July 2013, 07:09 PM   #2
diyAudio Member
Join Date: Mar 2007
Location: Canandaigua, NY USA
IMO, the scale of this isn't one of those things that lends itself to precise calculations, though the exact attenuation is. The factor of 10 rule may or may not be valid, depending on what the input impedance of the power amp really looks like. Specifically, does it change with frequency. Most of the amps I've ever measured had remarkably flat impedance over the whole audio range. That says they can be driven from a relatively high impedance without a problem, even if it violates the factor of 10 rule.

If you look at commercial power amps with level controls, it's nothing more than a pot wired as a level control on the input. The value is usually something like 25-100 kohms. Just to keep the load reasonable, when the control is full up it appears in parallel to the amp input. Thus, I'd want the parallel combination above about 20 kohms for solid state preamps, maybe higher for older tube stuff. It would be hard to go wrong with a 50 kohm pot. If you want to hard wire resistors, that should give you a pretty good idea of the range to be in. For a fixed divider 20 kohms for R1 and 4836 for R2 would be reasonable. The output impedance of the divider would be the parallel combination of the two if the preamp were near zero output impedance; a few hundred ohms is close enough, or factor that in too. BTW, put the attenuator at the amp, not the preamp, so cable capacitance doesn't add in there too.
I may be barking up the wrong tree, but at least I'm barking!

Last edited by Conrad Hoffman; 28th July 2013 at 07:21 PM.
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Old 28th July 2013, 08:25 PM   #3
diyAudio Member
Join Date: Oct 2004
Thank you, Conrad Hoffman.

The power amp designer suggested the 22K / 4K7 is the most optimal, but like you I'm wondering if there are benefits to going higher, for instance 47K / 10K for input Z = 57K. The designer does not share the schematic. He suggested the divider should be implemented by replacing existing R1 (887R) and R2 (40K2) with the new divider values (22K / 4K7).

Question - by replacing the input load of the amp (887R / 40K2) with my divider (i.e. 47K / 10K), I think this results in a new input Z of 57K for the amp. Is this correct?

Thank you.
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Old 28th July 2013, 08:41 PM   #4
DF96 is offline DF96  England
diyAudio Member
Join Date: May 2007
Some power amps have a non-linear input impedance, so lowish values for the attenuator resistors will reduce the resultant distortion. This needs to be balanced against the load on the preamp, which might have a non-linear output impedance. I would probably go for the 15k/3.3k combination, but possibly slightly higher for a valve preamp or lower for a solid-state preamp.
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Old 28th July 2013, 10:25 PM   #5
diyAudio Member
Join Date: Mar 2007
Location: Canandaigua, NY USA
You'll certainly have a new input Z if you replace the resistor. I never mess with that. When I need an attenuator I build it up on a small dual RCA card and either attach short cables and jacks, or just use very short RCA cables from a quad RCA card to the amp. IMO, it does make a huge difference in convenience, level matching and noise to reduce the signal to something more reasonable. That gets you higher up on the level control where the match is usually better. I find that most preamps with output stages have too much gain. Hmmm... about 15 dB too much, now that you mention it! BTW, if I'm using a solid state preamp that can drive a low impedance, say 600 ohms, I make my divider much lower in resistance, down as low as a couple kohms. That just swamps out all other issues with capacitance and flatness of input Z.
I may be barking up the wrong tree, but at least I'm barking!
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