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The ohm relationship between the preamplifier and power amplifier

post #1 of 19
9/12/07 at 10:52am
Thread Starter 
It is my understanding that the ohms input of the power amplifier should be substantially more than the ohms output of the preamplifier to get full frequency response. In my case the preamplifier is 200 ohms output and power amplifier is 20,000 ohms input which is a difference of 100x. Is there a simple way to know if you have the right ohms relationship?
Richard
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post #2 of 19
9/12/07 at 10:55am
10x is minimal. 100x or more is just fine.
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post #3 of 19
9/16/07 at 7:56am
Anything greater than 10x is considered "bridging" and should be fine.

That said, there is nothing wrong with matching impedances.
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post #4 of 19
9/16/07 at 9:13am
Thread Starter 
Quote:
Originally Posted by hphase View Post

Anything greater than 10x is considered "bridging" and should be fine.

That said, there is nothing wrong with matching impedances.

1-What do you mean by bridging if it's more than 10x?
2-It is my understanding that if you match the impedance you will not get full frequency response and it will be off by 3db. Please elucidate.
Richard
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post #5 of 19
9/16/07 at 10:56am
Quote:
Originally Posted by hphase View Post

Anything greater than 10x is considered "bridging" and should be fine.

That said, there is nothing wrong with matching impedances.

The question is whether you want to optimize voltage transfer or current/power transfer. If the former, a high ratio is what you want. If the latter, matching is the way to go.
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post #6 of 19
9/16/07 at 12:58pm
Thread Starter 
Quote:
Originally Posted by Kal Rubinson View Post

The question is whether you want to optimize voltage transfer or current/power transfer. If the former, a high ratio is what you want. If the latter, matching is the way to go.

Am I correct in stating that you want to optimize the voltage transfer between the preamp and the power amplifier?
Richard
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post #7 of 19
9/16/07 at 7:33pm
Quote:
Originally Posted by doxytuner View Post

Am I correct in stating that you want to optimize the voltage transfer between the preamp and the power amplifier?
Richard

Yes.
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post #8 of 19
9/18/07 at 3:09am
Take a look at this:
http://http://www.vtl.com/pages/syst....html#ampmatch
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post #9 of 19
9/18/07 at 6:28am
Thread Starter 
Quote:
Originally Posted by RabPaul View Post

Take a look at this:
http://http://www.vtl.com/pages/syst....html#ampmatch

Please correct web site.
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post #10 of 19
9/18/07 at 6:59am
Think of it this way:

In reality you have 2 signals the first is the signal that the pre-amplifier is trying to put out (V1) and the second is the signal that the 2nd stage (power amplifier) actually sees (V2). The relationship between these two signals is (I'm assuming purely resistive impedances):

V2 = R2/(R1+R2) * V1

So, if you match the impedances you will get V2 = 1/2*V1, which is half of the original signal (or roughly -3dB as you said).

As said above if you are concerned with Power transfer this is actually the point where power transfer is maximized, but the voltage signal is halfed. In the case of a pre-amplifier however you aren't concerned with power delivery but rather with signal preservation so you basically just want to make sure that the difference in impedances is enough that any attenuation is negligible. In the case of your power amplifier output to your speakers though you would want to match the impedances.

Edit: I was bored so here's the math for power transfer:

I=V1/(R1+R2)
P(R2) = V2*I = V2*V1/(R2+R1) = V1^2*R2/(R2+R1)

Take the derivative of P(R2) to find the relative maximum:
dP(R2)/dR2 = V1^2/(R1+R2)^2 + V1^2*R2*(-2)/(R1+R2)^3

solve for 0 for maximum:

-> V1^2/(R1+R2)^2 + V1^2*R2*(-2)/(R1+R2)^3 = 0
-> V1^2/(R1+R2)^2 = V1^2*R2*2/(R1+R2)^3
-> 1=2*R2/(R1+R2)
-> R1+R2 = 2*R2 -> R2=R1

Sorry if that seems unnecessary.
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post #11 of 19
9/18/07 at 9:56am
Thread Starter 
Quote:
Originally Posted by tkdee View Post

Think of it this way:

In reality you have 2 signals the first is the signal that the pre-amplifier is trying to put out (V1) and the second is the signal that the 2nd stage (power amplifier) actually sees (V2). The relationship between these two signals is (I'm assuming purely resistive impedances):

V2 = R2/(R1+R2) * V1

So, if you match the impedances you will get V2 = 1/2*V1, which is half of the original signal (or roughly -3dB as you said).

As said above if you are concerned with Power transfer this is actually the point where power transfer is maximized, but the voltage signal is halfed. In the case of a pre-amplifier however you aren't concerned with power delivery but rather with signal preservation so you basically just want to make sure that the difference in impedances is enough that any attenuation is negligible. In the case of your power amplifier output to your speakers though you would want to match the impedances.

Edit: I was bored so here's the math for power transfer:

I=V1/(R1+R2)
P(R2) = V2*I = V2*V1/(R2+R1) = V1^2*R2/(R2+R1)

Take the derivative of P(R2) to find the relative maximum:
dP(R2)/dR2 = V1^2/(R1+R2)^2 + V1^2*R2*(-2)/(R1+R2)^3

solve for 0 for maximum:

-> V1^2/(R1+R2)^2 + V1^2*R2*(-2)/(R1+R2)^3 = 0
-> V1^2/(R1+R2)^2 = V1^2*R2*2/(R1+R2)^3
-> 1=2*R2/(R1+R2)
-> R1+R2 = 2*R2 -> R2=R1

Sorry if that seems unnecessary.

My preamp is 200 ohms output and the power amp is 20,000 ohms input. Using your formula, I came up with voltage transfer of 200/202 which is 99%. Did I do the math right and what do you think of 99% voltage transfer and audio quality?
Richard
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post #12 of 19
9/18/07 at 10:26am
Quote:
Originally Posted by doxytuner View Post

Did I do the math right and what do you think of 99% voltage transfer and audio quality?
Richard

I think it's much ado about nothing. The impedance values you have quoted are not out of the ordinary to cause this much consternation.
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post #13 of 19
9/18/07 at 10:31am
I want to preface this by saying everything I write is from the perspective of an electrical engineer and music enthusiast not from the audioPHILE perspective.

Yup you did the math correctly. That sounds pretty reasonable to me. I'd be surprised if you could hear any artifacts. One thing to think about (and this may be getting more complicated than what you asked for) is if both your input and output impedances are purely resistive then you aren't going to introduce any frequency shaping at the junction so you will only have a constant voltage attenuation across the frequency band, which you won't hear anyway.

One caveat is that resistors are subject to thermal noise (to avoid over-complicating this just think of thermal noise as random noise that gets introduced by certain types of components), as opposed to say an inductive impedance that could introduce frequency distortion but wouldn't have thermal noise.

Long story short I think the setup you've described is fine.
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post #14 of 19
9/18/07 at 10:32am
Quote:
Originally Posted by tkdee View Post

In the case of your power amplifier output to your speakers though you would want to match the impedances.


Most amps I've seen have significantly lower output impedance than what load is presented by the speakers.

If memory serves me, having an output impedance on the amp that high (and matched to speakers would be considered very high) will result in very poor damping factor and and will cause the amp's frequency response to change as the speaker goes through it's normal impedance swings.
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post #15 of 19
9/18/07 at 11:21am
Thread Starter 
Quote:
Originally Posted by whoaru99 View Post

Most amps I've seen have significantly lower output impedance than what load is presented by the speakers.

If memory serves me, having an output impedance on the amp that high (and matched to speakers would be considered very high) will result in very poor damping factor and and will cause the amp's frequency response to change as the speaker goes through it's normal impedance swings.

I also had the same concerns that the speaker's impedance load fluctuated up and down in relationship to the output impedance of the power amplifier(At a couple of points my 4 ohm speakers went below 2 ohms.).To solve it I bought the NAD power amplifier which has a thing called ( Power Drive). This proprietary system keeps the voltage and current at an ideal proportion and in that way maximizes performance. I can't say that I heard any difference from my prior McIntosh power amplifier but having the Power Drive has a certain comfort level.
Richard
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post #16 of 19
9/18/07 at 11:47am
Quote:
Originally Posted by whoaru99 View Post

Most amps I've seen have significantly lower output impedance than what load is presented by the speakers.

If memory serves me, having an output impedance on the amp that high (and matched to speakers would be considered very high) will result in very poor damping factor and and will cause the amp's frequency response to change as the speaker goes through it's normal impedance swings.


Yeah what I said before is more suited to an idealized scenario where you have a very well quantified, resistive, input impedance. One way of looking at it is if you have a power amplifier with an output impedance of 0.01 Ohms then the maximum load (minimum impedance, maximum power out) would be if you hooked a 0.01 Ohms load out of the amplifier. If you have a speaker with 4-8 Ohms of input impedance you will not have any advantage of using a power amplifier with 4-8 Ohms of output impedance versus a power amplifier of 0.000001 Ohms of output impedance. Again not accounting for impedance fluctuations, both amplifiers will deliver the same current to the speakers. The difference is that the amplifier with the lower output impedance could handle a larger load (lower resistance speakers) and then deliver more power.

From the perspective of the amplifier if you give it a load equal to its output impedance you will maximize the amount of power that the amplifier is capable of delivering. From the perspective of the speaker if you give it a source with an output impedance LESS than its input impedance you will supply the maximum power it will accept.
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post #17 of 19
9/22/07 at 1:22pm
Quote:
Originally Posted by doxytuner View Post

1-What do you mean by bridging if it's more than 10x?
2-It is my understanding that if you match the impedance you will not get full frequency response and it will be off by 3db. Please elucidate.
Richard

"Bridging" means that the addition of the load will not reduce the voltage of the source.

You would need to explain a bit more about what you mean by full frequency response and how you determine when it is "off." Neither bridging nor impedance matching affect frequency response by themselves. Other factors do.

Amplifiers with very low output impedance seem like a good idea in theory, but they don't often occur in the real world. And they shouldn't. If you happened to short the output you would draw infinite current! Not a good idea, even if it were possible.
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post #18 of 19
9/23/07 at 10:46am
Thread Starter 
Quote:
Originally Posted by hphase View Post

"Bridging" means that the addition of the load will not reduce the voltage of the source.

You would need to explain a bit more about what you mean by full frequency response and how you determine when it is "off." Neither bridging nor impedance matching affect frequency response by themselves. Other factors do.

Amplifiers with very low output impedance seem like a good idea in theory, but they don't often occur in the real world. And they shouldn't. If you happened to short the output you would draw infinite current! Not a good idea, even if it were possible.

Thank you for helping me out on the bridging. I figured out my other question by reading the other posts.
Richard
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post #19 of 19
9/24/07 at 5:55pm
What do you consider very low output impedance?
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