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Old 23rd October 2007, 05:52 PM   #1
dre7 is offline dre7  United States
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Default Amplifier impedance and pot selection

Hi:

I could really use some help understanding the whole idea of input impedance and what role it plays in the selection of pot value. It's my understanding that the potentiometer value more or less determines the input impedance of the amp. I use a CD player as input, and I'm assuming the typical output impedance is around 500 ohms, probably no more than 1k. It seems most of the builders here go with 50k pots. Am I correct in saying that this would represent a 1:50 impedance match? Is this the "load" a CD player likes to work into? Is there a lower and upper threshold of values in this case? I see everything from (I believe) Carlosfm using 10k to Thorsten Loesch (sp?) using 1M law faked. Are the opamps of CD players hardy enough to handle such wildly varying values, with selection determined by how much rotational adjustment the user desires for his amp/speaker combo?

Ok, there's one more layer of complexity that's puzzling me. The amp I'm building uses an input transformer with a 1:10 ratio. Using a 100k pot I think I've settled on placing the pot from the input transformer's secondary to grid rather than amp input to transformer primary as I've seen recommended on the Sakuma site. Now, does this choice effectively let the CD player "see" a 10k load? And if in general the 50k pot is what most use, would I then need to use a 500k pot?

Help, please! As you can see I'm getting pretty confused.
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Old 23rd October 2007, 06:45 PM   #2
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Default Re: Amplifier impedance and pot selection

Quote:
Originally posted by dre7
Am I correct in saying that this would represent a 1:50 impedance match?

No match is essential or desired. It's rather simple in fact. CD players have various output stages and thus various driving capabilities. A larger impedance is generally easier to drive and the distortion will be less. An impedance which is too high is not an issue. Most output stages have a resistor in series with the output which effectively sets the output impedance. The load for least distortion is commonly in the kohm range, although some output stages have no problem driving a few hundred ohm.

For a preamp to be universally compatible with many cdps a minimum input resistance of 10k is pretty much the norm. If this value is higher, it's even better. It is not that critical anyway.

An input transformer changes the situation. As far as the cdp is concerned it reflects the actual input impedance of your amp by the inverse square of the transformer ratio. In your case this impedance (in paralle with the pot) will be divided by a 100. So, a 100k will become less than 1k load to the cdp - possibly a real issue.

Whether you can put the pot on the primary side of the transformer will depend on the primary inductance. It is not likely that a step-up transformer will enjoy being driven from a high impedance such as a pot.
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Old 23rd October 2007, 06:53 PM   #3
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And not to beat around the bush: if you are driving a 1:10 transformer you simply need a dedicated proper preamp; a cdp and pot won't really cut it.
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Old 23rd October 2007, 11:25 PM   #4
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Quote:
And not to beat around the bush: if you are driving a 1:10 transformer you simply need a dedicated proper preamp; a cdp and pot won't really cut it.

I agree. The CineMag CMMI-10C is (IMO) a nice 1:10 step up part. Notice that the trafo is to be driven by a source whose impedance is <= 150 Ohms. You can get that low source impedance by using an ECC99 cathode follower at the amp's I/P. With a CF at the I/P, a 100 KOhm pot. is a non-issue.
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Old 23rd October 2007, 11:49 PM   #5
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I thought the inherent capacitance of the cable connecting the CD player to the pot had varying undesirable effects (high frequency rolloff?) depending on the overall impedance (and instantaneous position) of the pot.
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Old 24th October 2007, 12:12 AM   #6
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Impedance matching is desired for power transfer - from an amp to a speaker, or a transmitter to an antenna. It's also important for best signal-to-noise ratio for very low-level signals, like an FM antenna or a phono cartridge.

But for stage-to stage connection, ESPECIALLY when a cable is involved, what you want is LOW source impedance, HIGH load impedance. The source has to drive the capacitance of the cable plus the input, and the lower the impedance, the better. The pot sits between the two, and should be low enough that it doesn't cause high-frequency roll-off when it's midrange. 50K should be low enough.

A 10:1 voltage step up transformer will be tough to drive - I don't think a CD player will do it - it multiplies the load capacitance by 100.
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Old 24th October 2007, 12:19 AM   #7
dre7 is offline dre7  United States
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analog_sa

Thanks very much for the reply. I just want to make sure I understand something. When you say

As far as the cdp is concerned it reflects the actual input impedance of your amp by the inverse square of the transformer ratio.

You're refering to impedance ratio? The input tx in question is a UTC A-12 with an impedance ratio of 600:80000 and a turns ratio 1:11. In light of that does your 1/100 multiplier still hold for me?

Also, I run my cdp through a Behringer DEQ2496 with analog out impedance of 100 ohms @ 1kHz. It also has a +12dBu boost switch which I haven't had to use with the amps presently in my system. I'm assuming these will be of some help whenever I get this amp off my workbench.

Topology is A-12 => 6s2s => 1:1 interstage splitter => PP6b4g. So far it's hum free and sounds ok hooked up to a Discman. Need to sort out a few bits before scoping it, though.

Thanks again for your help.

Andrew
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Old 24th October 2007, 12:46 AM   #8
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Quote:
You're refering to impedance ratio? The input tx in question is a UTC A-12 with an impedance ratio of 600:80000 and a turns ratio 1:11. In light of that does your 1/100 multiplier still hold for me?
the impedance ratio of a trafo is the square of the turns ratio. 112 = 121. So, your UTC A-12 presents approx. 661 Ohms to the driving circuitry, when loaded with 80 KOhms. A valuble rule of thumb requires the load be at least 10X the O/P impedance. As you can see, your Behringer DEQ2496 will be overtaxed. The capacitive multplying effect previously mentioned makes things WORSE.

Put the cathode follower in!
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Old 24th October 2007, 02:12 AM   #9
dre7 is offline dre7  United States
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Eli:

Just working this out a bit. So if I went "extreme" with a 2M pot. the reflected impedance would be 16528 Ohms? Why wouldn't I do that? I'm guessing high frequency rolloff from inductance?

I'll have to read up on cathode followers. Any reason against doing one with 1/2 a 6sn7? Available parts and space limitations are influencing me here.

Sounds like it may be easier all around to ditch the input tx altogether and go two stages.



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Old 24th October 2007, 08:13 AM   #10
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Quote:
Originally posted by dre7
So if I went "extreme" with a 2M pot. the reflected impedance would be 16528 Ohms? Why wouldn't I do that? I'm guessing high frequency rolloff from inductance?


You can. The high frequency roll-off will be caused by capacitance between the pot and the following grid. Btw, if there is a grid resistor in parallel to the pot you'll have to take it into account as well.

Subjectively, i don't like the sound of large pots. And you can't get anything of decent quality above 250k. Then again i'm also not a huge fan of CFs, especially the "Ugly Ms Piggy" variety.

Step-up transformers don't offer a free lunch.
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