Moving Coil Pickups for the Technically Curious

[Joe Gwinn]

You would not expect to find an IC for this application; it is pretty special. The requirement is for extremely low noise voltage; the noise current does not matter, of course, since the source impedance is so low. Junction transistors generally have lower nose voltage (and higher noise current) than FETS or tubes. Transistors with large areas, or multiple devices in parallel are two ways to go. This is not what you tend to find in ICs. But the transformer wins for simplicity. You can match normal devices, and the differential input is free.

I had to read it twice. The quoting got screwed up, so your point about differential inputs became part of the quote of my posting.

Transformers can be pretty good, and those current transformers that bbsailor uses ought to work.

The electronic dodge would be to use a matched pair of bipolar low-noise transistors in the common-base topology, with string inputs at the emitters and output from the collectors. Common-base amplifiers have very low input impedance, the better to match the strings (active and dummy).

[bbsailor]

I had to read it twice. The quoting got screwed up, so your point about differential inputs became part of the quote of my posting.

Transformers can be pretty good, and those current transformers that bbsailor uses ought to work.

The electronic dodge would be to use a matched pair of bipolar low-noise transistors in the common-base topology, with string inputs at the emitters and output from the collectors. Common-base amplifiers have very low input impedance, the better to match the strings (active and dummy).

Here is a web link for a common-base amplifier.
To use a Speaker as a Mike

Joseph Rogowski

[Steve Conner]

And like I said... Check out the INA217 from TI. Same idea as the THAT chip but without the minimum order.

[Mike Sulzer]

Sorry about the screw up.

I think the noise performance of common-base and common-emitter would be the same, so unless there is some specific reason for loading the source, either could be used.

I had to read it twice. The quoting got screwed up, so your point about differential inputs became part of the quote of my posting.

Transformers can be pretty good, and those current transformers that bbsailor uses ought to work.

The electronic dodge would be to use a matched pair of bipolar low-noise transistors in the common-base topology, with string inputs at the emitters and output from the collectors. Common-base amplifiers have very low input impedance, the better to match the strings (active and dummy).

[earthtonesaudio]

Too late, it's already overthought.

Oh, you have no idea. I only shared with you all a small fraction of my overly-complex musings on this subject. But I know I could overthink it further still!

[Joe Gwinn]

I think the noise performance of common-base and common-emitter would be the same, so unless there is some specific reason for loading the source, either could be used.

Impedance match is better with the common base configuration, so the resulting SNR is better, even though you are right about the inherent noise of the transistor.

[Mike Sulzer]

The source impedance is probably so low that even the common base configuration does not really load it significantly.

If loading is an issue, I think it is better unloaded, not loaded. It is the ratio of the signal voltage (developed at the input) to the input noise voltage that counts. The signal voltage is higher if it is unloaded. Matching would lower the voltage by a factor of two compared to a high impedance input.

Impedance match is better with the common base configuration, so the resulting SNR is better, even though you are right about the inherent noise of the transistor.

[earthtonesaudio]

The source impedance is probably so low that even the common base configuration does not really load it significantly.

If loading is an issue, I think it is better unloaded, not loaded. It is the ratio of the signal voltage (developed at the input) to the input noise voltage that counts. The signal voltage is higher if it is unloaded. Matching would lower the voltage by a factor of two compared to a high impedance input.

...But isn't the signal primarily a current source, rather than a voltage source? The ideal load for a current source is a short circuit, unless I'm recalling things incorrectly.

[Mike Sulzer]

The law of magnetic induction gives a voltage around a loop. You can construct a circuit model either way. I prefer to keep the model close to the physics.

[Joe Gwinn]

The source impedance is probably so low that even the common base configuration does not really load it significantly.

Yes. The common-base bipolar transistor amplifier usually has in input impedance of maybe 20 ohms, although this depends on collector current. The two strings will have an impedance well less than one ohm.

If loading is an issue, I think it is better unloaded, not loaded. It is the ratio of the signal voltage (developed at the input) to the input noise voltage that counts. The signal voltage is higher if it is unloaded. Matching would lower the voltage by a factor of two compared to a high impedance input.

The lowest noise impedance will often differ from the max-signal impedance. It may be useful to add a series resistor between string and common-base amplifier input. Having this resistor also helps with the anti-ESD circuit as well.

[Mike Sulzer]

Are you sure you are not thinking of the case where we are placing an impedance transformer between the source an preamp and adjusting the transformation ratio for the best SNR?

Here we are not doing that; we are using no transformer. The model used for the amplifier has a noise voltage and a noise current. We have low source impedance and so the effect of the noise current is negligible. In this case we get the best SNR by maximizing the voltage at the input which means we should not load the source.

Adding a resistor in series with the source can result in unwanted signal attenuation and additional noise from the resistor. That makes the SNR worse.

Yes. The common-base bipolar transistor amplifier usually has in input impedance of maybe 20 ohms, although this depends on collector current. The two strings will have an impedance well less than one ohm.

The lowest noise impedance will often differ from the max-signal impedance. It may be useful to add a series resistor between string and common-base amplifier input. Having this resistor also helps with the anti-ESD circuit as well.

[Joe Gwinn]

Are you sure you are not thinking of the case where we are placing an impedance transformer between the source an preamp and adjusting the transformation ratio for the best SNR?

No transformer involved.

Here we are not doing that; we are using no transformer. The model used for the amplifier has a noise voltage and a noise current. We have low source impedance and so the effect of the noise current is negligible. In this case we get the best SNR by maximizing the voltage at the input which means we should not load the source.

This maximizes the signal, and assumes that the SNR is thereby maximized. This may or may not be true.

Adding a resistor in series with the source can result in unwanted signal attenuation and additional noise from the resistor. That makes the SNR worse.

Not necessarily. Transistors are not linear devices, and the self-generated noise depends in a complex way on the source impedance presented to the input. So, it does not follow that the best impedance is zero, where the series-resistor noise is minimized and the signal voltage is maximized.

The theory is too complex to be worth the trouble to compute answers. Quickest way to tell is to experiment. But use low-noise (metal-film) series resistors, or a 50-ohm wirewound pot. (Do not use any form of carbon resistor.) It may also turn out that the series resistor value does not make much difference.

[Steve Conner]

This is an old myth. Padding the source with a resistor NEVER improves signal-to-noise ratio. The Johnson noise of the resistor always hurts you more than the "better impedance match" helps you.

Transistors generate noise voltage and noise current. For an application like this, we would get a bunch of large transistors and run them at a high collector current.

This gives you low voltage noise, because the transistor's internal noise-generating "resistances" scale down with increased die area and increased collector current. But it gives you high current noise, because the base bias current is increased. That's OK though, because the low impedance of the source swallows the noise current.

This is how you tune transistors for best noise performance. The higher the source impedance, the lower the collector current you need to run. Any good transistor intended for low-noise amplification will have graphs in its datasheet, showing noise voltage and noise current vs. collector current. You find the point on this graph where En/In = the impedance of your source, and you're done.

If you buy a commercial mic preamp on a chip, odds are that it's optimised for 200 ohms or 1k. Moving-coil phono head amps are designed for lower impedances, say 10 ohms. It gets harder to get any lower without a step-up transformer: ribbon mics are the lowest impedance transducers I know of, and I've never seen one that didn't use a transformer.

I think the physics of this pickup are about the same as a ribbon mic, so I still think the original idea with a current transformer was the best one.

[Joe Gwinn]

This is an old myth. Padding the source with a resistor NEVER improves signal-to-noise ratio. The Johnson noise of the resistor always hurts you more than the "better impedance match" helps you.

Well, I've seen the contrary analyses, so I don't know that it's a myth. But I don't know that the gain is enough to notice in practical systems. It all depends on how non-ideal the transistors are.

Transistors generate noise voltage and noise current. For an application like this, we would get a bunch of large transistors and run them at a high collector current.

This gives you low voltage noise, because the transistor's internal noise-generating "resistances" scale down with increased die area and increased collector current. But it gives you high current noise, because the base bias current is increased. That's OK though, because the low impedance of the source can swallow lots of noise current.

Yep. In common-base configuration, the input impedance depends on the total area of the emitters and the emitter current, and the larger these are the lower the impedance.

This is how you tune transistors for best noise performance. The higher the source impedance, the lower the collector current you need to run. Any good transistor intended for low-noise amplification will have graphs in its datasheet, showing equivalent input noise voltage and current vs. collector current. You find the point on this graph where noise voltage divided by noise current = the impedance of your source, and you're done.

This is equivalent to choosing the impedance at which the effects of current noise and voltage noise are equal, which is usually the best balance one can find.

If you buy a commercial mic preamp on a chip, odds are that it's optimised for 200 ohms or 1k. Moving-coil phono head amps are designed for lower impedances, say 10 ohms, and they all use a discrete front end: I've never seen one on a chip. It gets harder to get any lower without a step-up transformer: ribbon mics are the lowest impedance transducers I know of, and I've never seen one that didn't use a transformer.

Ribbon mics had their hay day in the 1950s, when vacuum tubes were king, so a transformer would have been essential. Ribbon mics have the same impedance level as the double-string pickups considered here.

It may be that the best solution is a transformer feeding a common-base amplifier, as the lower the transformer ratio the less the effect of various parasitics.

[Mike Sulzer]

I think we are getting a bit far from one important question: Is the SNR better with common base or common emitter. I say they are the same, you said above: "Impedance match is better with the common base configuration, so the resulting SNR is better, even though you are right about the inherent noise of the transistor."

I am not sure if you have changed your thinking on that, so let's review the situation. Start with a transistor stage with a resistor Rc in the collector, biased properly with both the base and emitter at ac ground. This circuit has some noise output. If we put an ac voltage source in series with either the base or emitter, the magnitude of the signal output is the same in the two cases. This situation does not change significantly if a low value resistor is inserted in series with the voltage source. Low means small compared to any resistor in the model of the transistor such as the input impedance looking into the emitter Clearly the SNR is the same in both cases. This is the case that applies to the situation under discussion, and so either type of stage can be used with no significance difference.

Now, consider the question of whether "the best impedance is zero". In demonstrating this, I used a model and some very simple analysis. To refute this, it is sufficient to show that either the model or the analysis is wrong. I believe that you have stated that the model is wrong; that is, that a transistor is too complicated to be represented in this way. Well, this is the model used by the maker of many commercial chips to describe the performance of their product, op amps and so on. If it is wrong, then the noise analysis engineers do is wrong.

No transformer involved.

This maximizes the signal, and assumes that the SNR is thereby maximized. This may or may not be true.

Not necessarily. Transistors are not linear devices, and the self-generated noise depends in a complex way on the source impedance presented to the input. So, it does not follow that the best impedance is zero, where the series-resistor noise is minimized and the signal voltage is maximized.

The theory is too complex to be worth the trouble to compute answers. Quickest way to tell is to experiment. But use low-noise (metal-film) series resistors, or a 50-ohm wirewound pot. (Do not use any form of carbon resistor.) It may also turn out that the series resistor value does not make much difference.