Amplifier transistor matching?

No, being 'informed' doesn't make an engineer pick a more difficult analytical approach to quantitative circuit design; quite the opposite. 90% of the time, a simple 0.6Vbe-plus-Beta model is the easiest and *most reliable* way to design transistor circuits.

This is sci.electronics.design, not sci.device.physics. If I had to analyze the physics of every component I use, I wouldn't get anything done. What matters is how they behave.

That's like saying that poison doesn't cause death because only heart failure *really* causes death.

John

Reply to
John Larkin
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Mr.T wrote in sci.electronics.design:

Perhaps because Tom Lehrer's text is different:

... If you're looking for adventure of a new and different kind And you come across a girl scout who is similarly inclined, Don't be nervous, don't be flustered(?), don't be scaaaared! Be prepared!

Anno

Reply to
Anno Siegel

Then I owe somebody about $200 million.

Because the beta model doesn't address current sharing of base currents when transistors are paralleled. Discrete circuit designers rarely (as in, never) hard-parallel bipolar transistors anyhow. If they ever elected to do so, they would have to analyze the situation properly.

Nobody can afford to design without regard to min/max beta specs. Not even you.

Well, maybe yours doesn't, but that's just the equations you prefer, to make your point.

So, if A causes B, and B causes C, you conclude that A does *not* cause C. OK, can't argue any more about that.

John

Reply to
John Larkin

Hello Jim,

The Sennheiser system does that nicely. Currently we have four but I can see it going to 6-8. However, we'll only do that with AA battery powered gear, no more 9V.

On our current gear the transmitters show their frequency on a little LCD, same for the receivers. Receivers also show RF level and audio level. The latter makes it really easy to see if a mike is active without the mixer pot turned on.

Regards, Joerg

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Reply to
Joerg

What, no attribution for Tom Lehrer?

MrT.

Reply to
Mr.T

The voltage controlled model is the simple model, and the correct one.

Not at all. This is not true in the slightest.

Please explain, as discussed in this thread, from a beta perspective, why a hfe mismatch of 2 can result in say, 10 times mismatch in current (or more) for || devices. Why is it not just a factor of 2?

Hint: Rbb' drops a voltage to Vibe, Ic = Io.exp(Vbi/Vt)

Its not about the physics is about understanding how to design transistor circuits correctly. To do this, one must understand that the transistor is a voltage controlled device. The why is irrelevant. Sure, for a simply switch one might just work out the base current needed to saturate the device, but for any design that isnt trivial, the beta model is, essentially, useless. The first order model for gain doesn't even have a beta term.

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To wit,

Av = RL/re = 40Vdc

Av max = Va/Vt.

No it isn't. Base current is not equivalent to poison, its equivalent to the body stinking after the heart failure. i.e. its just a nuisance.

This is really all getting a bit tiresome. I have explained this many times. Base current is an effect caused by an application of voltage. Without understanding this basic fact, it is impossible to *design* an amplifier correctly. All one can do is piss about under the illusion that there was a "design" performed. Period.

Kevin Aylward snipped-for-privacy@anasoft.co.uk

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SuperSpice, a very affordable Mixed-Mode Windows Simulator with Schematic Capture, Waveform Display, FFT's and Filter Design.

Reply to
Kevin Aylward

People believe they do.

As do the voltage controlled equations.

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It is the correct way. Once you get to really designing complex transistor level circuitry, it will become obvious. All basic ac understanding is based on gm, that is, vo=gm.vb

Its what real professional analogue designers do that actually understand transistor level design. Unfortunately, they are many that only believe they understand. They dont.

I have given the basics in the above link.

Kevin Aylward snipped-for-privacy@anasoft.co.uk

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SuperSpice, a very affordable Mixed-Mode Windows Simulator with Schematic Capture, Waveform Display, FFT's and Filter Design.

Reply to
Kevin Aylward

There is no practical example on that web page, and it surely does not look like a simpler or faster way to design a transistor stage.

--
 Roger J.
Reply to
Roger Johansson

Also, try to develop a connection with local hospitals and clinics. A friend of mine used to get cases of alkalines (9V, AA, whatever) free.

Apparently, for medical use, they must be discarded by law when they meet the expiry date. They were still good as new.

Mike Monett

Reply to
Mike Monett

You must be switching a lot of relays;-)

Thats still missing the point. No one is suggesting this.

But you cant do this without using the exponential nature of Ic verses vbe, i.e. the inherent voltage controlled nature of bipolar transistors. That's why the matching goes off tremendously. As I noted, 50ma at 5 ohms is 250mv. This is huge as the exponent of the exponential is

250/25.

That misses point as well. The voltage controlled model does not ignore beta in the slightest, so no, I don't ignore it. I have already pointed these interested to the relevant paper that shows how beta is included.

Indeed, in SS, worst case analyses is automatic with button presses. It has decent defaults for all main parameter variations, including max/min beta.

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Its the way the transistor works, and its what you need to do if you are designing non switching circuits.

Wrong analogy. I have explained many times that is *electric field* that causes charges to move. Period. It is why it is often referred to as an

*accelerating* potential. It tries to accelerate charges. Base current does not in any way try to cause an acceleration of emitter charge. This is so bloody simply. F=qE. End of story.

Look, this *is* how it is.

Apply a voltage to a diode. A current is generated, to wit:

Id = Io.exp(Vd/Vt)

This current will flow through that junction irrespective of how that voltage gets to the junction.

Now add a junction for the collector. *Nothing* bloody well changes. The voltage at the base is the same voltage as it was in the stand alone diode, therefore that junction current will still be set by that voltage. However, in this case, most of the emitter diode current gets sucked up into the collector, not the base. The base terminal is just a convenient way to impress a voltage across a diode junction, but without actually supplying the current for that junction.

The above is how one really needs to think about basic transistor operation.

Kevin Aylward snipped-for-privacy@anasoft.co.uk

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SuperSpice, a very affordable Mixed-Mode Windows Simulator with Schematic Capture, Waveform Display, FFT's and Filter Design.

Reply to
Kevin Aylward

Hello Laurence,

That would be 30c US. Really?

Regards, Joerg

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Reply to
Joerg

Agree. At the same gig where we've gone through literally tens of thousands of AAs in wireless mics with not a single problem, we've had probably a 1% failure rate in 9V batteries for the few wireless units we have that take those. Same thing as what you describe - they test okay, and then drain within minutes.

I assume it's because of the multi-cell internal construction of a 9V.

Reply to
Walter Harley

The theatre technicians here in London UK seem to be able to source alkaline 9v batteries at about a third of the retail price of a Duracell.

Reply to
Laurence Payne

Hello Mike,

That is a good idea. Although nowdays they can have over 5 years of shelf life. Yesterday I replenished the lab with AA and they expire in May 2012.

Another source that was mentioned here or on a.b.s.e. are photo processors. Apparently disposable cameras often contain an AA or AAA cell with lots of life left in it.

Regards, Joerg

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Reply to
Joerg

Hello Walter,

That may well be the reason. The first time I looked at how they are internally connected my confidence level dropped a lot. Statistically the six cells of a 9V must have a higher combined failure rate than two AA cells. But not by this much, there must be something else that causes failures.

The contacts on a 9V are a pain as well. How many times has something ripped out when disconnecting a 9V battery? A lot...

Regards, Joerg

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Reply to
Joerg

Which word?

MrT.

Reply to
Mr.T

A good reason is because a 9V can't source as much current as even a single AA cell for the same duration due to their much smaller individual cells. There are six seperate 1.5V cells connected in series to produce the 9 volts. The cells internal electrodes have significantly less surface area to interact with a much reduced quantity of electrolyte. If you still need 9 volts for a circuit try ganging six AA's in series. If weight is a concern, even AAA cells aranged in a battery of 9 volts will outlast most 9V's.

Regards, Chris

Reply to
BCTweaker

I once slapped together a textbook common-emitter amp because I needed a quick-n-dirty mic preamp. I used a constant .6V B-E drop and picked bias resistors based on beta, albeit swamped out my expected base current by about a factor of 10.

Sorry if I did it "wrong", but I got it done in 10 minutes, and it worked. ;-)

Cheers! Rich

Reply to
Rich Grise

And cell-phone manufacturers regularly plumb the depths of low-res to see how much we will put up with! :-)

Reply to
Richard Crowley

Years ago, I heard a demo of the human voice with 1-bit resolution. It was understandable, but not at all pleasant.

Best regards, Spehro Pefhany

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Spehro Pefhany

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