Re: Speaker Mic - Common Base Config

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Warren 14 years ago

Fred Bloggs expounded in news: snipped-for-privacy@d18g2000yqm.googlegr oups.com:

>> Fred Bloggs expounded

..

> Since current will affect the voltage across the resistor,
>> this 80% loss isn't correct AFAICS. If the 8.2 ohm
>> resistor was exactly 8 ohms, I'll bet the simulation will
>> more exactly show a difference of 50% in gains. I'll have
>> to try this tonight.
>>
>> Warren
>
> The small signal vbe is accounted for with rE. The speaker
> supplies all the signal current, and only 20% of it makes
> it into the collector- emitter path. The CB merely
> facilitates the transformation of signal current
> circulating in a low impedance input circuit to circulating
> in a high impedance output circuit thus providing voltage
> gain.

Ok, I see it now. So....

| |/ -| |>

| | .-. 8 ohms | | | | Rs '-' | | __ /| -| | | -|__| | | \| | GND

by putting the speaker in the emitter circuit, some signal is lost in its source resistance of 8 ohms (20%), while the remainder of the signal forms across re of 31 ohms (80%).

The difference is then 80% - 20% => 60%. So the coupled circuit should produce a gain of 60% less. This agrees with the difference in simulation gains of 78 (spkr inline) vs 46 (spkr coupled).

Cool. I like it when the calculator is right.

Warren

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Warren 14 years ago

Rich Grise expounded in news:j3pfoo$o9r$ snipped-for-privacy@dont-email.me:

You said something about not seeing that kind of circuit before (using speaker in CB config). I simply said implied that intercoms now use opamps. Where's the beef in that?

Warren

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Warren 14 years ago

NT expounded in news: snipped-for-privacy@l4g2000vbv.googlegro ups.com: ..

So you would advise putting 2 ohm speakers on your 8 ohm stereo then?

Or perhaps you'd rather have 32 ohm speakders on a 4 ohm stero?

No matter what you wanna call it, both scenarios are suboptimal.

Warren

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krw 14 years ago

Do you really think the output impedance of audio amplifiers is 8 ohms?

Do you really think the output impedance of audio amplifiers is 4 ohms?

Clueless.

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Warren 14 years ago

snipped-for-privacy@att.bizzzzzzzzzzzz expounded in news: snipped-for-privacy@4ax.com:

..

You're completely missing the point. I guess I have to lay it out for you:

Rs=0.75 ohms ___ |-----|___|------| | | | | | | --- | - 1.5V .-. . | | | | | RL | '-' | | |------|---------| | GND

Take a 1.5V battery with a source impedance of 0.75 ohms.

The optimal power transfer occurs when RL, i.e. the "load" is exactly the same as Rs, which is 0.75 ohms in this example.

The power transferred to the _load_ is 0.75 Watts. Do the math.

Now change RL to 1 ohm. P(RL) becomes 0.735 Watts.

Change RL to 1.25 ohms. P(RL) becomes 0.72 Watts.

The optimal power transfer to the load occurrs when RL = Rs = 0.75 ohms.

Impedance matching is the point when _power_ is involved.

I don't give flying fart if the speaker impedance is not exact 8 ohms at every frequency. Look at the higher principle being discussed.

Warren

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krw 14 years ago

Yes, and all completely irrelevant.

...and you're 100% wrong. Are you an engineer?

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John Fields 14 years ago

The concept of impedance matching as far as audio power amplifiers (and some others) goes is irrelevant because an audio amp is a voltage source with an output impedance of a few milliohms. As such, the load impedance is specified because the maximum power the amp can deliver will depend on how much current it can force into that load. For example, suppose an amplifier is rated to deliver 100 watts into an 8 ohm load. Then, since P = I²R, P 100W I = sqrt --- = sqrt ------ = 3.54 amperes R 8R and, since P = IE, the voltage needed push 3.54 amperes through that 8 ohm load will be: 100W E = ------- ~ 28V 3.54A Now, consider that same amplifier with a 16 ohm load on its output. Then the output into the load will be: E² 28V² P = --- = ----- = 49 watts, R 16R The reason being that since it's rated for 100 watts into 8 ohms, the output voltage won't go much beyond 28V. On the other side of it, if an 8 ohm load is connected to the amplifier and the amplifier is cranked, that 28 volt output will try to push 7 amperes into the load, but since the amplifier is only rated for 3.5 amperes into an 8 ohm load, the 4 ohm load will severely overload it.

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Jim Thompson 14 years ago

Which is why audiophools are properly named ;-) This source matching topic keeps recurring... it never sinks in :-( ...Jim Thompson

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NT 14 years ago

snipped-for-privacy@4ax.com:

Honestly, you've got it wrong here. Its easily demonstrated: take any transistor amplifier 1970s or later, put a signal thru it, measure Vout, then attach 8ohm speakers. Vout wont have changed.

NT

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Warren 14 years ago

snipped-for-privacy@att.bizzzzzzzzzzzz expounded in news: snipped-for-privacy@4ax.com:

Perfectly relevant- I'm talking about power transfer. Did you forget already?

Forget the audio output example- a bad example choice on my part. Shoot me for that, but the discussion you keep derailing is about POWER TRANSFER. Do I need to say it again?

THIS battery example does in fact work and is valid, and the numbers prove it.

If all you want to do is have a pissing contest, then I pull the plug.

Warren

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Warren 14 years ago

John Fields expounded in news: snipped-for-privacy@4ax.com:

...

..

Thanks for the thoughtful explanation. Indeed, I chose a bad example.

The 2nd example with battery is the idea I was attempting to discuss, until the constant derailing.

If that makes me an audioph00L, then so be it. ;-)

Warren

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krw 14 years ago

Wrong. You don't know *what* you're talking about.

You don't understand the first principles.

It has nothing to do with "power transfer". The subject is maximum power to the load. Assuming an 8-ohm (or whatever) load, the maximum power delivered to it is *NOT* when the source impedance is 8-ohms, rather *ZERO* ohms. With an 8-ohms source, *HALF* the power is dissipated in the output of the amplifier. The is a good reason why modern audio amplifiers have a (next to) zero-ohm output impedance.

It may prove "it", but "it" has nothing to do with the issue at hand.

You're wrong. Run in ignorance, if you must.

You didn't answer the question, so I'll assume others are right. You're not an engineer, rather an audiophool. If this is wrong, get a refund from your university.

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John Fields 14 years ago

It doesn't. :-) Another example of what you were talking about is an RF signal generator with a 50 ohm output impedance which, with no load attached, will output, say, 2VRMS. Then, when a 50 ohm load is connected to it, that voltage will drop to 1V, and the maximum power possible will be transferred to the load. Above and below that impedance, the power into the load will fall: Looking at the circuit like this: (View in Courier) EG | [RG] | +---EL [RL] | GND Makes it easy to see that it's just a 2 resistor voltage divider where EG * RL EL = --------- RG + RL With EG being equal to 2V, and by changing RL in 10 ohm steps from 10 to 90 ohms, the following table can be made: RG RL EL PL OHMS OHMS VOLTS WATTS 50 10 0.33 0.0109 50 20 0.57 0.0163 50 30 0.75 0.0188 50 40 0.89 0.0198 50 50 1.00 0.0200

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John Fields 14 years ago

--- Actually, he's right, since an amplifier with a near-zero output impedance will output maximum power into a near-zero load of the same value.

Looking at the circuit like this:

. EG . | . [RG] . | . +---EL . [RL] . | . GND

makes it easy to see that it's just a 2 resistor voltage divider where

. EG * RL . EL = --------- . RG + RL

With EG being equal to 1V, and by changing RL in 1 milliohm steps from

1 to 9 milliohms, the following table can be made:

. RG RL EL IL PL .MILLIOHMS MILLIOHMS VOLTS AMPERES WATTS . . 5 1 0.167 167.000 27.889 . 5 2 0.286 143.000 40.898 . 5 3 0.375 125.000 46.875 . 5 4 0.444 111.000 49.284 . 5 5 0.500 100.000 50.000

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John Fields 14 years ago

JF

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krw 14 years ago

Good God, you're obtuse today. No, I guess today isn't any different than any other.

Completely irrelevant bullshit. The assumption was an 8-ohm speaker.

You've fixed the output impedance, so you are looking for maximum power transfer. Of course, if you change the problem the answer is different. I wouldn't expect you to understand that any more than the audiophool.

Five is not zero. I can play stupid too, if that's what it takes to communicate with you.

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krw 14 years ago

Different question; different answer. What a dumbass.

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John Fields 14 years ago

If you wanna start name-calling, then all that means is that you know you've lost the argument and, instead of admitting defeat by acquiescing, taking the high road and taking what you've learned away with you gracefully, like a gentleman, you complain that your defeat was due to the chessmen being moved to identical places on a different board. But, with respect to the rest of your post; The man admitted that he had used a bad example earlier and submitted his battery-with-a-fixed-output-impedance-and-different-load-impedances example to prove the point he was making, and he was right. My example, above, was just to generate a few more data points in order to show that his claim was not in error, as it certainly seems not to be. Can you refute the data?

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Jim Thompson 14 years ago

"krw" has descended into Larkin-land :-( ...Jim Thompson

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krw 14 years ago

What a jerk.

No, if you knew anything about electronics, or logic for that matter, you would know that you're twisting in the wind.

Of course he used the bad example. He used the wrong "model" for the system of interest, too. He's hung up on "maximum power transfer", without understanding it. Just as you are.

It *IS* in error, dumbass! Maxumum power transfer has *nothing* to do with the question at hand.

The data is not the issue. Your poor grasp of electronics, and logic, is.

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