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Re: Speaker Mic - Common Base Config

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

Reply to
Warren
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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

Reply to
Warren

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

Reply to
Warren

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.

Reply to
krw

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

Reply to
Warren

Yes, and all completely irrelevant.

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

Reply to
krw 

--
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.
Reply to
John Fields

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

--
                  [On the Road, in New York]

| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

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

Reply to
NT

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

Reply to
Warren

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

Reply to
Warren

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.

Reply to
krw 

--
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
Reply to
John Fields

--- 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

Reply to
John Fields 

--
JF
Reply to
John Fields

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.

Reply to
krw

Different question; different answer. What a dumbass.

Reply to
krw 

--
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?
Reply to
John Fields

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

--
                  [On the Road, in New York]

| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

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.

Reply to
krw

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