Realizations with transistors and capacitors in an emitter follower

In playing with my simple emitter follower circuit trying to figure out why there's very faint output when connected to my 8 Ohm speakers, I came upon some important realizations.

My circuit is a simple emitter follower with one transistor. The base is biased to midpoint through a voltage divider and the input is capacitively coupled. The output is also capacitively coupled through a large capacitor to block DC. The emitter resistor is 5K.

I expected that with a large capacitor, even an 8 Ohm speaker would be suitable to place the corner frequency of the high pass filter low enough to pass audio frequencies and high enough to block DC. However I was faced with two anomalies. Whenever I connect my speaker after the capacitor, I hear a very faint and distorted sound. Whenever I connect the speaker before the capacitor (the NPN's emitter), I get an almost exact replica of the input (which is what the emitter follower is supposed to do). Simulation confirmed this behavior but I was puzzled with the result. Do I throw out of the window all I know about high pass filters and frequency response. Also do I throw out the importance of bias? Here we have a high pass filter that is not acting like a high pass filter. And we have a low resistance emitter resistor (after connecting the speaker in parallel with the 5K emitter resistor) that should throw the bias point way off midpoint.

After playing with simulation for a while, and replacing the transistor with equivalent circuit, etc, I hit upon one of my biggest realizations in electronics (mind me if it's too obvious for you). The models are correct so long as the transistor is ON and operating in its linear region. In such a case the high pass filter will act nicely and block DC and pass the audio signal. However, with such low resistance, If the output to follow the input faithfully, huge current will have to pass in the capacitor for the slightest negative output voltage, and since the current in the emitter resistor (5 KOhm) is very small compared, this current would have to follow in the transistor in the opposite direction, so the transistor turns off. That's my reasoning for why the transistor would turn off.

So now we have another mode of operation. Now it's another circuit. It's not a high pass filter but a capacitor discharging through a resistor. At a certain point the capacitor charge will be large enough to have the transistor turn on only very briefly and the output voltage change so little.

What caused my confusion is that in most electronic circuits, there are transistors but often the circuit description doesn't tell if that transistor will be ON all the time or not. Then analysis follows that replaces caps with shorts at signal frequencies etc. But that's valid only if the transistor stays on and hence acts as a linear device. I wonder if circuit designers even with all their experiences and rules of thumb take the extra step of analysis (or simulation) to make sure the transistor is always operating in the linear region. If they do, it's often not mentioned to us, less experienced people. In other applications, such as rectifiers, the charging and discharging is explicitly mentioned and used. This dual role of the capacitor always puzzled me, when to think of it as a charging/discharging device, and when to think in terms of frequency response and frequency dependent impedance.

The other question was why connecting the speaker directly to the output is OK. Why no blocking capacitor is needed? It turned out that the small impedance shifts the DC bias point very close to ground, and hence the output is a replica of the input but shifted up in DC level just a few millivolts above ground, which seems to not alter the speaker response much. That seems to be a problem if the gain is high and clipping will start to happen as you move away from midpoint, but when the gain is 1, no problem since the input signal is already a few millivolts around 0.

A possibility here is that the speaker also responds to changes in magnetic field and hence the DC component wouldn't matter, but i'm not sure about that one.

I apologize for anyone who took the effort and went through my long post. Am I right in my conclusions?

Thank you.

Reply to
M. Hamed
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Not at all. You just have to remember that those concepts apply to linear circuits, and your circuit effectively has a rectifier in series with the output (the base emitter junction).

Yes, the rectifier has switched the transistor out pf the output circuit.

That is incremental AC analysis or operation near a bias point. For many circuits, including this one, you have to check the extremes of the signal swing, to make sure the operating bias has not severely shifted.

Linear versus nonlinear circuits. Filters involve linear assumptions. Charge/discharge processes involve nonlinear processes. This changes the rules and the tools. Put away the paint brushes and get out the socket wrenches, so to speak.

Connecting the speaker directly to the emitter (as a DC load) adds DC current to the speaker, in addition to the AC current you intend. This distorts the position of the speaker cone, adds greatly to its heat production, and also heats the transistor or distorts its operating point. If you can find a tolerable compromise for all this, you might eliminate the output capacitor.

I think your understanding is much better than it was when you first built the circuit. You may now be in a mental state to appreciate why the complementary emitter follower was invented. (at bottom of:)

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--
Regards,

John Popelish
Reply to
John Popelish

Good grief, of course we do. It's second nature.

The emitter follower can only source current, not sink it. It can't conduct less than zero current, so there's nothing but the wimpy 5K resistor available to drive the speaker in the negative direction.

If you reduce the value of the emitter pulldown resistor, then you can get a little more current into and out of the speaker in both directions. But that's a losing game, which is why real power amps usually have push-pull output stages.

Connecting the speaker directly to the emitter works at low levels, but is inefficient and won't scale.

John

Reply to
John Larkin

[snip]
[snip]

Who is this "M. Hamed", some fricking amateur (that I've previously plonked)?

I even utilize macros built into PSpice to check that CMOS operation is in the saturation region, since it's often non-obvious.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave
Reply to
Jim Thompson

[snip]

This is the key concept ("the emitter follower can only source current...").

Read those paragraphs over and over until you think you understand them. Then, read them over and over again until you do understand them.

Bob

Reply to
BobW

In 40 years of electronics, I have never simulated anything. Unless you count breadboards as simulation :)

Reply to
Sjouke Burry

Of course you haven't. You also haven't designed a chip with 10,000 transistors on it either ;-)

I cut my teeth on paper (and breadboards)... pretty much why I'm so successful now... and why young bucks ask, "Why'd you do it that way?"

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave
Reply to
Jim Thompson

In the first case, the cap isn't discharging properly, so you end up with the input to the cap pegged at Vin(max)-Vbe. This means that the output transistor can't actually put any power through the cap, because it's voltage isn't varying.

The way to understand this is to figure out what the impedance is that the speaker sees when the transistor is on, vs the impedance it sees when the transistor is off. In the first case, it sees something like 1/10 the input impedance. In the second case, it sees 5k (since the transistor is off, so all you see is the emitter resistor).

As a consequence, the circuit acts kinda like a diode, where the cap can be charged when the transistor is on, but can't be discharged when the transistor is off. Current in, no current out means it pumps up to the maximum voltage.

No, this will happen with a real circuit too. The problem isn't the simulation, it is the circuit.

Right.

You need another transistor to pull down on the cap. The typical output driver needs both a way to pull up, and a way to pull down.

Connecting the speaker directly means you end up with a DC current through it all the time of (Vcc/2 - Vbe)/R

Where R is the DC resistance of the speaker. Not good. However, the transistor gets to vary the current on top of the bias, which causes the magnetic field to vary, causing the paper to move.

You are thinking too hard.

Try this circuit instead:

-------------------o------------. | | | | .-. | 470 | | | | | | '-' | | | | |/ o----------| NPN | |>

V | - | || | | || __ /| ------||--------o o----||--------| | | || | | || .--|__| | V | | \\| 1uF - | 10uF | | |< | o----------| PNP | | |\\ === | | GND | | .-. | 470 | | | | | | '-' | | | ------------------o------------' (created by AACircuit v1.28.6 beta 04/19/05

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It still sucks, but it sucks far less than your circuit.

Filters are interesting, but that isn't your problem here. Notice that when the input goes down below the midpoint, that causes the NPN to turn off, and the PNP to turn on. This is also a 'follower' of a kind, it just has a more balanced impedance in both directions.

Regards, Bob Monsen

Reply to
Bob Monsen

If you want it to work rather better, Use 2 input caps, commoned at the input side and connect one to each base.

Stabilising the quiesecent current with some emitter Rs wouldn't go amiss either (use an extra diode in the biasing network).

Graham

Reply to
Eeyore

We have been properly chastized for not mentioning this to less experienced people. Let's try to be more careful in the future.

John

Reply to
John Larkin

Simulation can be neat, when nonlinear behavior exceeds reasonable math analysis, or to train your instincts as regards causalities.

Breadboarding is good when device models are unavailable or flat wrong, but can bite you on production tolerances.

John

Reply to
John Larkin

Sno-o-o-o-ort ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave
Reply to
Jim Thompson

We're all amateurs in the school of life, aren't we? one way or another.

At this point, I don't give a **** (use your imagination) what you do with your PSpice

For an attitude like yours, I wouldn't post my company's name. Not good for the business.

Reply to
M. Hamed

You probably couldn't get online without using one of Jim's 'Pspice' IC designs. What have you done, other than act like a spoiled brat?

--
Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
Reply to
Michael A. Terrell

That's unrelated. You're linking the amount of knowledge some people have about a certain topic to the amount of abuse they can give others.

I come from a Software/FPGA background. I only had one Electronics course in school and the teacher was crappy and I am self teaching myself about Analog Electronics. I could tell you what I have done in my areas of specialization and how many people I've helped, but that doesn't add to my point. Give me 40 more years and I'll have more to brag about.

Being around long enough to know how long this thread can grow wasting thousands of people's time, this is my last word on the subject.

Reply to
M. Hamed

M. Hamed wrote: (snip)

I suggest that you just ask your next question in the basics group, and a lot of this attitude you are getting will evaporate. The same people watch both groups, but often behave differently in them. The design group was created as a place where designers hash out esoterica and discuss grand design issues (or at least that is what they like to think). Posting there implies that you think you are a designer. It is sort of a club. Do you see?

--
Regards,

John Popelish
Reply to
John Popelish

If you use a 5K speaker, the thing might work ok, but not with an 8 ohm speaker and 5K emitter resistor. How can you ever get much current, or power through a 8 ohm speaker in series with a 5K resistor? Might work at 200 volts.

Solution is to use a transformer in place of the 5K resistor and bias the stage so the idle current and supply voltage represents about twice the output power. Or, maybe that's 4 times the power, I forget. Anyway, for a 1 watt output using a 6 volt supply, the bias current will be somewhere around 500mA. The transistor will dissipate most all the voltage times the current, or maybe 3 watts, so you need a heat sink. The secondary of the transformer is connected to the speaker. It's called a 'class A' amplifier and has very good audio quality. Play around with the bias current to see how much power you can get out.

-Bill

Reply to
Bill Bowden

AoE stated that the output impedance of an emitter follower is almost independent of the emitter resistor. It's hard for me to think of another model in which the 5k plays a role. Perhaps a voltage source in series with a diode then the source & diode combo in parallel with

5k & 8Ohm resistors.

How do I know the output power? The speaker I want to drive?

Could you elaborate on how you came up with this value? I'm thinking the transformer would make and impedance transformation and so the 8 Ohm across the secondary would appear higher across the primary, and current will drop. 500mA sounds to me like a very high current.

3 watts, so you need a heat
Reply to
ShamShoon

Last thing I did in my quest for learning about simple amplifiers was the common emitter amplifier. The output is opposite of polarity to the input. I thought that would be fun to try to have the voice reversed. However after trying it I realized that speakers don't really have a polarity (despite the clear + & - signs on the speaker) which makes sense. The output of the CE amp was very good. I had a blast when I added a potentiometer in place of the collector resistor, and used it to amplify my little Mp3 player signal before it gets into my car. It used to be not powerful enough to drive the car sound system and I had to crank the volume all the way up.

No more! this simple circuit worked and it worked beautifully! Now I have an amplifier with volume control and very simple, one transistor, one OpAmp, and a few resistors and capacitors, and a battery. Sound quality is not bad but certainly much better than I expected with such a simple configuration. The only thing I need now is to figure out how to reduce the power (actually I need to calculate it), because I don't want it to drain the battery quick!

Reply to
ShamShoon

Go learn what class B or class D amplifier is.

Bob

Reply to
BobW

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