Realizations with transistors and capacitors in an emitter follower

M. Hamed · 2008-02-24T19:21:59+00:00

In playing with my simple emitter follower circuit trying to figureout 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 baseis biased to midpoint through a voltage divider and the input iscapacitively coupled. The output is also capacitively coupled througha large capacitor to block DC. The emitter resistor is 5K.I expected that with a large capacitor, even an 8 Ohm speaker would besuitable to place the corner frequency of the high pass filter lowenough to pass audio frequencies and high enough to block DC. HoweverI was faced with two anomalies. Whenever I connect my speaker afterthe capacitor, I hear a very faint and distorted sound. Whenever Iconnect the speaker before the capacitor (the NPN's emitter), I get analmost exact replica of the input (which is what the emitter followeris supposed to do). Simulation confirmed this behavior but I waspuzzled with the result. Do I throw out of the window all I know abouthigh pass filters and frequency response. Also do I throw out theimportance of bias? Here we have a high pass filter that is not actinglike a high pass filter. And we have a low resistance emitter resistor(after connecting the speaker in parallel with the 5K emitterresistor) that should throw the bias point way off midpoint.After playing with simulation for a while, and replacing thetransistor with equivalent circuit, etc, I hit upon one of my biggestrealizations in electronics (mind me if it's too obvious for you). Themodels are correct so long as the transistor is ON and operating inits linear region. In such a case the high pass filter will act nicelyand block DC and pass the audio signal. However, with such lowresistance, If the output to follow the input faithfully, huge currentwill have to pass in the capacitor for the slightest negative outputvoltage, and since the current in the emitter resistor (5 KOhm) isvery small compared,...

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John Popelish 18 years ago

An NPN emitter follower turns off if its emitter is not pulled negatively enough by its load to keep the base-emitter junction forward biased enough to keep current flowing through that junction.

When M. Hamed's input signal swung negative, the output capacitor held the emitter very near the zero signal voltage, while the base swung negatively. This base swing eliminated the base-emitter junction forward bias, and turned the transistor off, leaving only the feeble 5k pull down resistor to slowly discharge the output capacitor, while passing a tiny current through the 8 ohm speaker in series with them.

Regards, John Popelish

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Jim Thompson 18 years ago

Naaah! It was probably the hole current ;-)

...Jim Thompson

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lcorbin 18 years ago

Thanks very much to John and Jim for their kindly, helpful replies. I only realized belatedly that I should have posted to a beginner's list (I'm not really up to speed even with Yahoo groups). Later, after seeing some of the incendiary and even profane posts, I thought I was going to be rudely dismissed. So thanks again for the civility and help!

Yours, Lee Corbin

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John Popelish 18 years ago

Much of the rudeness was in fake posts (posted by someone other than the sender shown) in an attempt to cause trouble. You always have to watch out for trouble makers who get no other pleasure from life.

Regards, John Popelish

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John Larkin 18 years ago

This newsgroup can be a bit coarse, but it's mild compared to some. Anyone, no matter how big a jerk or fathead, can post here. So don't take it personally.

John

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John Larkin 18 years ago

That's a minority opinion.

John

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ShamShoon 18 years ago

Hi John. If I may interject here, correct me if I'm wrong. The problem to me seems more current related than voltage related. The base of the transistor never goes negative because it's riding on top of the DC bias signal. The same for the emitter, it never goes to 0, unless you were talking in small-signal terms. So it seems to me that once the emitter goes below the bias point, the transistor can not pass the current in the reverse direction (neglecting the small current passing in the 5k), so as the output voltage goes very slightly below zero, the emitter current decreases gradually (by KCL) and that causes Vbe to decrease (by Ebers-Moll relation), until it's totally zero. So my understanding here is that in this case, it's the current decrease that led to forward bias decrease until disconnection.

Does that make sense?

Regards,

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Jim Thompson 18 years ago

...Jim Thompson

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John Popelish 18 years ago

They are interconnected. Voltage drives current.

The base voltage swing is driven by the input signal. The emitter voltage is a result of the current through the transistor dropping voltage across the emitter load. Once the base emitter ceases to be forward biased, no current passes through the collector-emitter path, so the emitter load is left floating (with the transistor effectively disconnected from it) and does not follow the base voltage any more negative.

Regards, John Popelish

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Realizations with transistors and capacitors in an emitter follower

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