Realizations with transistors and capacitors in an emitter follower

Naaah! It was probably the hole current ;-)

...Jim Thompson

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

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.

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

That's a minority opinion.

John

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,

...Jim Thompson

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.