Realizations with transistors and capacitors in an emitter follower

On Feb 24, 11:21 am, "M. Hamed" gave a very clear description of a phenomenon he had observed. I was wondering if you could answer a probably easy basic question for me. The clipping that you observed was related to the transistor turning off? If an NPN transistor, in an emitter follower configuration say, does not get enough current, will that alone cause it to turn off? In other words, perhaps the emitter voltage can go so low that not enough electron current is drawn through the emitter resistor to keep it on?

Thanks very much.

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.

Naaah! It was probably the hole current ;-)

...Jim Thompson

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.