> From: Peter Chadwick <Peter.Chadwick@gpsemi.com>
To: <amps@contesting.com>
> Date: Wed, 3 Sep 97 14:02:52 +0000
> I still need convincing.
The important part is agreeing the Thevenin and Norton models can't
be used to desribe anything inside the source, excepot what it looks
like at its terminals.
> Let's take a class A transistor to start with. Assume that the collector
> volts can swing from 12 down to 1 and up to 23 because of the
> transformer. That's 22 volts peak to peak, or 7.78 volts rms. Now assume
> that the load that the transformer refers back to its primary is 10
> ohms. The output (assuming the xfmr is lossless) is 7.78 squared
> divided by 10, or 6.05 watts. The collector current is 2.2 amps peak,
> and we'd bias the baby at 1.1Amps in theory, and about 1.25 Amps in
> practice. DC wise, that puppy looks like 12/1.25 amps or 9.6 ohms, but
> we're not interested in DC. The output impedance of the transistor,
> especially if we're down at low frequency, will be high - typically 50
> or a 100Kohms, depending on the early voltage. The transistor equivalent
> circuit is a current generator of magnitude hfe times ib in parallel
> with the output resistance, and maximum output appears when all the
> current goes through the load, rather than the shunt output resistance -
> which you can transform to a series out resistance and a voltage
> generator if you like. So there's an optimum output load impedance
> (resistance is a particular case of impedance) which is where the device
> can provide the volts and amps.
I'm not sure what you mean by the "output impedance of the
transistor". How was that parameter determined?
73, Tom W8JI
--
FAQ on WWW: http://www.contesting.com/ampfaq.html
Submissions: amps@contesting.com
Administrative requests: amps-REQUEST@contesting.com
Problems: owner-amps@contesting.com
Search: http://www.contesting.com/km9p/search.htm
|