darrenheywood
Newbie level 4
early voltage of bjt
Hello there,
Greetings all, I am new here so I have a few nagging questions about BJT operational characterstics.
According to electrical theory, a current source can be constructed and understood by taking a high voltage source and placing a resistive potential divider across this source, in this way the top resistor is made very high whilst the bottom load resistor is made relatively much lower perhaps 3 decades lower. With this arrangement, it is possible to emulate a approximate perfect current source. There's no problem with that, Nortons theory privides a switch between Thevenin's voltage source and a current source. The only problem I have with that, is Thevenin's generators provides inherent boundaries whilst Norton's current generator does not.
Assume now that I have a ce bjt set up on the bench, biased and ready to go. I purposely make the base bias source resistance very low so as to avoid reverse voltage transfer effects and to maximise the early voltage. I then take a differential slice measurement of the output, i.e., delta Vce/ delta Ic across the bjt's non saturated active region.With this, I can yield output resistance and further calculations can yield the early voltage. Now, if memory serves me correct, a 2N3904 will yield about -100V in series with about 20k Ohms. So, looking into the bjt's collector is as though I have -100V source generator, in series with 20k Ohms feeding RL. So I have a approximate current source as first outlined. Now, here's the kicker, the power being dissipated in the bjt should be many orders higher than what the bjt is actually dissipating, which would violate Ohms law, as such there is no violation of Ohms law, as the power being dissipatated in the BJT is simply Vce*Ic and not (-100V+Vce)^2/(20k).
How does the BJT or FET etc, manage to pull this one.
Look forward to your reply
Hello there,
Greetings all, I am new here so I have a few nagging questions about BJT operational characterstics.
According to electrical theory, a current source can be constructed and understood by taking a high voltage source and placing a resistive potential divider across this source, in this way the top resistor is made very high whilst the bottom load resistor is made relatively much lower perhaps 3 decades lower. With this arrangement, it is possible to emulate a approximate perfect current source. There's no problem with that, Nortons theory privides a switch between Thevenin's voltage source and a current source. The only problem I have with that, is Thevenin's generators provides inherent boundaries whilst Norton's current generator does not.
Assume now that I have a ce bjt set up on the bench, biased and ready to go. I purposely make the base bias source resistance very low so as to avoid reverse voltage transfer effects and to maximise the early voltage. I then take a differential slice measurement of the output, i.e., delta Vce/ delta Ic across the bjt's non saturated active region.With this, I can yield output resistance and further calculations can yield the early voltage. Now, if memory serves me correct, a 2N3904 will yield about -100V in series with about 20k Ohms. So, looking into the bjt's collector is as though I have -100V source generator, in series with 20k Ohms feeding RL. So I have a approximate current source as first outlined. Now, here's the kicker, the power being dissipated in the bjt should be many orders higher than what the bjt is actually dissipating, which would violate Ohms law, as such there is no violation of Ohms law, as the power being dissipatated in the BJT is simply Vce*Ic and not (-100V+Vce)^2/(20k).
How does the BJT or FET etc, manage to pull this one.
Look forward to your reply