I was reading the beginning of Ch.2 of Art of Electronics, and I thought I understood how transistors behaved. The first example they gave was a 10 V source that connects to 1) a 100 Ohm lamp that goes to the (npn transistor) Collector and 2) a 1 kOhm resistor that goes to Base. (Emitter at ground.) You assume the Base-Emitter junction drops 0.6 V, so you have 9.4 V through the 1k resistor, hence 9.4mA. Assuming Beta=100, the collector then tries to drive 940 mA, but it can't because the 100 Ohm lamp limits current to 0.1 A for a 10 V source. Cool.
They reinforce this by going on to describe the Transistor Man that looks at the Base-junction circuit acting independently and then sets the effective resistance of the Collector in order to achieve I_C = Beta * I_B. (Up to transistor saturation at 0 Ohms, or rather 0.2 V drop.) Also fine.
The next thing they discuss is the emitter-follower. 10V at the Collector, Base receives variable V_in=V_base, and Emitter passes through resistor R to ground. V_out=V_emitter measured between Emitter and resistor.
My initial analysis (using mostly their equations):
V_out=V_in-0.6 (their equation)
dV_out = dV_in (theirs)
dI_out = dV_out / R = dV_in / R (theirs)
Depending on how you look at it, I either implicitly assumed that I could treat the Base-junction circuit separately or, as I describe below, I may have implicitly assumed that the current:voltage relationship across the Base junction continued to hold such that the voltage drop was 0.6 regardless of current. Anyway, I then did:
dI_in = d(V_in-0.6) / R = dV_in / R
Of course, if dV_out = dV_in (theirs) my assumption implies bad things like dI_collector=dI_out - dI_in = 0, which would be a good indication that I shouldn't do transistor analysis this way...
By their calcuation,
I_out = I_in + I_collector = (1+Beta)*I_in
Hence, dI_in = dI_out / (1+Beta)
Since dI_out = dV_out / R and dV_in = dV_out,
dI_in = dV_in / [(1+Beta)*R]
(rather than my dI_in = dV_in / R)
This was confusing the *crap* out of me. After banging my head against the wall for awhile, I realized that my assumption about the fixed 0.6 V drop may be the problem. A diode has an exponential I:V relationship, so you can usually just assume a 0.6V (or 0.7) drop and then treat it like a short circuit. However, maybe that's not the case here; if I treat the collector as a super-diode that has I:V relationship 100 times that of the base (and is magically dependent on the Base-Emitter voltage drop), then it seems like that might somehow be able to save me.
At the outset, I thought I could just calculate i_Base by converting the Base junction to a diode and independently calculating the current of that circuit assuming a 0.6V drop across the diode. Then take the current multiplier, and the Collector either delivers that current or delivers as much as it can under saturation. (As with the lamp-example.) But if I treat the Base-junction voltage drop as fixed 0.6V, then I seem to get bad answers.
If it isn't fixed at 0.6V, what voltage drop should I be using in order to calculate the Base current?? Something dependent on the Beta multiplier? Would that approach deal properly with transistors in saturation?
Or did I make some other silly mistake and I'm off in left field?
Sorry for the long post...thanks for reading it.
They reinforce this by going on to describe the Transistor Man that looks at the Base-junction circuit acting independently and then sets the effective resistance of the Collector in order to achieve I_C = Beta * I_B. (Up to transistor saturation at 0 Ohms, or rather 0.2 V drop.) Also fine.
The next thing they discuss is the emitter-follower. 10V at the Collector, Base receives variable V_in=V_base, and Emitter passes through resistor R to ground. V_out=V_emitter measured between Emitter and resistor.
My initial analysis (using mostly their equations):
V_out=V_in-0.6 (their equation)
dV_out = dV_in (theirs)
dI_out = dV_out / R = dV_in / R (theirs)
Depending on how you look at it, I either implicitly assumed that I could treat the Base-junction circuit separately or, as I describe below, I may have implicitly assumed that the current:voltage relationship across the Base junction continued to hold such that the voltage drop was 0.6 regardless of current. Anyway, I then did:
dI_in = d(V_in-0.6) / R = dV_in / R
Of course, if dV_out = dV_in (theirs) my assumption implies bad things like dI_collector=dI_out - dI_in = 0, which would be a good indication that I shouldn't do transistor analysis this way...
By their calcuation,
I_out = I_in + I_collector = (1+Beta)*I_in
Hence, dI_in = dI_out / (1+Beta)
Since dI_out = dV_out / R and dV_in = dV_out,
dI_in = dV_in / [(1+Beta)*R]
(rather than my dI_in = dV_in / R)
This was confusing the *crap* out of me. After banging my head against the wall for awhile, I realized that my assumption about the fixed 0.6 V drop may be the problem. A diode has an exponential I:V relationship, so you can usually just assume a 0.6V (or 0.7) drop and then treat it like a short circuit. However, maybe that's not the case here; if I treat the collector as a super-diode that has I:V relationship 100 times that of the base (and is magically dependent on the Base-Emitter voltage drop), then it seems like that might somehow be able to save me.
At the outset, I thought I could just calculate i_Base by converting the Base junction to a diode and independently calculating the current of that circuit assuming a 0.6V drop across the diode. Then take the current multiplier, and the Collector either delivers that current or delivers as much as it can under saturation. (As with the lamp-example.) But if I treat the Base-junction voltage drop as fixed 0.6V, then I seem to get bad answers.
If it isn't fixed at 0.6V, what voltage drop should I be using in order to calculate the Base current?? Something dependent on the Beta multiplier? Would that approach deal properly with transistors in saturation?
Or did I make some other silly mistake and I'm off in left field?
Sorry for the long post...thanks for reading it.