Close, but the base current is the emitter current divided by the gain. For high gains, the difference is pretty insignificant, but power transistors have low gains to begin with and if you are driving a transistor into saturation (using it as a switch, for instance) then you can easily get into current gains in the range of 5 or even less.Sorry, I'll make sure to add units
Assuming the gain is 40.
Base current is: emitter devided by 40
118 mA / 40 = 2.95 mA
True -- under conditions where the current gain is sufficiently high.And as far as i know the collecter current is approximately the same as emitter.
Yes and no. Remember that the transistor has three possible states -- cutoff, linear, and saturation. Having a forward-biased base-emitter junction with non-zero base current is required in order to be in either linear or in saturation.Doesn't the base current determine if the transistor is closed or open?
Right idea except that the emitter of Q3 isn't at 0 V. What is the voltage at the top of the zener diode D1?R3 would be 0.6V which is just enough to activate Q3.
The zener diode require 5.6V but i thought it would take what it needs from the R2 connection?Right idea except that the emitter of Q3 isn't at 0 V. What is the voltage at the top of the zener diode D1?
Correct.If all voltage required in the Zener diode is to be taken from Q3 then the emitter has to be 5.6 V like the zener diode.
Q3 base is base-emitter diode added to emitter
5.6 V + 0.6 V = 6.2 V
6.2 V is the Q3 base (which is what you want)Uout is Q3 emitter times 2.5
6.2 V x 2.5 = 15.5 V
But, for Q2 to be on, doesn't Q2 base need to be 0.6 V higher than Q2 emitter?Q1 base is Uout subtract 2x diodes
15.5 V - 0.6 V - 0.6 V = 14,3 V
Haha guess it was late last night
I had to ADD on the diodes to find Q1 base, not subtract them.
15.5 V + 0.6 V + 0.6 V = 16.7 V
I generally stop reading as soon as we get to something that is incorrect, since it commonly messes up much of what follows, and then focus on getting that point corrected.In my post above i calculated the currrent through R4 and R3 but i have a feeling its wrong.
R4 also has what ever current is flowing to the base of Q3, which we are assuming is negligible.The current that runs to R3 has to run through R4. In which case current in R4 is Uout devided by R4?
Depends on what you are including in "total current" -- is it the total current leaving the output of this regulator, or the total current draw from the supply? Need to work on being explicit and clear in your terms.So the total current would be Uout devided by R4 parallel with RL?
Most of these are good, but look at R2 more carefully, remembering that Ohm's Law requires the use of the voltage ACROSS the resistor -- you can't just pick the voltage on one end of it.
by Aaron Carman
by Jake Hertz
by Aaron Carman
by Jake Hertz