Is there any other information given? For example, was the Vbe for the transistors used provided as part of the problem statement? hgmjr
Can you explain how/why it's wrong? Generally, answers can be right or wrong, but questions are usually good or bad. Although, occasionally a question is so bad that it is reasonable to call it "wrong". I think the question is worded oddly and requires a number of assumptions to be made, but it seems to ask a question that can be answered. But, it's hard to comment further on that without your reasoning explained. My opinion is that this question needs to be answered using the exponential relationship for a transistor emitter current versus base-emitter voltage. However, the question is vague about the type of transistor so you would have to make some assumptions about transistor type, reverse saturation current and effective thermal voltage nVT. In addition, you need to assume that an output current path is available. Ie=Is*(exp(Vbe/nVT)-1) would be a reasonable approximate formula to use.
I try some calculation like this for the current across the driver I= E/R I= 150mV/2.2 I= 68.2mA how about the current for Output BJT's
Why don't like the one I suggested? Ie=Is*(exp(Vbe/nVt)-1) If you invert this, you get the following. Vbe=nVt*ln(Ie/Is+1) now using Kirchoff's voltage law 0.15V=Ie*(0.47ohms)+nVt*ln(Ie/Is+1) This is one equation and one unknown, hence you can get an answer for emitter current Ie. However the question asks for voltage on the .47 ohm resistor, which is found using ohms law, or the following. V=0.15V-nVt*ln(Ie/Is+1) In order to get a numerical answer, you need to make some assumptions about the transistor. Unless otherwise stated, it's usual to assume a silicon BJT at room temperature. Hence, nVt=30mV and Is is in the 1-100 femtoAmp range (use 10 fA if no value is given). Note that the answer is transistor-dependent and temperature-dependent.
thanks steve what if I used the Ohm law for calculating the Voltage across 0.47 ohm resistor, what will be be the formula for this?
Ohms law is just V=IR, so R=.47 Ohms and I=Ie as found in solving the equation 0.15V=Ie*(0.47ohms)+nVt*ln(Ie/Is+1) using a numerical technique such as Newton's method, or search algorithm.