The weed-eater circuit
Figure 1 The weed-eater circuit
Rather than trying to describe the tortuous paths that were traversed by some interview candidates, I will simply present the assumptions that I expected would be made and the ensuing analysis. The starting assumptions are that the transistors are silicon and will exhibit base to emitter voltages of 0.6V and that the ß values of both transistors are very high so that their base currents are virtually zero.
Figure 2 The first step of the analysis
With essentially zero base current to the NPN, R1 and R2 voltage divide the +12V rail voltage to +4V at the NPN base. With a Vbe of 0.6V, the NPN emitter is +3.4V and the current flowing in R3 will be 3.4 mA.
The next issue is, how does that 3.4 mA get shared by the NPN emitter and R5?
Figure 3 The second step of the analysis
The Vbe of the PNP is 0.6V so that the current flow in R4 is 0.06 mA or 60 µA. With the PNP base current being virtually zero, that 60 µA becomes the collector current of the NPN which in turn becomes the emitter current of the NPN because of the NPN's extremely high ß value.
The current flowing through R5 must be the difference between the R3 current of 3.4 mA and the 0.06 mA current of the NPN emitter. That value is 3.4 - 0.06 = 3.34 mA.
Figure 4 The third step of the analysis
The voltage drop across R5 is then 3.34V which when added to the 3.4V at the top of R3, puts the top of R5 and the PNP collector at +6.74V.
Easy, huh?? Well, yes, it is easy but even so, this circuit weeded out a lot of unqualified candidates.
John Dunn is an electronics consultant, and a graduate of The Polytechnic Institute of Brooklyn (BSEE) and of New York University (MSEE).
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