"Mower" circuit diagram? This is not the circuit diagram required to design the weeder. However, when the author interviews a new circuit designer, it is used to judge whether the applicant is really an “examination question†for an engineer who can analyze the analog circuit, and through such an examination question, the dissatisfied “miscellaneous†can be removed from the problem. grass"…
This is a "question" that an interviewer thinks, and it will remove anyone who is really not good at handling analog circuit analysis. The details are as follows: there are two transistors (transistor), one NPN and one PNP, and the connection mode is as shown in Fig. 1.
figure 1
Here is a brief introduction to the assumptions I am expected to make, and the subsequent analysis.
The initial assumption was that the transistor was silicon (Si) and showed a base voltage of 0.6 volts (V) to the emitter voltage, and the ß values ​​of the two transistors were very high, making the base current almost zero.
figure 2
For a base current where NPN is essentially zero, the voltages of R1 and R2 divide the +12V turn-on voltage to +4V on an NPV basis. When Vbe is 0.6V, the NPN emission is extremely +3.4V, and the current flowing in R3 is 3.4mA.
The next question is, how does the NPN transmitter and R5 share 3.4 mA?
image 3
The Vbe of the PNP is 0.6V, which makes the current in R4 0.06 mA or 60 μA. In the case where the PNP base current is almost zero, since the ß value of NPN is very high, 60 μA becomes the collector current of the NPN and also becomes the emitter current of the NPN.
The current flowing through R5 must be the difference between the 3.4 mA current of R3 and the 0.06 mA current of the NPN emitter. This value is 3.4-0.06 = 3.34 mA.
Figure 4
The voltage drop across R5 is 3.34V. When applied to 3.4V at the top of R3, the top of the R5 and PNP collectors are placed at + 6.74V.
Is it easy? That's right, this circuit is very easy. Even so, this circuit helped me to remove many unqualified candidates.
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