I would like some advice on troubleshooting the mnemonic reading device in the analog circuits book?

早起的达仙僧

I would like some advice on troubleshooting the mnemonic reading device in the analog circuits book? [Copy link]

I have recently been reading a self-study manual called "Analog Circuits and Digital Circuits" by Cai Xingshan, which mentions troubleshooting of a mnemonic device (involving transistors). Why is the base voltage greater than 0.8V when the transistor emitter is open and R5 is open?

chunyang

When the emitter or R5 is open, the transistor base potential is the voltage-divided value of R2 and R3.

maychang

"Why is the base voltage greater than 0.8V when the transistor emitter is open and R5 is open?"

This statement is wrong.

When the transistor emitter is open and R5 is open, the base voltage will not be greater than 0.8V.

When the transistor emitter is open, there is no current in the base, and the base voltage is the voltage divided by R2 and R3. The power supply is 6V, and the voltage drop on R10 is ignored, so the voltage across R3 is 0.545V.

早起的达仙僧

chunyang posted on 2019-8-11 14:50 When the emitter or R5 is open, the base potential of the transistor is the voltage divided by R2 and R3.

Well, according to this explanation, the content in the book is probably wrong. The base voltage should be about 0.54 V at this time. In addition, if the transistor is normal, can the base voltage also be approximately replaced by the voltage divider value of R2 and R3?

Because the book says, "In the circuit, the transistor base current Ib is much smaller than the current flowing through R2, and the base voltage Ub is basically determined by the voltage divider of R2 and R3." I wonder if this statement is advisable in practical applications?

早起的达仙僧

maychang posted on 2019-8-11 15:20 "Why is the base voltage greater than 0.8V when the transistor emitter is open and R5 is open?" This statement is wrong. Transistor emitter open...

Yes, I calculated for a long time but couldn't figure out how 0.8V could be so large. The content in the book is wrong.

chunyang

Posted by Daxian Monk who got up early on 2019-8-11 21:11 Well, according to this explanation, it is estimated that the content in the book is wrong. The base voltage should be about 0.54V at this time. In addition, I would like to ask, if the transistor is positive...

That is not necessarily the case. For example, if the emitter of a transistor is directly grounded, then the base potential is about 0.7V no matter how much you divide it. Even if it is as shown in the original poster's picture, this statement also requires that Ib must be much smaller than IR2. Whether this is true must be calculated. In practical applications, this statement is basically useless.

早起的达仙僧

chunyang posted on 2019-8-12 15:10 That is not necessarily true. For example, if the emitter of a transistor is directly grounded, then the base potential will be around 0.7V no matter how much you divide it. Even if it is as shown in the original poster's picture, this statement is also true...

Got it, specific analysis of specific issues, thank you for your answer