Analog Electronic Circuit Problem 10, Analog Electronic Circuit Problem

Analog Electronic Circuits Answer...

Give me your question, big brother.

Okay, that's all there is to say, if the landlord doesn't understand anything, you can qq me....

<>1) Static working point.

T1: Base voltage, UBQ = VCC*RB12 (RB11+RB12);

Emitter voltage, UEQ = UBQ - UBE, then IE = UEQ RE1;

collector voltage, UCQ = VCC - IC*RC1 = VCC - IE*RC1;

t2: the same way;

t3： vcc = ib*rb3 + ube + ie*re3 = ib*rb3 + ube + ib*(1+β)re3 ；

Thus we get IB, then UBQ = VCC - IB*RB3; ueq = ib*(1+β)re3；

Let's write about it here;

R2 and R3 serve to limit the current of a single tube, and R3 provides a voltage signal to the lower amplifier.

Co-emission circuitry.

It is to transmit the signal to the buzzer through C3.

187957968 This group is helping comrades who are interested in making achievements in the field of analog electronics and hardware design to better communicate and ask for help and guidance.

1） ic = ib*β ie1 = 1+β)ib ; ie = 2*ie1 ； uc1 = vcc - rc*ic；

2) When calculating the static working point, take ui1 = ui2 = 0;

Then: ib*rb + ube + ie1*rw 2 + ie*re = ui1 - vee;

Thus finding ib;

As for aud, rid and rod according to the textbook, just apply the formula;

1. Not only can it not, but it is also possible to burn out the voltage stabilizer level 2 tube. The regulated 2-stage transistor is a nonlinear resistor that is higher than the breakdown voltage and the resistance decreases rapidly. This feature is used to stabilize the voltage. Therefore, it is also necessary to connect a resistor of suitable size in series on the dry circuit to limit the current.

2. It can be within a certain range. The voltage regulator 2-stage tube has a shunt effect, which can stabilize the voltage within the specified range.

3. Not absolutely. If the load is too heavy, the voltage will be too low. The voltage regulator circuit of the 2nd stage tube can only keep the voltage stable when the load does not exceed the rated value. If the voltage is too low due to the heavy load, it cannot be stable.

4.I don't understand this!

a1 is an inverting proportional op amp with u1 = -ui1*r1 r2;

A2 is the subtractor circuit and has (imaginary short).

According to the false judgment, there is.

i =（u1-un）/r4 = （un-uo）/r5；

According to the imaginary shortness, there is up=un; and up = ui2*r5 (r4+r5);

uo = (ui2 - u1)*r5 r4;

Okay, let's do the rest for ourselves.

If you don't understand this sentence or two, you might as well read the book a few more times!!