Trouble the master to help me answer this question about mold electricity.,Urgent.,Thank you.

1. VT is a one-way thyristor, when the effective trigger voltage is added between the cathode K and the gate G (the trigger voltage of different types of thyristors is not necessarily the same), VT will be triggered and turned on, after VT is turned on, as long as the working voltage between the anode A and the cathode K of VT remains unchanged, even if the trigger voltage disappears, VT will still maintain the conduction state.

vu is a dual-base transistor (also called a single junction transistor): when the voltage applied by its emitter is its peak voltage, vu is on, and when the voltage applied by the emitter is its valley voltage, vu is cut off.

2. The voltage stabilized filter circuit is composed of R1, VD3 and C4: when the voltage at both ends of VD3 VD3 breaks down voltage, VD3 is on, and the higher the voltage at both ends of VD3, the stronger VD3 conduction, so that the current flowing through R1 is also larger, so that the voltage drop on R1 is higher, so that the output voltage UO=U-UR1 remains basically unchanged.

3. The relaxation circuit is composed of VU, RP1, R2, C3, R3, RP2, R4, VD2 and R5: the DC voltage U+E after half-wave rectification and voltage stabilization filtering is charged to C3 through RP1 and R2, and the VU is supplied with power after RP2 partial voltage, and the other is charged to C3 at the same time through RP2, R4 and VD2; When the voltage at both ends of C3 is charged to UR5, VD2 is cut-off, and the charging circuit of C3 only leaves the branch of RP1 and R2, and the charging speed of C3 slows down at this time, and when the voltage at both ends of C3 is before the peak voltage of VU, VU maintains a cut-off state, and there is almost no voltage drop on R3 at this time, that is, UR3 0V; When the voltage at both ends of C3 reaches the peak voltage of VU, VU is turned on, R3 discharges C3 through the conductive VU, R3 and RP2 form a voltage divider circuit at this time, UR3 charges C2 through R6, when the voltage at both ends of C2 rises to VT to trigger conduction, when VT is turned on, the relay KA is engaged, and the KA contact is switched from left to right; C3 discharges the residual voltage to 0V through the closed contact of KA, during the discharge process of C3, when the voltage at both ends of C3 is the valley voltage of VU, VU is turned on to cut-off, during this period C2 is discharged through R6 and R3, when the power supply is cut off, C1 is discharged to VT cut-off, the contact of KA pick-up is disconnected, and when it is powered on again, the whole process begins to repeat again.

There are really masters, Diodes 1, salute!!

This multiple-choice fill-in-the-blank question can be chosen like this:

1. Decrease; 2. Stability;

3. broadening; 4. Decrease;

5. Output current;

6. Enlargement; 7. an output voltage;

8. Decrease.

1.The open-loop differential-mode voltage amplification factor AU0 of an ideal integrated op amp can be considered infinity, the input impedance RID is infinity, and the output impedance R0 is 0.

2.The cut-off frequency fh or fr of a first-order low-pass or high-pass active filtering circuit is related to a passive low-pass or high-pass RC circuit, and its relation is expressed as 2.

3.A circuit that converts a sinusoidal alternating voltage into a unidirectional ripple voltage using a conductive element of the unit is a rectifier circuit.

4.Series DC regulator circuits include sampling, amplification, regulation, and adjustment.

5.In order to satisfy the phase equilibrium condition of the oscillation, the phase difference between the feedback signal and the input signal should be equal to (d).

a) 90 degrees b) 180 degrees c) 270 degrees d) 360 degrees.

6.The regulated power supply mainly requires that the output voltage is basically unchanged when the input voltage and output current change.

a=2/;a (1+af)=1, so the feedback coefficient f=, feedback depth 1+af=2

This diagram is a non-inverting proportional operation circuit, and his formula is.

If R1 is open, it means that R1 is infinite, then the value in the parentheses is close to 1, the coefficient in front of U1 is close to 1, then U0=U1, if R1 is short-circuited, it means that R1 is close to 0, then the value in the parentheses is infinite, but U0 cannot be equal to infinity U1, he will be equal to the maximum output voltage of the op amp, the voltage you give is 14V, and you should be able to calculate it next, just substitute the value into it.

The four diodes in this bridge rectifier circuit are all left yang and right yin.

The 7810 belongs to the 78 series of three-terminal positive voltage integrated voltage regulators, and the last two digits in its model number indicate the regulated output voltage value, and the input and output voltage difference is δu when working. In the circuit of this problem, UI=13V, UO=10V, and the differential voltage meets the requirements.

The output current of this circuit is IO=UO RL=10 RL, and the power consumption of 7810 is P=ΔU·IO.

The K with a circle on the left replaces the J with the box on the top, and the one on the right replaces the switch J1 on the top

RL = when the maximum power is 10W, 50% of the power emitted by the power supply is transmitted to RL

Turn the current first.

4a, 5 ohms and 5=

Then RL = Euclidean time power is the maximum, and the maximum is half.

w=u*i;

i=(20V-U) 5 ohms-u 5 ohms;

substitution, w=(20v-2u)*u 5 ohms;

When u=5, the power is maximum.

At this point, rl = ohne.

The power of the power supply is 60W.

The resistor power is 10W.

The power emitted by the power supply has 1 6 transmitted to the RL.

If you don't know the specifics, you can ask me again.