Ask the master of modular electricity, 12V input in the circuit

This calculation process is a lot, don't think that just a few components are simple.

This depends on the resistance of the load carried by 78L05 and the current it wants to use, which determines the size of your R1, the maximum output current of 78L05 is 100mA, if you want to use such a large current, then, the current through the C pole of the triode should also reach more than 100mA, and then, according to the amplification of the triode to calculate the required B pole current, assuming that the current through the B pole of the triode is IB, and the current through D2 is ID, then, the current IR through R1 is.

IR IB+ID, and the current of the load directly affects these two currents, so it is difficult to calculate, I personally recommend that you use or, as for energy consumption, 12V power supply will care so much.

Hello:— 1. R1 in the circuit diagram has two functions: , providing voltage regulator current for the voltage regulator diode; , provides bias for the regulator transistor.

2. The resistance value of R1 is large, and the voltage stabilization performance of the voltage regulating diode is poor; The large resistance value of R1 also affects the bias current of the Zener transistor and affects the output current.

The output current of L05 is not large, and the input terminal does not require a stable voltage, so 10K can meet the requirements, and replacing it with or resistor has basically no effect on the output voltage of 78L05.

The value of R1 comprehensively considers the HFE parameters of the transistor, the maximum output current of 78L05, and the lowest voltage difference at the maximum output current.

In terms of power consumption, there is room for the installation of the first stage of voltage regulation, and it is enough to change the 7805 in the TO220 package

Whoever suggested it, ask whom, I think the 10K power consumption is small.

Roughly speaking, it is an amplitude modulation circuit, but if you look closely at the T1 secondary stage being short-circuited, it loses its meaning, and it should be a wrong circuit.

It is recommended that you personify the parts that you can understand into several lines, and mark the current, direction, and function so that you will get used to it.

According to me, the landlord actually hasn't really figured out the unit circuit (a single small circuit) yet. When the unit circuit is really understood, the complete large circuit can be understood. For example, the computer power supply, starting from the 220V incoming line, there will be insurance, buffering, electromagnetic filtering, power factor improvement, rectification, filtering, 110V 220V conversion, switching power supply, output rectification, output filtering, boot circuit ......and so on, especially the switching power supply part is the most complex, and it can be subdivided into several parts.

The first figure: according to the polarity of the voltage, the diode is in the cut-off state, and the voltmeter measures the voltage of the series resistor, so u04=2v. Second picture:

The power supply connection method turns the diode on, and the voltmeter measures the voltage of the series diode, u05=. The third figure: the diode is in the cut-off state, the same as the first figure, u06=2v.

The first means the same as the third.

(1) When the model is not ideal, let the voltage drop of the diode forward tube be.

2) It has been drawn in Figure (b) and is not in the solution.

At this time, when the UI < 3V, the diode is cut off, and U0 is the waveform of the UI. When UI>3V, the diode is turned on and the output U0 is limited to the voltage output of UEF=3V.

The output voltage must be smaller than the power supply voltage, so the input voltage multiplied by the magnification must be less than the power supply voltage.

<> inverting input and output are open, just like a negative inverted stupid feed op amp circuit with an infinite resistance resistor r connected to both ends, and the op amp output UO'=-ui x r r1=- v, touch the pants and know the saturated output voltage of the op amp = -12v, dz anode 0v > cathode full with noisy blue foot conduction condition, the voltage at both ends is the same = 0, uo = 0.