Explain the principle of the circuit and explain the principle of the circuit

At the moment of closing, the transistor is turned on, the C pole has a current through, L1 and L2 form a step-up transformer, and the L2 coil has a high-voltage pulse output. However, the switch cannot be closed continuously, and if it is closed continuously, there will be no high-voltage output. It should be constantly turned on, off, on, off, on, off.

You must have noticed that Rebao was liquid before use, and after use, it became solid (strictly speaking, it is estimated to be glassy); If it is to be reused, it must be heated back to a liquid state. Its heating principle is very simple, that is, it uses the heat of phase change (the energy required for phase change) Ice melts into water to absorb heat, and in turn water freezes to release heat. The liquid in the hot treasure becomes solid and also exothermic.

The role of the small metal piece is to induce this phase transition. Generally speaking, matter has a stable phase at a certain temperature, such as water below 0 degrees will freeze, and ice will melt above 0 degrees, this is because below 0 degrees ice is more stable than water, and above 0 degrees water is more stable. The stable phase of the liquid in Rebao at room temperature is solid, but there will be substable supercooled liquids.

When you bend a small piece of metal, the local temperature rises, which induces the supercooled liquid to solidify and gradually release heat. From the above analysis, it can be seen that the core of Thermogemon is its medium, which must meet the following conditions: The melting point is less than 100 degrees (so that it can be restored to a liquid with hot water) There is a relatively stable metastable state at room temperature Higher melting heat The curing rate is relatively slow (so that the heat can be gradually released).

It is best to find an expert who will give you a satisfactory answer.

This is very good to calculate, its equivalent resistance can be calculated in this way, the two 80 electric resistance on the left side of the buried wax are connected in series with the 60 resistance in parallel, and then the series 20 resistance is connected in series with 30 resistors in series, and the 80 resistance on the right is not calculated because it is short-circuited by the wire, so the equivalent resistance is calculated as follows, first calculate the resistance of the two 80 resistors in series is 160, and then the resistance value after the parallel 60 resistance is 9600 220=, so the equivalent resistance in the above figure is 20 + 30 +.

The loop through which the current flows is called a circuit, also known as a conductive loop. The simplest circuit is composed of components such as power supply, load, wires, and switches. The on-circuit is called a path.

Only in the path can there be an electric current passing through the circuit. A break in a circuit is called an open circuit or an open circuit. If there is no load between the positive and negative poles of the power supply in the circuit and it is directly connected, it is called a short circuit, and this situation is never allowed.

Another type of short circuit is when the two ends of a component are directly connected, and the current flows through the direct connection without passing through the component, which is called a short circuit of the component. An open circuit (or open circuit) is allowed, while the first type of short circuit is never allowed, because a short circuit of the power supply will cause the power supply, electrical appliances, and ammeter to be burned out.

An electrical circuit or electronic circuit is a total that is connected by electrical equipment and components in a certain way to provide a path for the flow of charges, also known as an electronic circuit or electrical circuit, referred to as a network or circuit. Such as power supply, resistor, capacitor, inductor, diode, transistor, IC and electric key, etc., constitute the network and hardware.

Negative charges can flow in it. I am a home appliance maintenance master of China United Warranty, if you still don't understand, you can continue to ask.

Think like a signal amplifier made of MOS tubes or an output level controlled! The principle is very simple, IRF7309 has an upper tube and a lower tube, and the input signal on the left is amplified on the right!

The basic includes: simple resistance circuits, analysis methods and circuit theorems of linear resistance circuits, nonlinear resistance circuits, first-order circuits, second-order circuits, step response, impulse response, convolutional integration, phasor method, impedance and admittance, frequency response, filters, resonance, circuits with mutual inductance, transformers and three-phase circuits, etc.

Circuits are divided into series and parallel.

Then if we want to solve the problem, we must first know the formula.

Simply provide you with all the electrical formulas [Electrical Part] 1. Current intensity: i q quantity t 2. Resistance: r = l s 3. Ohm's law:

I u r 4, Joule's law: (1), q i2rt universal formula) (2), q uit pt uq electric amount u2t r (pure resistance formula).

Hope it helps.

Hello:-1, this is oneDelay circuits。One to be exactDelay release circuit

2. In a stationary stateThe switch is disconnectedQ1 and Q2 are cut-off, and there is no voltage output at the output JP

When the switch is closed, there are two changes at this time, capacitor C1 charging, voltage is 12VV is provided by resistor R1 to the triode (Q1 and Q2 form the Darlington triode).The transistor is turned on, and the output voltage is about 12V from the output terminal JP

When the switch is disconnectedThe voltage of C1 is still 12V, and the triode is provided with bias current through R1, and the transistor continues to turn onA delay of a certain amount of timeThe voltage of C1 decreases graduallyWhen it drops to the triode cut-off voltageThere is no voltage output at the output JP

Hello: At this time, a driving circuit, the driven JP socket should be connected to a relay or other inductive load;

1. When the switch is closed, due to the large capacitance of C1, there will be a charging delay (the delay time is relatively short), after the C1 voltage reaches the base conduction voltage of Q1, the Q1 and Q2 composite tubes are turned on (the current driving ability of the composite tube will be strong), and the load of JP is like relay engagement;

2. When the switch is disconnected, the presence of the remaining charge of C1 will delay the cut-off of Q1 and Q2, and the delay of JP will be disconnected.

As for the C1 role:

C1 has a short delay time, which is mainly to remove the messy jitter of the relay contact closure and disconnection caused by the jitter of the switch, and protect the contact (to prevent the contact from burning black by electric sparks) to stabilize the switched circuit;

If you don't add C1, the above problems will occur, and if you add it, there will be basically no such problems because of the delay, because the switch jitter time is very short, you can not use RC delay (that is, you can not use series resistance on the switch), of course, you really want to string resistors on the switch, but the resistance can not be too large (it is recommended to be tens of ohms), otherwise the delay time will be too long, resulting in the slow response of the relay.

This circuit should be connected with an appropriate resistor R2 at the upper end of the switch to form a delay (delay) switching circuit.

After the manual switch is closed, C1 is charged by R2 to make Q1 base reach about, Q1, Q2 is turned on, and the relay will be connected to JP at the action, when the switch is disconnected, the voltage of C1 will be maintained for a period of time, so that the state of Q1, Q2 and JP will be maintained for a period of time, until C1 is discharged to below the initial state through Q1 and Q2.

This is a delay circuit, when the switch is closed, the composite transistor composed of Q1 and Q2 is turned on, and JP is energized to work; When the switch is disconnected, the capacitor C1 can make the triode continue to be turned on for a while, and after a period of delay, it will be cut off, and the JP will stop working when the power is lost.

This is a delay shutdown circuit, after the switch is opened, JP immediately gets power to work, and C1 is fully charged at the same time, when the switch is disconnected, because C1 is discharged through R1, Q1, Q2, and leads to JP to lose power after a while after the switch is disconnected.

1. R4, R5, D1, D2 and N2 constitute a half-wave precision rectifier circuit.

When the input is positive, the output is 0, and when the input is negative, the output is positive.

2. R6, R7, R8 and N3 constitute the inverting adder circuit.

3. The above two parts of the circuit together constitute a full-wave precision rectifier circuit

The adder is 1 times the input plus 2 times the half-wave rectification, that is, when the input is negative, the half-wave rectification output is positive, and the negative input of 1 times plus 2 times the rectification output is equal to 1 times the positive input; When the input is positive, the half-wave rectified output is 0, and the additive output is 1 times the positive input, so that the output is a full-wave rectified waveform.

4. Capacitor C3 plays the role of low-pass filtering, and capacitor C2 has the role of AC coupling.

To sum up, the function of this circuit is to carry out full-wave rectification of the AC part and filter the smooth output, which is generally used for AC and DC voltage conversion, and is often used for AC signal measurement.

This circuit seems to be a precision rectifier circuit in front of the op amp, this circuit can eliminate the rectifier voltage error caused by the diode junction voltage drop, and the back op amp is the average value-RMS conversion circuit, this circuit is generally a circuit that converts the RMS value of AC voltage into DC voltage, and is generally used to measure AC voltage in instruments such as universal meters.

Circuit diagram of automatic conversion of positive and negative motors.

1. AO3148 is an N-channel MOSFET, which is equivalent to a low-side switch. Its quick on/off provides support for later LC charging and discharging.

2. R1 is a feedback start resistor. In addition to providing a power conversion path, it also transmits to the chip's CS startup circuit when an input voltage is present.

3, C5 and T1 form the LC charge-discharge circuit, and the voltage is increased.

4. D1 is the high-frequency power supply rectification, which rectifies the oscillating alternating voltage and outputs it.

5, R4, R5 are output voltage feedback resistors. Its precision determines the accuracy of the output voltage.

Pin 6,2 is a switching frequency regulator, and the resistance on it determines the frequency level. Pin 3 is the compensating pin, and the RC network on top of it provides compensation for the oscillating loop.

7. After powering on, PT1304 begins to vibrate internally, and AO3148 is driven on and off by 7 pins, and the input power supply generates higher self-inductive high-frequency voltage at both ends of T1 in the LC circuit composed of C5 and T1, and is output after rectification by D1. Precise selection of R2 and feedback resistors R4 and R5 can stabilize the output voltage at 5V with an accuracy of better than 2%.

1) PT100 application principle.

First of all, R1, R2, pt100, rp, these 4 resistors form a bridge circuit, when R1=R2, Pt100=rp, the circuit balance, that is, U1=U2, U1-U2=0;

Only Pt100 varies with temperature, so when Pt100 ≠ rp, U1-U2 ≠ 0;

e.g. pt100 > rp ==> u1-u2 < 0;pt100 < rp ==> u1-u2 > 0；

Then there is the amplification circuit, which amplifies the difference between U1 and U2 for subsequent AD conversion and other circuits.

2) There seems to be a problem with the circuit, parameters, and design of your diagram, and I don't know where I copied it;

According to the theory, pt100=rp is required, obviously rp belongs to the order of k, and pt100 is only 100 at a temperature of 0 °C, and 138 at 100 °C, which cannot meet the balance condition of the bridge;

In addition, the two-stage op amp circuit (ignoring the U2 signal first) attenuates and then amplifies U1, and the total gain = 1, that is, there is no amplification. At the same time, u1 is a stable reference voltage, so there is no need to deal with it this way;

Just remove the U1A op amp circuit;