Some problems with electrical design! I hope you will give some guidance! Thank you!

I'm in the field of electrical control, and your problem is simple, some of the most basic ones. I'll just keep it simple.

1.To make a control circuit, to determine whether the sensor is in place is nothing more than to judge whether it is in place through the switching signal of the sensor, (such as normally open to normally closed, normally closed to normally open) if there is no digital such as what the opening is simpler, only open and closed two states to judge. What should I do?

Just add some programs on the industrial computer. How to choose a sensor? It is necessary to consider the signal of the sensor, what kind of signal it is, weak current or strong current.

AC or DC.

2.Do strong currents, weak currents, and signal lines have to be separated? Separate, of course.

The technique should be beautiful and reasonable. The layout of electrical components should be reasonable. Pay attention to the color of the line.

Black, 380VAC, Red, 220VAC Blue and Brown are 24VDC and the flowers are grounded. You can check it out online. **Upstairs has been given.

3.Because there are too many serial ports behind the industrial computer, the leader said to make a distribution board, connect the serial port to a board, the internal space of the machine is tight, try to make the board smaller, I don't know if you can replace the big DB9 head with another head? Of course.

You can change it according to the space.

4.If you want to recharge, look for it on the Internet. Don't be afraid of hard work. Learning is a matter of self-effort. When I first graduated, I couldn't do anything when I arrived at the unit, haha.

There are a lot of domestic sensors. It's cheaper. In fact, the sensors I use are all bad. Omron's positioning is just pretty average.

Automation Network: China Industrial Control Network:

Automated Learning Network:

Automation Forum:

Here are some of my collections**. Estimates are useful.

Strong electricity, weak electricity, and signal lines must all be separated Your answer can be.

China Industrial Control Network has a very detailed introduction, and it is worth reporting to you.

The following diagram provides data for the user:

First, let's calculate the theoretical value of the load:

1. When all the lamps are full, p total =; u total =

2. When all the lights are full, p total =; u total =

3. Each LED light piece is in between, less than 3W

2. Drive power supply parameters: input 24V; The output voltage is 28V 36V, the output current is 1500mA, and the output power is 50W

1. Output power (calculated value): 42W 54W

2. Rated working voltage (calculated value): 50

3. Input current (according to the total calculated value of theoretical u, assuming efficiency:

4. Characteristics of driving power supply: a. Keep the current of 1500mA unchanged when the output voltage is 28V 36V; b. When the load voltage is lower than 28V, the output current will be forcibly increased, so that the output reaches 28V or the output is stopped, and the fault is reported; C. When the output voltage reaches 36V, regardless of whether the output current reaches 1500mA, the voltage remains unchanged and will not continue to rise.

4. Analysis. According to the above data, it can be seen that:

1. The driving power supply may not match the voltage of the actual LED string light.

2. If it is normal, as long as the input side cable is more than a square millimeter of copper wire, it can meet the use and will not burn.

5. Improvements. 1. Replace the driving power supply so that its output voltage is between 23V and 30V.

2. Or the string lights are changed to 3 and 10 strings.

3. Increase the cross-sectional area of the input side wire.

The LED load does not match the LED driver, either replace (adjust) the drive power supply, or replace the light bar. Not only is the power mismatched (the load power is too high), but the voltage is also mismatched (the driving voltage is too high).

Adding cement resistor is equivalent to limiting the current and dividing the voltage of the load, and the resistance consumption is about (, the cement resistor must be heated badly, and it is likely to burn out in the end. And for a 50W power supply, the efficiency is greatly reduced.

It is recommended to change it to 2 and 9 strings to make do with it.

The driver power 50W and the LED load 72W do not match the difficulty of not burning, you series cement resistance is the practice of reducing the load current, but the efficiency is lost, which is equivalent to paying 100 yuan of electricity and 30 yuan so that the resistance is consumed.

50 watts drive 70 watts of electrical appliances, which is a high load.

Consider enlarging the drive and thickening the wires.

If that doesn't work, add a few more wires.

Willing to work with you**. It's best to draw a diagram.

Can't you go to the next search software to do this kind of question?

The resistor is changed to 10 ohms, and the wire is replaced by a thick wire!

Try changing the resistor to 10 ohms.

There is no need to change the current for anything.

Function introduction Three-terminal voltage regulator integrated circuit is also known as three-terminal voltage regulator tube, it looks like an ordinary transistor, and the three-terminal voltage regulator integrated circuit commonly seen in electronic products has 78 series of positive voltage output and 79 series of negative voltage output. As the name suggests, a three-terminal IC means that the integrated circuit used for voltage regulation has only three pinouts, which are the input, output, and ground. Orient the identified side of the component towards you, in the case of the 78 series, the three pins are the input, ground, and output; In the case of the 79 series, the three pins are ground, input, and output.

There are very few external components required to use the 78 79 series three-terminal regulated IC to form a regulated power supply, and there are overcurrent, overheating and adjustment tube protection circuits inside the circuit, which is reliable, convenient and cheap to use. The number after 78 or 79 in this series of integrated regulator IC models represents the output voltage of the three-terminal integrated regulator circuit, such as 7806 indicates that the output voltage is positive 6V, and 7909 indicates that the output voltage is negative 9V. Sometimes there is an M or L after the number 78 or 79, such as 78M12 or 79L24, which is used to distinguish the output current and package form, etc., among which the maximum output current of the 78L series is 100mA, the maximum output current of the 78M series is 1A, and the maximum output current of the 78 series is 1 5A.

They are also available in a variety of packages, which vary depending on the placement of the components. In practical applications, a sufficiently large heat sink should be installed on the three-terminal integrated voltage regulator circuit (of course, not in the case of low power). When the temperature of the regulator tube is too high, the regulation performance will deteriorate or even be damaged.

Where to find the topic?,It's so difficult.。。。

1. Connect the low water level signal of the water level gauge (generally with a normally open electric shock) to the starting circuit of the water pump as a starting signal.

2. Connect the high water level signal of the water level gauge (if there is a normally closed electric shock) to the back of the automatic transfer switch and the front of the start signal.

3. String together the water level of the underground water pool and the high water level signal. Normally closed electric shock, that is, when the water level is higher than the lowest water level, normally closed closed.

4. The rest of the manual part is also simple. Draw your own.

The negative feedback amplification circuit that forms the circuit into a non-inverting terminal input is the one;

Left loop. Set clockwise to u1, u2, u3, u4, Kirchhoff's voltage law:

6=u1+u2+u4

Again, r1 = r2 total = r3. u=

2V is the same as the reference direction!

The answer is supplemented. R2 is always the sum of the resistors of two resistors connected in parallel.

The answer is supplemented. Superposition method: only when the voltage source acts, the current source is open, Kirchhoff's voltage law:

6 = U1 + U2 + U4 because, r1 = r2 total = r3 so. u=2v is the same as the reference direction!

Only when the current source acts, the voltage source is short-circuited, and there is a resistance relationship, and the current is 2A, U=2*2 ohms=4V, so U=2+4=6V

The answer is supplemented. An inactive current source is open (open), and a non-acting voltage source is short-circuited.

1. The active power of all mercury lamps is 250W, kx=, and the mercury lamp current is I=P (active) (U*COS)=250 (220*, the apparent power of each mercury lamp is P=UI=220*, the incandescent lamp is pure resistance, the active power and the apparent power are the same, both are 1kw, and its working current is 1000W 220V=.

The working current of the line is the sum of the currents of each lamp.

The total power factor of the line is the ratio of the total active power to the total apparent power, the total active power is, the total apparent power is, and the total power factor is.

What is your load, the average peak load, is the load active power, reactive power, or complex power, it also depends on the power factor of each villa, calculate the load capacity of each villa, plus a certain margin, which is your transformer capacity.