Nominal diameter: DN10-DN1200 DN150-DN6000 (optional insertion electromagnetic flow meter)
Nominal pressure: 0.6 ~ 4.0Mpa (special pressure can be customized)< /p>< p>Accuracy: ±0.2%, ±0.5%
Measuring range: 0.02～8m/s
Measuring medium (conductivity) : 〉20μs/cm
Medium temperature: -20℃～180℃
Ambient temperature: -10℃～160℃
Ambient humidity: ≤95 % (relative humidity)
Power mode: Built-in lithium battery 3.6V/DC, continuous working time 5-10 years (3.6V, 12V)
< p>Electrode material: 316L stainless steel, HC, HB, titanium, tantalum, platinum, iridium alloy< /p>Coating material: neoprene, polyurethane rubber, PTFE, F46, PFA
Connection method: flange, plug-in, clamp (optional)< /p>
Structural form: integrated type, split type (optional)
Protection level: IP68 , IP65 (optional)
Display mode: large LCD liquid crystal screeninstantaneous bit, flow, (pressure), forward and reverse cumulative total and alarm prompt
Output signal: 0.0001-10 pulse output, 1-1000HZ frequency output can be set arbitrarily (passive optocoupler output)
Communication method: GSM/GPRS Wireless data transmission (data packet in the form of SMS)
What is the principle of voltage transformation of a cell phone charger?
Is it possible to charge a 7.5V rechargeable battery with a 5V charger?
Answer: Yes, but the circuit needs to be slightly modified!
The 7.5 V battery in question 1 consists of two 3.7 V lithium batteries connected in series. Look at the picture below, I removed a battery from under an old device. is exactly the same as what the other person said. When the battery is discharged it is 7.5V, but when fully charged it is 8.4V.
2. Nothing moves. Simply remove the series connection from both batteries and change them to parallel charging. I know this type of lithium battery does not have protection board, but when charging, you just need to connect them in series. With just one diode, you can rest assured that you can safely plug it in and charge it all day and night!
Question 3 said to use a 5V charger to charge. It turns out that you are talking about a cell phone charger, because the commonly used ones in the world are 5V1A, so you just need to find the charging data. There are only two wires inside, one red and one black, one positive and one negative. You can cut the output plug and connect a diode directly to the 3.7V battery connected in parallel. simple.
If your electrical devices really need 8.4V as power supply, then you can add an inteswitch to the battery wiring and turn it to the right for the battery. Parallel charging is 4.2V and turning to the left provides 8.4V power for both batteries in series.
The method above is a simple way to touch just the battery without touching anything else!
3 If you need to use 8.4V to charge the batteries in series, you can only open the cover of the 5V1A charger and connect the isolation output end of the charger high frequency transformer ( optocoupler Connect a resistor of several hundred ohms to a stable 5 V diode to increase the output to 8.4 V. Specifically, you need to test the size of the resistor. You can rotate a variable resistor to get 8.4V, then measure the resistance, just choose a fixed resistor and install it I use this method to adjust the 5V1A charger to any te.voltage between 3 and 12V.
The picture below is an adjustable charging source that I modified using a 5V1A cell phone charger as the positive voltage terminal output of the optocoupler is connected to the negative terminal. the reference voltage, a resistor is connected in series. Simply adjust the resistance during measurement
Thanks for reading!
Generally, as long as the voltages are the same, cell phone charger transformers mainly use small switching transformers. Its working principle is as follows: 220V AC current enters a rectifier circuit composed of four diodes through a fuse resistor, obtains a DC voltage of 300V through a 400V2.2U filter capacitor, and then enters the transistor of switching via a resistor to reduce the voltage to be obtained. primary oscillation to satisfy the work of the circuitit primary. Then the AC voltage of over nine volts is output through the transformer and then stepped down and filtered to obtain a DC voltage of around 5 volts for battery charging. In cooperation with the triode and the integrated circuit to detect when the current reaches the saturated state, the electronic switch is closed to complete charging. Pay attention to measuring whether the pin coils are the same, do not look at the color of the transformer.