Packaging Details
- Unit Type:piece
- Package Weight:0.5kg (1.10lb.)
- Package Size:10cm x 10cm x 10cm (3.94in x 3.94in x 3.94in)
First, functional features
1Simple and easy to use, convenient connection, knob operation, immediate response output after adjustment, without delay.
2The power supply voltage range is wide: the input power is DC5-28V DCIt is recommended that the power supply be12-24V DC.
3The protection function is perfect: with the input voltage reverse protection, the power supply reverse, this instrument does not work, not because of reverse burning.
4Standard current output range: the potentiometer on the product can be generated by manual adjustment4-20MACurrent output for analog current4-20MAOutput occasion.
5The current regulation is very stable: the winding potentiometer is used to adjust the current and adjust smoothly to avoid the sudden change of resistance, such as the common carbon film potentiometer.
6With current meter, accurate display.
7With dial dial, it can quickly locate and recognize the running state of the equipment according to the scale.
8It can fix the current output value: the output current can be automatically and constant according to the regulation of the customer. At the setting point, it can also be used as a constant current source. When the constant current source is used, the output current is not greater than the maximum20MA.
9The golden ratio of the body is easy to grip and turns the work into a pleasure.
10Test output is normal, you can use multimeter (DC current file) short connection“The current is positive and the current is negative”The terminal compares the display value to the multimeter reading.
11Products can be used for: signal source production, valve adjustment, frequency converter control,PLCDebug, instrument test, LED test, analog transmitter output.
Two, about4-20MAThe problem of output impedance matching of current generator
The internal resistance of the output signal must meet the following requirements
According to formula:I=U/R(Electric current=Voltage divided by resistance)
So your power voltage divided by your signal internal resistance is equal to the maximum current you can generate at the receiving end of your signal.
In addition, our current generator has a certain internal resistance.
Thus a formula is derived:
Supply voltage/Signal impedance of customer+The internal resistance of our current generator)=Constant current output
The power supply of our module is4-35VVoltage, so if the supply voltage is low and the internal resistance of the customer is large, the constant flow generated by the module will not be enough.
Here is a signal matching requirement for our module:
24VWhen the power supply, the signal internal resistance shall not be greater than1000Ohm.
12VWhen the power supply, the signal internal resistance shall not be greater than500Ohm.
5VWhen the power supply, the signal internal resistance shall not be greater than200Ohm.
If your signal internal resistance is greater than the above parameters, you can increase the power supply voltage or reduce the internal resistance of this signal2A method to solve the problem of insufficient current supply of constant current source. It should be noted that the supply voltage can not be raised to over28VAbove.
Three, product real shot:
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