At present, LED luminescent diode not only but also uses as a kind of light source as the instruction of the mains switch.
A simple circuit that this text describes can test the luminescent diode rapidly. And its type of differentiation is under-current or higher power. The luminescent diode of the under-current only uses the electric current of 1-2mA, but can only offer darker light; And the heavy-duty luminescent diode should produce more high luminance with 10 mA or people’s electric current.
In addition, when there is a plurality of circuit of LED in one at the same time, requires these luminescent diodes when a particular electric current, can send out the same luminance. This requirement can be examined with this test circuit too: Stand up in series two or more luminescent diode, it is convenient for you to compare and choose luminance getting same in the they.
In this circuit Refer to fig. 1 ,Begin to use an adjustable current source.
Enable, adopt first luminescent diode Or two cascade connected luminescent diodes Electric current heighten since O to 20mA,luminance on according
Current transducer to luminescent diode, when the voltmeter is varied from O to the highest volt of values. You can judge which type your luminescent diode is. The luminescent diode of a under-current will be normal and shiny in a relatively trickle, and will not become brighter when the voltmeter reading continue rising; On the contrary, the heavy-duty luminescent diode will continue increasing the luminance with voltage rise slowly.
If want to choose the luminescent diode of the same luminance, you can connect two even more luminescent diodes in series. For example, examine the luminescent diode of red, adopt the voltage when the power of 9V, you can even connect 4 luminescent diodes in series, in order to make, select result to be easy to find out. Need it have,it can if you can’t last mains voltage, get by maximum value 15 V it is two battery of 9V to connect in series! . One that use herein transport, it puts TLC271 to be 16V in mains voltage in acceptable maximum. Under this voltage, you can compare 6 to 8 light-emitting diode tubes green, yellow and red . Test actually the maximum number
Battery monitor of the luminescent diode depends on testing the forward voltage drop of the luminescent diode, the pressure drop of the white light emitting diode is probably 3 . 6V, so you can only measure 3 at the same time at for 15V mains voltage.
Circuit diagram 1 is formed by a classical current source This current source is made up of a transistor-resistor logic and a operational amplifier . Transport and put IC1 and presume good voltage comparison of pressure drop and slide wire variable-resistance device P1 of the emitter resistance R5 of T1. The output voltage of operational amplifier drives the base electrode of T1 after R3 and R4 voltage divider. The reason for choosing this voltage divider lies in reducing the danger of a potential trouble situation: For example, the output that in IC1 goes beyond and supplies the range, can’t become too high through the electric current of T1 exceed at most: A little of 20mA .
But need to pay attention to! If you have improved the mains voltage of the whole circuit, will increase destructively through the maximum current of T1.
Employ the dipole tube D1 of voltage regulation to obtain a reference voltage, make so that the voltage of P1 does not depend on the service voltage. Through about 1mA, enable the steady voltage of Dl 4 current control of Dl only. 2 V, is worth 4 instead of being nominal. 7V . The voltage of Pl is making an appointment for 1V now.
Before assembling this resistance, should pay attention to the actual numerical value of the electric potential device P1. The potentiometer of this kind often has a error of 20.
If your potentiometer deflection has gone beyond 5 of nominal value, then you can adjust the value of R2 in the same proportion.
4 . Voltage stabilizing diode of 7V D2 and D3 It is parallel to examine the luminescent diode with each one. The function of these voltage stabilizing diodes is double. On one hand, when a luminescent diode is removed. Can keep flowing through other paths of current of LED. On the other hand, when instead one the intersection of light-emitting diode and pipe joint, 2 of voltage regulation can prevent the voltage of the luminescent diode from exceeding the reverse cutoff voltage of maximum. This voltage is generally 5V, but may be lower than this value sometimes.
Comments: It puts up the intersection of circuit and method preferably this to be to use one knob omnipotent circuit board, because less component and they on-line to install while being very easy. For convenience but fast insert and remove the luminescent diode, preferably use 2 spring hole scokets as the interface unit.
The maximum current consumption of the circuit prototype is low in 23mA, the electric current which passes R1 is 1mA. Operational amplifier pass The foot is connected to the positive pole of voltage Set as the low-powered mode : The electric current that it consumes is only each now.
If hope to test more luminescent diodes Safety Ground at the same time, can use one to be single, more high voltage is that a bunch of luminescent diodes supplies power, but please notice, can’t exceed the withstand voltage of the crystal tube T1. If is necessary, when T1 voltage is very high, can use the power triode and install the heat sink. However, don’t forget for each light-emitting diode pipe coupling and a voltage stabilizing diode, in this way, will be much safer. Fig. 2 is the material object photo of this circuit.
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