The Safe constant current source.
Figure the top is a Safe constant-current source circuit, how it works?
A cmos op-amp (number ICL 7611) controls the input current through a P-channel Hexfet power transistor (No. IRF 9520), then to keep up a constant voltage across the R1.
As they are connected in a serial form, so use the together current by: I = VREF /R1 , while the Vref to be defined by the IC2 is 1.25V.
The advantage of this outline are:
1. The load current is limited by R1 when the load is too heavy.
2. The op-amp and Hexfet there are the overhead voltage very low.
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This is a Constant Current Source LED Driver, When the LED driver Upper-NPN is driven by a voltage thru 4.7K the LED lights up. Assume that the Lower-NPN at bottom is absent. The current via LED and NPN is limited by R. 20mA may be ok 15mA even better. Or LED blows even transistor goes.
source: http://schematics.blogspot.com/2007/02/constant-current-source-led-drive.html
Thank you.
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Current source A current source is an electrical or electronic device that delivers or absorbs electric current. A current source is the dual of a voltage source. Current sources can be theoretical or pratical. I will handle only the practical with the use of a LM317.
Current adjustment:
Now we know the needed input voltages but still don’t have a constant output current. For this we gone abuse the DC Voltage Regulator. We place a resistor in series with the LM317 and the output device (ex. a led) and connect the adj pin over the resistor. Because the LM317 will regulate the voltage on the adj input always to 1.25V, we become a constant current through the resistor and connected device
How it works: Over the resistor there is always a voltage present of 1.25V
This means when the current decrease, normaly the voltage over the resistor will be lower also but what happens now: the regulator lets increase his output voltage to adjust a constant voltage over the resistor of 1.2V
So we can calculate with ohms law what resistor is needed to get a specific current.
R = U / I
R = 1.25V / I
Example: We will supply 3 lumileds 1W power rated in serie with a 12V battery. The nominal current for the leds is 0,35A We can find the proper resistor value with the formula above:
R = 1.25V / I
R = 1.25V /0,35A = 3.57 ohm
Thus we need a resistor of 3.57 ohm but will not find one with this value. To solve this problem take a value that’s higher. Here in the example we will take one of 3,9 ohm.
The real current will be then I = U / R = 1,25V / 3.9ohm = 0,32A what not will be a problem (it extends the lifetime of the leds)
Power rating of the resistor:
It’s easy because:
P = U * I
P = 1,25 * 0.32A = 0.4W
In real we take a resistor with a 10% higher power rating: here we find 1/2 Watt
Source:http://users.pandora.be/davshomepage/current-source.htm
Thank you.
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