led driver

What it is
The circuit allows a precision regulated drive current to be set to drive an LED, and in response to a TTL level signal, the LED is switched on and off with rise and fall times of less than 500 nanoseconds and less than 7% overshoot. This was designed for biological experiments to study the photo repsonse of cone cells rabbit eyes. This circuit may also be useful in systems that require a stable pulsed light source, such as those  that measures light transmission (turbidity and densitomitry insturments,  for example) and for the transmission of low speed optical data (to 200 kHz). The components are all easy to obtain and low cost.
How it works
The circuit consists of an adjustable regulated current source (the op amp and Q2), a differential amplifier (Q3 and Q4) driven as a switch, and a level shifter (Q1) to shift the TTL input signal to levels necessary to drive the differential pair. The current from the current source is switched to either the +5 volt supply to the LED by the differential amplifier.
The current current source (explained assuming the current is set to 50 ma):
The current source establishes the current that drives the LED. Voltage on the wiper of R6 is set to 0.5 volts, and this voltage is the reference for the op amp. By virtue of the feedback from the emitter of Q2, the op amp maintains 0.5 volts across R9, which is a 10 ohm resistor. The current through this resistor is 0.5V/10 ohms = 50 ma.
The emitter current is equal to the current through R9 because the op amp’s inverting input does not draw any appreciable current. Q1’s emitter current is shared between the base and collector of Q2, and because of Q2’s high current gain (greater than 100), more than 99% of the emitter current is passed to the collector, and this ratio is stable. Thus, the voltage set by R6 determines the collector current in Q2. The collector of Q2 is connected to the emitters of differential pair Q3 and Q4.
The 4.7k resistor, R8 is used to compensate for input bias drift. It is approximately equal to the resistance seen at the other op amp terminal.
C4 is to reduce the pickup of noise on the noninverting input of the op amp. On my breadboard, a couple of microseconds of ringing was seen on the output of the circuit until I put this capacitor in place. In your particular layout, you might need something like is for the inverting in put of the op amp. If you do, put a small capacitor across R8 instead of from the inverting input to ground because to put it to ground would induce a phase shift that could cause the circuit to oscillate.

led driver

Differential pair Q3 and Q4:
Transistors Q3 and Q4 are connected as an over-driven differential pair. The emitters are supplied current from Q2. Of the two transistors, the one with the highest base voltage will conduct all of the emitter current, which is equal to the collector current on Q2.
The all-or-nothing switching action results from the large difference in base voltages between the two transistors. When Q3 has the high base voltage, current from Q2’s collector goes to the +5 volt regulator. When Q4 has the higher voltage, current from Q2’s collector goes through the LED then to the +12 volt power supply.
When the LED is off, Q3’s base is pulled to about +5 volts while the base of Q4 remains at +2.5 volts, which makes Q3’s emitter more positive than that of Q4, and so reverse biasing Q4’s emitter so it cannot draw any of the current from Q2.
When the LED is on, the base of Q3 is switched to less than 1 volt while the base of Q4 is kept at +2.5 volts, thus assuring that the emitter of Q3 is reverse biased and all the current from Q2 goes through the emitter (and therefore the collector) of Q4.
A small amount of the current from Q2 is used as base drive on Q3 (when Q3 is conducting) and Q4 (when Q4 is conducting). As is the case with Q2, this fraction of the total current is less than 1% of the total and is stable.
Voltage divider R10 and R11 create a 2.5 volt bias current for the base of Q4, while the voltage on the base of Q3 is switched by Q1 to well above and well below 2.5 volts. The 2.5 volt reference is also used to bias the base of the level shifter, Q1.

Level shifter, Q1:

Q1 is a common-base amplifier. Its is essentially the same as a TTL input stage, with the exception of the 1000 pf capacitor across the base resistor, R4. 

When the input signal is greater than 2 volts (the LED is off), the base of Q1 is biased to +2.5 volts and no significant current flows through Q1’s collector. As a result of this condition, the collector voltage and therefore the base of Q3, is held at +5 volts -the condition that causes all the current from Q2 to be dumped into the +5 volt power supply.

When the input signal is lower than 0.4 volts, the emitter of Q1 is essentially grounded and base current is drawn through the 10k resistor causing Q1 to saturate and in doing so, pulling the base of Q3 low. This condition corresponds to all of the current from Q2 being passed through Q4 to the LED.

When the input signal switches from low to high, energy stored in C3 is used to develop a reverse bias across the base-emitter junction of Q1 to deplete the stored base charge, thus reducing storage time and allowing the collector signal to switch within hundreds of nanoseconds rather than several microseconds. C3 needs to be adjusted for the specific type of transistor being used. Too large a value for C3 could result in increased storage time or possibly damage to Q1, therefore, it is important to use the minimum capacitance that provides acceptable storage time (storage time is not the same as rise and fall time).

Read more: FAST PRECISION LED DRIVER using microcontroller

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