{"id":271,"date":"2013-11-22T17:48:17","date_gmt":"2013-11-22T17:48:17","guid":{"rendered":"http:\/\/www.analysir.com\/blog\/?p=271"},"modified":"2016-02-14T22:15:23","modified_gmt":"2016-02-14T22:15:23","slug":"constant-current-infrared-led-circuit","status":"publish","type":"post","link":"https:\/\/www.analysir.com\/blog\/2013\/11\/22\/constant-current-infrared-led-circuit\/","title":{"rendered":"Constant current infrared LED emitter circuit"},"content":{"rendered":"

Recently, we have been asked several times about driver circuits for IR LEDs, particularly with good range and coverage. Most of us start off driving infrared leds directly from an Arduino or other MCU using a basic resistor in series with the LED connected to a digital output pin of the MCU.<\/p>\n

\"IRsignaltraceRectifierDiodeIN4148\"<\/a>
An oscilloscope snalpshot showing a trace of Ve from the constant current circuit shown below.<\/figcaption><\/figure>\n

 The next step is to progress to a circuit being driven by an NPN transistor, which is more than adequate for >90% of hobbyists. But what if you need that extra bit of range and quality from your Infrared signal?<\/em><\/p>\n

Here we show a constant current circuit for driving one or two IR LEDS from a circuit supplied with 5V.<\/p>\n

\"Constant<\/a>
Constant current IR LED circuit
(Click for larger images)<\/figcaption><\/figure>\n

We have presented two almost identical circuits above with the only difference being that the one on the left is driving 2 x IR LEDs and the one on the right is driving just one. It is important to note that because the current is constant [ I(R1) ~= I(R4) ] you get (almost) twice the IR radiance from the circuit on the left for the equivalent amount of power – nice. Of course you must use a second IR LED. Before reading on please see the Caveat at the bottom of this post.<\/em><\/span><\/p>\n

So let us try to explain the magic. The key to the constant current through R1\/R4 is to keep the voltage Ve constant. This is achieved with the help of the diode pairs (D1\/D2 or D3\/D4). When a modulated IR signal is applied to the circuit the voltage at Vb is either 0V or the sum of the forward voltage for the diode pair (provided sufficient current is flowing through R3).<\/p>\n

So let’s say the forward voltage of the diode is 0.6V then Vb will be 1.2V. It follows then that once the NPN transistor is activated that Ve will be determined by the formula Ve=Vb-Vbe. So if Vbe is 0.7V then in our ideal case Ve will be 0.5V. In fact it will be oscillating between 0V and 0.5V because of the modulation. Next we look at R1\/R4, which at 2.5ohms means that the current going through the resistor will be 0.5\/2.5 = 0.2A or 200 mA. This is of course at the peak of the square wave and the average current @ 50% duty cycle will be 100mA.<\/p>\n

Great that it is so simple in theory, but in an actual circuit you will see some variations with different components, temperature etc. So here are the real world results we obtained this week on a solder-less breadboard:<\/p>\n