A simple task:
Given a battery, a Photovoltaic relay (PVT312) with datasheet, and choice of resistor value, how to select the resistance such that the internal LED current is at a value that ensure output switch portion of the relay is ON properly? Assume Vdd is 2V.
Background study of the problem
1. This is an Photovoltaic relay, to properly switch it “ON”, we must have a min LED current.
2. From the datasheet, the switch characteristic is stated at test condition of 5mA LED current, so 5mA is the desired current, and given Vdd, we need to pick a Rin.
3. From the datasheet, we know that the voltage drop across pin1, 2 (the internal LED) can vary a lot, at 5mA, it can be at 1.5V is the device is operated at -40degC, and only 0.8V at operating temperature of +85degC
4. Since this blog is only meant for beginner, let's just assume our device is only to be operated at room temperature (about 25degC), and thus we only consider the typical curve which has about 1.2V of diode drop at LED current level of 5mA.
Solution method#1 (direct calculation)
1. From the curve, we know that at desired LED current of 5mA, the voltage drop of LED is about 1.2V, so Rin = (Vdd - 1.2V)/5mA, if Vdd = 2V, then Rin = (2V-1.2V)/5mA = 160Ohm.
2. This is really quite simply, and everybody can do it, but there is a better way that gives you a lot more "insight", the load line way...
Solution method#2 (graphical approach – load line)
Just by drawing the load line, we can easily estimate Rin = 2V/13mA = 153Ohm. And it is not much different from the method#1. The difference is about 100*(160-153)/160 = 4%. The slightly different values is due to approximation when reading the graph scale, but in actual world it hardly matters, as when the time passes (aging), or different batch of IC being use, some variances are inevitable.
On the bonus side, using load line we can easily see at extreme operating temperature the actual current about 4mA to 7mA. Without doing any calculation!!!