# Photo-transistor

A photo-transistor optocoupler contains a photo-transistor on the output side. The photo-transistor can be either NPN or PNP.

Unlike when you use a transistor in a conventional manner, you do not always connect the base of the photo-transistor! When the base is used, it is usually connected through a high-value resistor. This configuration is often used to control the sensitivity of the photo-transistor.

We will look at a 4N35 optocoupler and I will demonstrate how to design and build a circuit that uses two independent power supplies, so that you can see the isolation of two circuits in operation.

To understand how to connect the 4N35, take a look at Figure 1. Pin 1 is the anode of the input photo LED. Pin 2 is the cathode of the input photo LED. Pin 4 is the emitter of the output photo-transistor. Pin 5 is the collector of the output photo-transistor. Pin 6 is the base of the output photo-transistor.

As with LEDs that emit light in the visible spectrum, we still need to use a resistor to protect out photo LED. The calculations are exactly the same. I am going to use 3V to power the input circuit. Looking at Figure 2, which is a portion of the datasheet of the 4N35, we can see what the forward voltage and current for the input photo-LED is. The current that we want to flow through the photo-LED is 10mA and the voltage drop across the photo-LED is 1.3V. Using a combination of Ohm’s law and Kirchhoff’s current law, we can calculate the correct resistance required. We are using a 3V supply and therefore we need to subtract 1.3V (the voltage drop across the photo-LED) from our supply: 3V – 1.3V = 1.7V. We know that the current that flows through the photo-LED (10mA) will be the same through the resistor. We can now calculate the required resistance by using Ohms law:

R = V / I

First I convert 10mA to A: 10 / 1000 = 0.01A

1.7V / 0.01A = 170Ω

180Ω is a common value, so I will use that one.

I will also use the same value for the output resistor as the calculation can be considered the same. Technically there would be a small voltage loss from the opto-transistor but when the transistor is switched on, it will be like a wire, with very little resistance. For this circuit we can consider this loss as negligible.

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