Parallel Port Interface

Here I am going to discuss the many ways of attaching high current devices to your parallel port interface

Such items include:

Motors
Lights (Low and High Voltage)
Any household device

I will assume that you have already constructed a working interface, if you haven?t then go here to learn how to build one.

The output from your interface should be 5v, it really depends on the buffer chip used in the construction of your interface, if it is not 5v then you will have to make some alterations to allow digital devices to be connected (more detail later)

Analogue connection:

If your interface is giving you a maximum of 4.5 ? 4.9v then don?t worry, as there are other ways of powering your devices. (If you have a 5v relay then you could try connecting it directly to your buffer chip if it fails to work then you will need to follow these easy steps below)

1) Transistors are very versatile they come in many different types and power ratings, the more common general-purpose transistor is the 2N2222, and it will switch about 600mA, which is perfect for a relay. Attaching the collector leg to a power supply not exceeding 12v and the emitter leg to your relay, now as for the base leg you will need to do a calculation to work out the resistance for your relay.

You will need to know:

· The maximum current used by your relay

· The voltage that your interface is giving out

To find the Current (I)

Your relay will have come with some technical specifications more specifically the resistance of the coil, use this resistance with the driving voltage of the relay to find the current needed.

Calculation:

Voltage (V) = Current (I) x Resistance (R) Rearranged to = Current (I) = Voltage (V) / Resistance (R)

E.G. Resistance or my relay = 450W, Supply voltage = 12v

12v / 450W = 0.0266 Amps or 26mA

Back to connecting your base leg, so you have now got a figure for your Current (I), now another calculation will find the resistance needed for the base leg.

This resistor is important without it too much current will flood into the transistor and destroy it.

The current needed by the base leg is = 10% of the current flowing through the transistor at any one time. In my case 10% of 26mA = 2.6mA needed by the base leg, so the calculation is as follows:

Resistance (R.) = Voltage (V) / Current (I)

E.G. Voltage required by the transistors base leg = 0.69 volts, Current needed by the base leg = 0.0026 Amps or 2.6mA

0.69v / 0.0026A = 265W

So a 265W resistor will be needed but this value will be very difficult to get, so a resistor of similar value will need to be substituted.

Continued soon???..