need help again

hello there...

some questions here.

1. I want to modify a transmitter for remote control to be connected to the parallel port of a computer. i am thinking of using a relay, so that when there is a 5V at the output pin of the parallel port, the relay will trigger the transmitter circuit. but what i heard is that there is not sufficient current coming out from the parallel port. is that true? if so, is there any solution?

2. i came across a circuit configuration yesterday, which is shown in the attachment. can some one explain.. the explanation i've heard seems to be a quite wierd.. just to confirm.
it is a circuit that uses serial port to provide input to a hex inverter. so the output of this circuit is to be input for the inverter IC.

thanks

hi,
The PC's parallel port is not suitable for directly driving relays or motors.

You require a transistor to drive motors/relays.

The diagram you posted, is a voltage clamp.
The voltages on a RS232 serial line can swing between -15v and +15v, which is too high for a standard TTL ic.

The circuit clamps the voltage swing to near 0V and +5Volts.

Try to pick up the forum thread that 'JobbinJP' Controlling Switches and Motors via Parallel Port started, its got information regarding the parallel port.

Eric.

ericgibbs said:

hi,
The PC's parallel port is not suitable for directly driving relays or motors.

You require a transistor to drive motors/relays.

Click to expand...

Can you post me a circuit diagram- a simple one will do. what transistor i can use? is 2N2222 suitable. i think thats the only one that i have spare....

ericgibbs said:

The diagram you posted, is a voltage clamp.
The voltages on a RS232 serial line can swing between -15v and +15v, which is too high for a standard TTL ic.

The circuit clamps the voltage swing to near 0V and +5Volts.

Try to pick up the forum thread that 'JobbinJP' Controlling Switches and Motors via Parallel Port started, its got information regarding the parallel port.

Eric.

Click to expand...

+15 ~ -15V??? so it not like parallel port which output is 0V and 5V is it? because that is what my friend told me

one of my lecturer actually explained it in this way:

when the voltage between the resistor is bigger than (5+0.7)=5.7V, then the upper diode will be forward biased, and hence the diode is "bypassed" and the +5V "appears" at that point.
when the voltage between the resistors is smaller than 0V(means negative voltage), then the lower diode will be forward biased, and the 0V "appears at that point.

is that true?

hi,

When the serial input swings to say -15v the bottom diode is forward biassed , so it conducts and clamps the voltage at the junction of the two diodes to -0.7v.

When input swings to say +15v, the top diode conducts and the voltage is clamped to +5.7v.

The resistors are used to limit the diode currents.

Look at the information on the other thread I mentioned.

Attached dwg.

Eric