Hi.

Imaginary scenario is as follows:

I have an IC (something like 4000 series), with a reset pin. This resets the IC when the pin is taken low (unusual, I know, but bear with it)

So, I have the pin held high, through a 10K resistor. It is connected low through a transistor (something common, like a BC548).

The transistor will turn on when there is voltage at the base, and take the pin low. That means that when the input goes 'high', it will take the pin low (avoiding the need for an inverter).

1) Is this right?

2) What sort of resistance will the transistor have (collector to emitter), when the voltage at the base is around 6v?
3) Is there any easy way of working this out?
4) Is it roughly the same for all transistors?
5) Could I use the same method to take a pin high?
6) Is there anything I have forgotten to take account of here?

Thanks very much

Tim

 

Yes, make sure you feed the base of the transistor via a resistor though. One minor point, this isn't "avoiding the need for an inverter", the transistor is an inverter.

It doesn't have 'resistance' as such, so it's not a valid question - if you used an FET those do effectively switch a resistance in circuit. But a bipolar transistor draws current, depending on the Hfe of the transistor and the base current.

Base current * Hfe.

No, it depends on the Hfe of the particular transistor. Generally you would try and design it to drive the transistor fairly hard on, and the pull-up resistor would be reasonably high to give a good reliable logic low. The 10K you specified only requires 0.5mA to take it as low as possible.

Yes, use a PNP transistor and a pull-down resistor.

No!.

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lol, ok, by inverter I meant a NOT gate type thing, which would have required an IC of some sort, so this avoids the need for this extra IC. As soon as I posted the questions, I realised that it was a slightly silly question.

Thanks for all the answers, they have really cleared a few things up.

Tim