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One answer to this question suggests the following circuit to drive a LED from an open collector output of a 74LS47:

schematic

simulate this circuit – Schematic created using CircuitLab

What is the purpose of the R1 resistor? What would happen if R1 would be missing?

I did a simulation, but could not find a different behavior regardless whether or not R1 was present.

(I could have added this question in the original question as a comment to the answer, but I did not want to make the already excessively long list of comments even longer. Therefore I created a separate question. I hope that is OK.)

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I agree with Brian's answer +1. –  Spehro Pefhany Jun 25 at 16:04

4 Answers 4

up vote 9 down vote accepted

That open collector output ... when off, it will have some leakage (possibly microamps) depending on temperature. That can supply enough base current to turn Q2 on, at least partially. The simulation might not model that leakage accurately.

R1 pulls Q2 base to 5V against that leakage, ensuring Q2 is fully off.

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The data sheet mentions a max. Io(off) of 250 µA. This indeed is enough to turn Q2 on. The value of R1, however, is too high to prevent that from happening. I guess a value of 1k or 2k would be more suitable for this worst-case leakage current. –  user36113 Jun 25 at 15:10
    
That's a worst case CYA figure for Vo(off)=15V according to my reading of the datasheet. Actual figures are likely to be a lot lower and 10K would probably be fine outside of a high reliability application. But strictly speaking you are correct (note : another part has a builtin 2k pullup). –  Brian Drummond Jun 25 at 15:22
    
@BrianDrummond It's a feature .. if the LED segments come on then the board is too hot. –  Spehro Pefhany Jun 25 at 17:33
    
I'm sorry, designing a circuit to the maximum leakage current and expecting predictable behavior are mutually exclusive –  Scott Seidman Jun 25 at 21:35

If R1 were missing, the base of the transistor would float when the OC output were off. With the resistor, the base is at 5Volts when the output is off, and very near 5V*1k/11k when the output is on (voltage divider between 5 volts and near ground).

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I'm aware that you should not have floating FET gates (e.g. in CMOS devices), because they would pick up tiny charges from the environment. But what is the reason why you should not have a floating base on a BJT? –  user36113 Jun 25 at 14:05
    
You need to set V_BE properly to turn the transistor on and off. –  Scott Seidman Jun 25 at 14:37

Besides Brian's answer another reason for a pullup resistor is to increase the speed of turnoff. If the LED is just for humans to look at it wouldn't be necessary, but if the LED is for communication purposes then it will increase the maximum rate of signalling.

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Due to the neutralization of charge contained in the base of the BJT, right? I never fully understood this concept but only had a vauge idea. Why does such charge exist and why is it located in the base? I recall it's actually located in the junction of the doped regions, but why does it exist like this? –  sherrellbc Jun 25 at 17:37

The base-emitter junction of Q2 has a finite capacitance. When you turn off the gate (collector current stops), then that capacitance needs to discharge before the BJT turns off. This process will happen much faster when there is a leakage path through the resistor R1.

In fact - if the "off" gate is not quite off, then the transistor Q2 would act as an amplifier of the leakage - with a beta of at least 100, you get a 100x multiplier of the leakage current. But when you have the pull-up resistor, it will ensure that the leakage current does not turn on the transistor, and all is well.

So - the pullup ensures a shorter, sharper turn-off.

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