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Pull up on the output of an LS series IC

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fabiodl:
Just a confirmation on the role of R19 in this schematic.
Is it to assure that the transistor Q1 is off when the output 1 of IC6 is logically high, but on a voltage low enough to have Q1 still conduct?
Shouldn't the values of R18/R19 be swapped to assure that when the output is 2V?

retiredfeline:
If you look at the totem-pole output of the LS02 you will see that there is no current path to ground when the output is logic 1. So R19 is to ensure rapid switch off by draining any base charge to the +ve rail.

MrAl:

--- Quote from: retiredfeline on August 10, 2022, 09:28:37 am ---If you look at the totem-pole output of the LS02 you will see that there is no current path to ground when the output is logic 1. So R19 is to ensure rapid switch off by draining any base charge to the +ve rail.

--- End quote ---


Hi,

That's the same impression i get when i consider the max output of the LS02 gate for a logic high.
The combination of 4.7 and 1k like that only raises the output by about 0.26 volts, which does not seem like enough.  To turn that transistor completely off, the base voltage has to be close the +Vcc.  If the supply is 5v that would mean the base should be about 4.7 volts as a minimum i would say.  Any higher and the transistor will begin to conduct.

The other thing though is that you have to ensure that the transistor can turn on fully too, and if you swap the two resistors the only resistor that can turn it on would be the 4.7k so that would be a reduction in base emitter current of about 1/5 what it was, roughly, plus the 1k would pull it up a lot so there's a chance that it may not turn on at all which is also not good.

What you could do is assume the output of the gate goes from 0.5v for a low to 3.5v for a high and go from there.  Unfortunately we dont know the load at the collector of the transistor so it's a bit impossible to specify any exact values for the two resistors that would ensure both proper turn on and proper turn off.  If you can find that information that would help. If it is around 1ma then we probably dont have to worry too much, but if it has to power a small LED then we'd have to think a little more.  It may be necessary to place a silicon diode in series with the emitter of that transistor, and then the output would look more like a logic high anyway once we get the resistor values right.

Swapping values may just barely allow the transistor to turn on, but changing the values a bit could be better, and of course adding that diode would ensure proper operation with the right resistor values.
Maybe swap but also change the 4.7k to a smaller value, like 3.9k or maybe 3.3k.

We can get more exact if you like.





retiredfeline:
You have to do your calculations on the basis that Vbe is around 0.6V when the transistor is on so the two resistors do not form a pure voltage divider but one that's loaded by the be junction. The base current will be about 4.4/1k minus 0.6/4.7k. BJTs are current amplifiers. It's a mistake to consider voltages only.

As for the 2V or 2.4V for logic 1, this is the voltage required to turn off the current coming from a LS gate input to be considered logic 1. It doesn't mean the totem-pole will rise to this voltage. Although there is a limit to which it can rise without a pull-up due to the upper part of the totem pole. Do this experiment: put a pull-up (within the limit of 10 LS loads) on a totem-pole output and take it to logic 1. You will find that it will rise to 5V because the bottom transistor is turned off, and current will not flow back into the top transistor.

David Hess:

--- Quote from: fabiodl on August 10, 2022, 08:38:16 am ---Just a confirmation on the role of R19 in this schematic.
Is it to assure that the transistor Q1 is off when the output 1 of IC6 is logically high, but on a voltage low enough to have Q1 still conduct?
--- End quote ---

TTL outputs, including LS, only actively rise to the supply voltage minus two Vbe junctions, or about 3.8 volts, which would not be enough to guarantee that Q1 turns off reliably, so R19 pulls the output up to the +5 volt supply.


--- Quote ---Shouldn't the values of R18/R19 be swapped to assure that when the output is 2V?
--- End quote ---

I would have placed the pull-up resistor directly at the output of the LS gate, but the way shown is arguably better.

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