Author Topic: Pull up on the output of an LS series IC  (Read 4852 times)

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Offline fabiodlTopic starter

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Pull up on the output of an LS series IC
« 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?
Shouldn't the values of R18/R19 be swapped to assure that when the output is 2V?
« Last Edit: August 10, 2022, 08:48:42 am by fabiodl »
 

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Re: Pull up on the output of an LS series IC
« Reply #1 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.
 

Offline MrAl

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Re: Pull up on the output of an LS series IC
« Reply #2 on: August 10, 2022, 12:25:00 pm »
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.


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.





 
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Re: Pull up on the output of an LS series IC
« Reply #3 on: August 10, 2022, 01:11:18 pm »
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.
« Last Edit: August 10, 2022, 01:25:55 pm by retiredfeline »
 
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Offline David Hess

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Re: Pull up on the output of an LS series IC
« Reply #4 on: August 10, 2022, 11:36:35 pm »
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?

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?

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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Offline MrAl

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Re: Pull up on the output of an LS series IC
« Reply #5 on: August 11, 2022, 01:55:40 am »
Hello again,

Here are some calculations that illustrate the problems with the circuit and how they can be corrected.  The main idea is that the transistor has to turn fully on and fully off with the respective gate output voltage limitations.


Starting with the two equations for the circuit which includes the two resistors,
the logic gate output Vout, and the transistor base emitter voltages:
VbeH=(Vcc*R2)/(R2+R1)-(VoutH*R2)/(R2+R1)
VbeL=(Vcc*R2)/(R2+R1)-(VoutL*R2)/(R2+R1)

where VbeH is Vbe when the logic gate output is high at the voltage VoutH, and
where VbeL is the 'target' Vbe when the logic gate output is low at the voltage VoutL.
R1 is the resistor connected to the output of the gate, R2 is the other resistor
shown as 4.7k on the schematic as it is now (R19).
We assume Vcc=5 volts.  We assume VoutH=3.5 volts and VoutL=0.5 volts which is
typical for a logic gate output.

First we can set Vcc=5,VoutH=3.5,VoutL=0.5 and get:
VbeH=(1.5*R2)/(R2+R1)
VbeL=(4.5*R2)/(R2+R1)

Now we see that if VbeL was equal to 3 times VbeH, these two would be exactly the same,
and that would also be a decent ratio for the on/off Vbe voltages.  So this would mean
we set VbeL=3*VbeL in those two and of course get:
VbeH=(1.5*R2)/(R2+R1)
3*VbeL=(4.5*R2)/(R2+R1)

and now we can concentrate on the first equation:
VbeH=(1.5*R2)/(R2+R1)

Since we have two unknowns and only one equation, we have to pick one value and solve for the other.
If we choose R1=1-R2 we can solve for ratios instead of actual values and so we end up with:
VbeH=1.5*R2

and now solving for R2 we get:
R2=(2*VbeH)/3

Now if we set VbeH=0.3 volts for the turn off Vbe, we get:
R2=0.2

and since R1=1-R2 then:
R1=0.8

keeping in mind these are ratios.  Taking them both together:
[0.8,0.2]

and multiplying by 5000, we get:
R1=4000 and R2=1000.

Inserting these back into the original equations we get:
VbeH=0.3 and VbeL=0.9

and that seems good.  Note however the actual VbeL will be lower but all we need to do is make sure it is lower than about 0.75, but not as close as say 0.8 due to temperature variations.
The only thing left now is to calculate the expected base current.
Since when turned on Vbe is 0.9 volts and the output of the gate is 0.5 volts,
the voltage across R1 is 5-0.9-0.5=3.6 volts, which means the base current
is 0.9 milliamps.  It will actually be higher because of the actual Vbe voltage so that is good.
If we expect the transistor to remain in saturation, we can only draw about
9 milliamps from the collector, so we'd have to see if the load requirement
would match that criterion.  If it is just a logic output, it probably will
be ok, but if it has to drive anything significant we'd have to reexamine.
The actual current will be a little higher because of the actual Vbe voltage when turned 'on'.

The question that comes up of course is does 1k and 4.7k work instead of
1k and 4k.  The answer is that they might, but the turn on Vbe voltage
would be just barely enough and may either limit the collector output
current too much or may not even turn it on enough at all.
It's simple enough to calculate, just use those values in the first set
of equations instead of the newly calculated values.  Doing that we
find that the turn on Vbe is barely enough, if it works at all.



« Last Edit: August 11, 2022, 01:58:04 am by MrAl »
 
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Re: Pull up on the output of an LS series IC
« Reply #6 on: August 11, 2022, 04:36:44 am »
You could have simplified all that by applying Thevenin's Theorem.
 

Offline fabiodlTopic starter

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Re: Pull up on the output of an LS series IC
« Reply #7 on: August 11, 2022, 07:01:25 am »
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.

It's around 300 ohms. This resistor from the output of a logic gate and this additional logic gate input
 

Offline MrAl

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Re: Pull up on the output of an LS series IC
« Reply #8 on: August 11, 2022, 08:56:45 am »
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.

It's around 300 ohms. This resistor from the output of a logic gate and this additional logic gate input

Hello again,

Ok then the transistor should have a Beta of around 20 then when in saturation or near that, and it probably will if you use a modern transistor.
 

Offline Peabody

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Re: Pull up on the output of an LS series IC
« Reply #9 on: August 11, 2022, 03:12:25 pm »
It seems to me that the LS02's output can't sink current when it's high, so only the 1K resistor is needed - as a base resistor for when the LS02 goes low.  When high, the output is essentially sourced from Vcc through a resistor and two diodes.  If there's no path to ground, then no current can flow through the transistor base, and the transistor will be off.  It will turn on only if the LS02 output goes low.  When the LS02 is high, it would be pulled up to 4.4V through the emitter of the PNP if the 4.7K is not present.  If it is present, then the output will be pulled up to 5V.  But no current will flow in either case, and the transistor will be off.  The 4.7K does affect the transistor turn-on and turn-off times, but I think that's all it does.

 

Offline David Hess

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Re: Pull up on the output of an LS series IC
« Reply #10 on: August 12, 2022, 12:38:52 am »
It seems to me that the LS02's output can't sink current when it's high, so only the 1K resistor is needed - as a base resistor for when the LS02 goes low.  When high, the output is essentially sourced from Vcc through a resistor and two diodes.  If there's no path to ground, then no current can flow through the transistor base, and the transistor will be off.  It will turn on only if the LS02 output goes low.  When the LS02 is high, it would be pulled up to 4.4V through the emitter of the PNP if the 4.7K is not present.  If it is present, then the output will be pulled up to 5V.  But no current will flow in either case, and the transistor will be off.

I am just shooting from the hip here.  There is some leakage current through the schottky diodes used as Baker clamps in the LS process.  The bottom transistor's schottky leakage gets multiplied by its current gain overwhelming the leakage from the top baker clamp diode.  The top transistor is in cutoff so does not amplify its Baker clamp leakage.

Quote
The 4.7K does affect the transistor turn-on and turn-off times, but I think that's all it does.

But I agree that the 4.7 kilohm resistor has the greatest effect on the turn-off time and leakage of the transistor itself.  It should be included with any transistor switch which is not actively pulled off.
 

Offline MrAl

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Re: Pull up on the output of an LS series IC
« Reply #11 on: August 12, 2022, 10:41:58 am »
It seems to me that the LS02's output can't sink current when it's high, so only the 1K resistor is needed - as a base resistor for when the LS02 goes low.  When high, the output is essentially sourced from Vcc through a resistor and two diodes.  If there's no path to ground, then no current can flow through the transistor base, and the transistor will be off.  It will turn on only if the LS02 output goes low.  When the LS02 is high, it would be pulled up to 4.4V through the emitter of the PNP if the 4.7K is not present.  If it is present, then the output will be pulled up to 5V.  But no current will flow in either case, and the transistor will be off.  The 4.7K does affect the transistor turn-on and turn-off times, but I think that's all it does.

Hello there,

You would think that the emitter of an NPN transistor with the emitter 'pulled up' to +5v would have to go to at least close to +5v (or above), but a spice simulation shows the output to be 3.4v with no pullup and 3.48v with a 1k pullup.  So there may be something else at work internal to the LS02 that we cant see on the data sheet.
It could also be that the spice model is not that accurate, and to find out we would have to measure the output with a 1k resistor pullup using an actual real life gate package.
Of course for these tests we ground both inputs so the output is a logic high.
I used 3.5v in my calculations because that is a typical output for the high state.
 

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Re: Pull up on the output of an LS series IC
« Reply #12 on: August 12, 2022, 10:59:35 am »
It seems to me that the LS02's output can't sink current when it's high, so only the 1K resistor is needed - as a base resistor for when the LS02 goes low.  When high, the output is essentially sourced from Vcc through a resistor and two diodes.  If there's no path to ground, then no current can flow through the transistor base, and the transistor will be off.  It will turn on only if the LS02 output goes low.  When the LS02 is high, it would be pulled up to 4.4V through the emitter of the PNP if the 4.7K is not present.  If it is present, then the output will be pulled up to 5V.  But no current will flow in either case, and the transistor will be off.  The 4.7K does affect the transistor turn-on and turn-off times, but I think that's all it does.

Hello there,

You would think that the emitter of an NPN transistor with the emitter 'pulled up' to +5v would have to go to at least close to +5v (or above), but a spice simulation shows the output to be 3.4v with no pullup and 3.48v with a 1k pullup.  So there may be something else at work internal to the LS02 that we cant see on the data sheet.
It could also be that the spice model is not that accurate, and to find out we would have to measure the output with a 1k resistor pullup using an actual real life gate package.
Of course for these tests we ground both inputs so the output is a logic high.
I used 3.5v in my calculations because that is a typical output for the high state.

Those voltages are absolutely standard and to be expected: look at the data sheet and the Voh parameter limits. It only has to be >2.0V, since that is the Vih of an LSTTL input connected to the output.

Other TTL families of similar vintage have similar parameter limits. CMOS 74 logic (e.g.74ACxxx) with the same pinout has the symmetrical output voltage that is familiar in modern logic families, and that can drive LSTTL inputs directly. But the asymmetric 74LS outputs cannot drive 74AC inputs; if you want to do that, you have to use 74ACT devices, where teh T indicates the input parameters are compatible with TTL outputs.
There are lies, damned lies, statistics - and ADC/DAC specs.
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Offline Peabody

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Re: Pull up on the output of an LS series IC
« Reply #13 on: August 12, 2022, 05:46:30 pm »
I breadboarded this part of the circuit using a 74LS32, 1K and 4.7K resistors,  a 2N3906, and a 100R collector resistor as the load.  Here's what I got:

Code: [Select]

LS Output State          High            Low

1K and 4.7K
-----------

Vcc                     5.11V           5.10V

LS out voltage          5.11V           228mV

Base voltage            5.11V           4.17V

Collector voltage          0V           4.74V

Collector current          0mA          48.9mA


1K only
-------

Vcc                     5.11V           5.10V

LS out voltage          4.59V           227mV

Base voltage            4.59V           4.16V

Collector voltage          0V           4.80V

Collector current          0mA          49.1mA


The LS32 had a date code of 8016, so barely broken in.  Anyway, this shows you have to have base current to get collector current.  And when the LS output is high, it can't sink any current, and the transistor will stay off - with or without the 4.7K pullup.


« Last Edit: August 12, 2022, 05:49:47 pm by Peabody »
 
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Re: Pull up on the output of an LS series IC
« Reply #14 on: August 13, 2022, 03:27:51 am »
The one thing that's surprising is the Vbe of the transistor when hard on, about 1V. I suppose it's possible. Could you measure between b and e if you still have the breadboard set up?
 

Offline MrAl

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Re: Pull up on the output of an LS series IC
« Reply #15 on: August 13, 2022, 08:17:03 am »
I breadboarded this part of the circuit using a 74LS32, 1K and 4.7K resistors,  a 2N3906, and a 100R collector resistor as the load.  Here's what I got:

Code: [Select]

LS Output State          High            Low

1K and 4.7K
-----------

Vcc                     5.11V           5.10V

LS out voltage          5.11V           228mV

Base voltage            5.11V           4.17V

Collector voltage          0V           4.74V

Collector current          0mA          48.9mA


1K only
-------

Vcc                     5.11V           5.10V

LS out voltage          4.59V           227mV

Base voltage            4.59V           4.16V

Collector voltage          0V           4.80V

Collector current          0mA          49.1mA


The LS32 had a date code of 8016, so barely broken in.  Anyway, this shows you have to have base current to get collector current.  And when the LS output is high, it can't sink any current, and the transistor will stay off - with or without the 4.7K pullup.

Hello,

That's nice of you to do that test.

There is something that does not look right though with the measurements.
That is, the data suggests that with a 5.7k pullup the output of the LS is 5.11v but with a 1k pullup it's only 4.59 volts.  Is that correct?
Maybe you could post the schematic.  in the mean time, i'll try this test with the spice model too.
If that data is correct as i understand the circuit, then the spice model is a bit inaccurate in some respects.  It could be that the company that made the spice models (2 different simulators) did not expect anyone to connect a pullup on the output of an LS gate.  That also brings up the question should we actually be using a pullup on a gate that does not have an open collector output.
It would be great if we could find an internal diagram that is not overly simplified too.
So if you could post a small schematic that would be good.

Nice of you to do the test this is all we can do sometimes; perform some real life tests.
 

Offline Peabody

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Re: Pull up on the output of an LS series IC
« Reply #16 on: August 13, 2022, 03:10:14 pm »
The one thing that's surprising is the Vbe of the transistor when hard on, about 1V. I suppose it's possible. Could you measure between b and e if you still have the breadboard set up?

I don't still have it set up, but perhaps someone else could give it a try.  You're thinking Vbe should be lower?
 

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Re: Pull up on the output of an LS series IC
« Reply #17 on: August 13, 2022, 03:37:06 pm »
Yes, normally when the transistor is on, Vbe is in the range 0.6 - 0.8 V. Well, the datasheet says about 0.95V at Ic = 50mA and Ib = 5mA so I guess that's ok. https://www.onsemi.com/pdf/datasheet/pzt3906-d.pdf
 

Offline Peabody

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Re: Pull up on the output of an LS series IC
« Reply #18 on: August 13, 2022, 03:52:02 pm »

There is something that does not look right though with the measurements.
That is, the data suggests that with a 5.7k pullup the output of the LS is 5.11v but with a 1k pullup it's only 4.59 volts.  Is that correct?

It's not a 1K pullup.  The 1K goes to the base of the PNP.  The 4.7K goes from the base to Vcc.  I'll attach a schematic.  It's the same circuit as the original post.

With the 4.7K in place, since there is nothing to sink current, the LS output voltage will just be Vcc.  Without the 4.7K, the voltage is simply the one-diode drop across E-B when there's no load.  The resulting voltage is higher than the LS would produce on its own if nothing were connected to it, so the nominal LS output voltage when high is irrelevant.

I should say that in fact the 4.7K probably should be directly connected to the LS output, not to the other side of the 1K resistor.  As it is, you have a voltage divider at the base,  which normally you would not want.  But it probably doesn't matter in this case.


Quote
Maybe you could post the schematic.  in the mean time, i'll try this test with the spice model too.
If that data is correct as i understand the circuit, then the spice model is a bit inaccurate in some respects.  It could be that the company that made the spice models (2 different simulators) did not expect anyone to connect a pullup on the output of an LS gate.  That also brings up the question should we actually be using a pullup on a gate that does not have an open collector output.
It would be great if we could find an internal diagram that is not overly simplified too.
So if you could post a small schematic that would be good.

Nice of you to do the test this is all we can do sometimes; perform some real life tests.

I would expect the Spice model to produce the same result.  As for the pullup question, I'll repeat what I said earlier that unless it is needed to speed up transistor turn-off, or slow down turn-on, I don't think the pullup does anything useful.  I know that many here would always use a base-emitter resistor to bias the transistor off, but I don't do that, and don't think it's needed.  If would be different for a mosfet, but these are bipolars.  In any case, if it's not needed for speed adjustment, then it could be a much higher value, like 100K.
 

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Re: Pull up on the output of an LS series IC
« Reply #19 on: August 13, 2022, 04:47:08 pm »
You don't need the pull-up resistor R19 here, the totem-pole output stage of the 74LS device ensures that Q1 will be turned on and off properly.

Offline MrAl

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Re: Pull up on the output of an LS series IC
« Reply #20 on: August 13, 2022, 05:56:20 pm »

There is something that does not look right though with the measurements.
That is, the data suggests that with a 5.7k pullup the output of the LS is 5.11v but with a 1k pullup it's only 4.59 volts.  Is that correct?

It's not a 1K pullup.  The 1K goes to the base of the PNP.  The 4.7K goes from the base to Vcc.  I'll attach a schematic.  It's the same circuit as the original post.

With the 4.7K in place, since there is nothing to sink current, the LS output voltage will just be Vcc.  Without the 4.7K, the voltage is simply the one-diode drop across E-B when there's no load.  The resulting voltage is higher than the LS would produce on its own if nothing were connected to it, so the nominal LS output voltage when high is irrelevant.

I should say that in fact the 4.7K probably should be directly connected to the LS output, not to the other side of the 1K resistor.  As it is, you have a voltage divider at the base,  which normally you would not want.  But it probably doesn't matter in this case.


Quote
Maybe you could post the schematic.  in the mean time, i'll try this test with the spice model too.
If that data is correct as i understand the circuit, then the spice model is a bit inaccurate in some respects.  It could be that the company that made the spice models (2 different simulators) did not expect anyone to connect a pullup on the output of an LS gate.  That also brings up the question should we actually be using a pullup on a gate that does not have an open collector output.
It would be great if we could find an internal diagram that is not overly simplified too.
So if you could post a small schematic that would be good.

Nice of you to do the test this is all we can do sometimes; perform some real life tests.

I would expect the Spice model to produce the same result.  As for the pullup question, I'll repeat what I said earlier that unless it is needed to speed up transistor turn-off, or slow down turn-on, I don't think the pullup does anything useful.  I know that many here would always use a base-emitter resistor to bias the transistor off, but I don't do that, and don't think it's needed.  If would be different for a mosfet, but these are bipolars.  In any case, if it's not needed for speed adjustment, then it could be a much higher value, like 100K.

Hello again,

Yes that's interesting because the raw transistor model would do that too.
But i suspect that the design point should be more like 3.5 volts, as that is what they specify.  I dont know the reason for that though, but i do know that in the raw transistor model it would take a resistor around 500 Ohms to get the output to go to 3.4 volts or there about.

So it is unclear why the spice model is not showing what you are measuring but it does show what the data sheet specifies.  I'd have to look deeper into this issue but not sure i will have that much time and dedicate that much energy into it :-)

This wouldnt be the first time this happened though, there are specs that are not spec'd on some data sheets and you dont find that out until the breadboard phase of the design.
 

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Re: Pull up on the output of an LS series IC
« Reply #21 on: August 13, 2022, 10:45:40 pm »

Yes that's interesting because the raw transistor model would do that too.
But i suspect that the design point should be more like 3.5 volts, as that is what they specify.  I dont know the reason for that though, but i do know that in the raw transistor model it would take a resistor around 500 Ohms to get the output to go to 3.4 volts or there about.

I don't understand what you're referring to.  Do what too?  500 ohms placed where?  3.4V under what conditions?  Who specifies 3.5V for what?  The original schematic in post #1 has Vcc at 5V, which is what you would need for 74LS.   Are we talking about two different circuits?
 

Quote
So it is unclear why the spice model is not showing what you are measuring but it does show what the data sheet specifies.

Could you be more specific?  Which measurement are you referring to?  What does the spice model say?

« Last Edit: August 13, 2022, 10:49:17 pm by Peabody »
 

Online retiredfeline

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Re: Pull up on the output of an LS series IC
« Reply #22 on: August 13, 2022, 11:21:46 pm »
I don't understand that obsession with 3.5 V and simulation. Just looking at the totem-pole circuit it's can be seen that when the output is logic high, it's effectively open circuit because current can't flow out of the top transistor, so the pull up makes transistor base rise to the 5V rail. The 3.5 V (3.6, 3.7 whatever) would happen if there were a load to ground due to the 2 diode drops. But there isn't a path to ground because the lower transistor is off.

If the transistor load is something like a bulb then turn-off time doesn't matter. Then if the transistor is perfect you don't need a pull-up. But in case you encounter a slightly leaky cb junction, the pull-up ensures that the transistor will never be partly on.
« Last Edit: August 14, 2022, 01:33:20 am by retiredfeline »
 

Offline gamalot

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Re: Pull up on the output of an LS series IC
« Reply #23 on: August 14, 2022, 04:51:08 am »
I don't understand that obsession with 3.5 V and simulation. Just looking at the totem-pole circuit it's can be seen that when the output is logic high, it's effectively open circuit because current can't flow out of the top transistor, so the pull up makes transistor base rise to the 5V rail. The 3.5 V (3.6, 3.7 whatever) would happen if there were a load to ground due to the 2 diode drops. But there isn't a path to ground because the lower transistor is off.

If the transistor load is something like a bulb then turn-off time doesn't matter. Then if the transistor is perfect you don't need a pull-up. But in case you encounter a slightly leaky cb junction, the pull-up ensures that the transistor will never be partly on.

If you think the leakage current of transistor Q1 needs to be considered, then you should not think that the output stage of the LS02 is completely open circuit.

The OP's circuit diagram is incomplete, I found where the /JOYSEL signal goes, it is connected to a 2.2K pull-down resistor and the input of a 74LS32 logic gate.

So in this specific application, removing the pull-up resistor R19 from the base of Q1 will not cause any problems.
« Last Edit: August 14, 2022, 04:57:20 am by gamalot »
 
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Offline fabiodlTopic starter

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Re: Pull up on the output of an LS series IC
« Reply #24 on: August 14, 2022, 08:42:18 am »
The OP's circuit diagram is incomplete, I found where the /JOYSEL signal goes, it is connected to a 2.2K pull-down resistor and the input of a 74LS32 logic gate.
So in this specific application, removing the pull-up resistor R19 from the base of Q1 will not cause any problems.
Thanks!
 


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