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Open-Collector Comparator

Thread starter #1
Hello to all,

have the following question,
many comperators has open collector ouput,
im a bit confused about this,

because i tought, open collector means, if the iput signal is high-the output is open and could be pulled up to lets say V+.
And if the output is not pulled up to a Voltage, what happens then ?

lets say wh have an open collector comparator, the positiv input has 5V and the negativ input has 3 V, and the ouput is not pulled up , what would be the output, 0V ?

would it be better to use an cmos type?
if i use a cmos type, then the output switches internaly from +V to 0V, right?

Thankss
 

Nigel Goodwin

Super Moderator
Most Helpful Member
#2
If it's an open collector (or open drain) output then you MUST have some kind of pull-up, otherwise the output will always be zero.

An open drain CMOS device would be no different, but in both cases you simply add a pull-up resistor.
 

alec_t

Well-Known Member
Most Helpful Member
#4
lets say wh have an open collector comparator, the positiv input has 5V and the negativ input has 3 V, and the ouput is not pulled up , what would be the output, 0V ?
Yes.
would it be better to use an cmos type?
Better in what way? It depends what you want to do with the output.
if i use a cmos type, then the output switches internaly from +V to 0V, right?
Almost 5V and almost 0V (depending on what the ouput load current is), if the supply rails are 5V and 0V.

The LT1716 datasheet says :-
"The output stage includes a class “B” pull-up current source, eliminating the need for an external resistive pull-up and saving power. Output voltage swings to within 35mV of the negative supply and 55mV of the positive supply, which makes the comparator a good choice for low voltage single supply operation. The output stage is also designed to drive loads connected to a higher supply than the LT1716"
 
Thread starter #5
Thank you for your answer,
Yeah and this is the part wich i could not understand well,
it seems to work like a cmos type rigth?:D

if i make the supply rail +5V and 0 V, it has on the output 5v and 0V without external pull up, because it has one internal, right?
 
Last edited:
#8
Open collector allows you to use it in current mode and use it to drive one leg of a current mirror. In this configuration you would not want a pull-up resistor.
 

eTech

Active Member
#9
Hi

What everyone in this thread is implying is that an open collector device allows the output to be used anyway the designer wishes.
It is a non-committed output, meaning it is not tied internally to the supply of the device. For example, in the case of comparator with an NPN open collector, the output could be pulled up thru a resistor to a different +V supply rail than the comparator +V supply rail.
The non-committed CMOS devices work in a similar way except they are called open-drain output.

Hope that helps..

eT
 

MikeMl

Well-Known Member
Most Helpful Member
#10
An open-collector/open-drain output pin usually has its output driver transistor emitter/source internally connected to the device's Gnd pin, meaning that the voltage it pulls down to is pre-defined (near Vgnd), while the voltage it pulls-up to depends on whatever Voltage source the top of the external pull-up resistor is tied to.
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
#11
The hard part with any comparitor datasheet is figureing out how it functions. This https://datasheet.octopart.com/LT6700HS6-3#TR-Linear-Technology-datasheet-130860.pdf#page=1

makes things very messy.

I think it works this way:

Make A, the inverting input and B the non-inverting input.

the comparitors do an A-B (some big number). This is basically what an OP amp does. It subtracts (A-B)*G and multiples the result by the open loop gain

(400 mV-A); the same is true fr both of these comparitors. This is (inverting input)-(the input)
>400 mV; then the output is positive which is really an open. It can't be positive, but pretend it can be.
if <400 mV, the output is negative or the open drain FET is ON.

What's confusing me, is that I want to think of it as the output transistor/FET being on

I think this is the way the logic works.

it's not easy to figure out what the datasheet is trying to tell you.

Anybody think this is a good way or have another way to help figure it out.

if it were an OP amp as a comparitor, i think it would be easier to understand. The open collector and open drain just complicate things.

No where in these datasheets do they make the operation clear IMHO

Anybody else have any ideas?
 

MikeMl

Well-Known Member
Most Helpful Member
#12
Using the "test jig" for the LT6700 built right in to LTSpice:

6700.png

The built-in reference and trip point is 400mV. In this simple test, the pull-up resistor is returned to 5V, but it could be returned to other voltages (see data sheet). Here it is again, this time with the pull-up resistor returned to 3.2V instead of 5V:

6700a.png
 
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MikeMl

Well-Known Member
Most Helpful Member
#14
Except you used the wrong part. It's an LT6700-3.

See what i mean that this datasheet is HARD to understand.
Only difference is that the upper comparitor has its inputs swapped, so V(a) and V(b) are in-phase instead of out-of-phase.
 
Thread starter #15
Thank you all for the answers,
im comig from the software field so its a bit hard to understand the things for me..

first, my question was about lt1716 ond not lt6700
But you have used for the lt 6700 pull up resistor R1 and R2 and my question was if this is also neccesary for the lt1716 ?

Lets say, i use a lt1716 with single supply (V+=5V and V-=0V), and on the positiv input is 8V and in the negative input is 3V, what is the output going to be?
 

eTech

Active Member
#16
The hard part with any comparitor datasheet is figureing out how it functions. This https://datasheet.octopart.com/LT6700HS6-3#TR-Linear-Technology-datasheet-130860.pdf#page=1

makes things very messy.

I think it works this way:

Make A, the inverting input and B the non-inverting input.

the comparitors do an A-B (some big number). This is basically what an OP amp does. It subtracts (A-B)*G and multiples the result by the open loop gain

(400 mV-A); the same is true fr both of these comparitors. This is (inverting input)-(the input)
>400 mV; then the output is positive which is really an open. It can't be positive, but pretend it can be.
if <400 mV, the output is negative or the open drain FET is ON.

What's confusing me, is that I want to think of it as the output transistor/FET being on

I think this is the way the logic works.

it's not easy to figure out what the datasheet is trying to tell you.

Anybody think this is a good way or have another way to help figure it out.

if it were an OP amp as a comparitor, i think it would be easier to understand. The open collector and open drain just complicate things.

No where in these datasheets do they make the operation clear IMHO

Anybody else have any ideas?
There are three different devices on that datasheet. Which one are you referring to?:D
 

eTech

Active Member
#17
Lets say, i use a lt1716 with single supply (V+=5V and V-=0V), and on the positiv input is 8V and in the negative input is 3V, what is the output going to be?
The output is internally referenced. Since, in your scenario, the -input voltage never exceeds the +input voltage the output will be within 55mv of 5V.

eT
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
#18
There are three different devices on that datasheet. Which one are you referring to?:D
I subtlety used the #PAGE=1 directive to open the PDF file to a specific page. https://datasheet.octopart.com/LT6700HS6-3%23TR-Linear-Technology-datasheet-130860.pdf#page=1 This https://datasheet.octopart.com/LT6700HS6-3#TR-Linear-Technology-datasheet-130860.pdf#page=3 will open to PDF page 3.

But, your right, I was somewhat referring the the example of PDF page 1. I apologize for that.
 

MikeMl

Well-Known Member
Most Helpful Member
#19
Thank you all for the answers,
im comig from the software field so its a bit hard to understand the things for me..

first, my question was about lt1716 ond not lt6700
But you have used for the lt 6700 pull up resistor R1 and R2 and my question was if this is also neccesary for the lt1716 ?

Lets say, i use a lt1716 with single supply (V+=5V and V-=0V), and on the positiv input is 8V and in the negative input is 3V, what is the output going to be?
1716.png 1716o.png

Because the + input is more postive than the - input, the output will switch high. How high depends on the load, and if the 1716 is sinking current, or sourcing current, in which case you would have to refer to the graphs on Page 6 of the datasheet.

If the load current is very small as it would be into a voltmeter or scope probe, then you will see just a few mV below whatever pin 5 (Vdd) is tied to.
 

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