# Opamp question

#### swty_todd

Joined Aug 3, 2008
82 The ciruit in question is as above.We are supposed to find the output if the input is a sine wave with a peak of 10V.This is what I think.
The circuit is that of a window comparator.
The input at the non-inverting terminal of the top opamp is 12*9.1/(7.5+9.1) =6.578 V
The input at the inverting terminal of the down opamp is 12*1/(8.2+1)=1.304 V
As far as I understand the output of the comparator will be positive only if the outputs of both the opamps are positive.
The output should be postive saturation only if 1.304 < Vin <6.578
The Blue wave is my Input and the Red is my output.
Am I correct? #### studiot

Joined Nov 9, 2007
4,998
Yep you got it, positive half cycles only, except shouldn't the red line go to ground, not negative volts? i.e. Vo is zero when off.

#### SawabyPlus

Joined Feb 27, 2009
14
i think it's ok. Vo saturates at some value near -Vee.
i am assuming open-collector outputs and the 1k to be a pull-up resistor, otherwise, if not, when each Op-Amp tries to force o/p node to be at different voltage, short-circuit protection will get activated and Vo will be at about zero volts, as studiot has said.
open-collector is widely used for wired AND applications.

#### DonQ

Joined May 6, 2009
321
Does it say somewhere that they are comparators. I see the word "op amp" used a lot in the original post. Since that would make the question invalid, we will assume open collector outputs on a comparator.

There are comparators that have dedicated output transistors with a separate emitter/source pin on the chip. Since it is not listed on the schematic, this also needs to be assumed to be not so. The only other option is the emitter/source connected to the Vee pin.

So how many assumptions did we have to make for this problem? Too many. Does everyone know the engineering definition of "assume"? (It's an old joke, tell me if you don't know it!)

#### swty_todd

Joined Aug 3, 2008
82
i think it's ok. Vo saturates at some value near -Vee.
i am assuming open-collector outputs and the 1k to be a pull-up resistor, otherwise, if not, when each Op-Amp tries to force o/p node to be at different voltage, short-circuit protection will get activated and Vo will be at about zero volts, as studiot has said.
open-collector is widely used for wired AND applications.
Which short circuit protection gets activated? I have to confess that the only thing that i know about window detectors is that output of any one opamp going low brings Vo to zero or low.How that works I fail to understand.
Finally during negative cycles should i take the output to negative saturation or ground as studiot says.

And DonQ the question that was given , i have put it in the first figure.

#### SawabyPlus

Joined Feb 27, 2009
14
as DonQ says, there are many possible answers. the problem comes from different characteristics of available comparators. you should use the characteristics you are familiar with.
as you know, for open-collector comparators, output will be hi-impedance, i.e. a floating open-circuit, for some range of the input above/below some reference, the 1k resistor will pull that open-circuit to Vcc when both comparators are hi. but when one comparator goes low, it will introduce a short-circuit to the negative supply or ground and the resistor can't pull the node up any more.
but how much is that low in voltage? this depends on the comparator itself, for normal cases, other than the one DonQ referred to, it would be the negative supply -Vee.

so output should saturate at some value near Vcc when hi, and some value near -Vee when low. for your problem, it will be hi only when input is between the two reference voltages.

also it doesn't matter if the input singal is negative or positive as long as it doesn't cross the reference voltages.

so i think your answer is correct. also never mind about that short-circuit protection thing, i assumed a general purpose Op-Amp not a comparator.

#### studiot

Joined Nov 9, 2007
4,998
You are correct the swing will be between +V and -V, I misread the original diagram, thinking the negative supply was grounded. Sorry.

#### swty_todd

Joined Aug 3, 2008
82
You are correct the swing will be between +V and -V, I misread the original diagram, thinking the negative supply was grounded. Sorry.
Happens sometimes... #### swty_todd

Joined Aug 3, 2008
82
as DonQ says, there are many possible answers. the problem comes from different characteristics of available comparators. you should use the characteristics you are familiar with.
as you know, for open-collector comparators, output will be hi-impedance, i.e. a floating open-circuit, for some range of the input above/below some reference, the 1k resistor will pull that open-circuit to Vcc when both comparators are hi. but when one comparator goes low, it will introduce a short-circuit to the negative supply or ground and the resistor can't pull the node up any more.
but how much is that low in voltage? this depends on the comparator itself, for normal cases, other than the one DonQ referred to, it would be the negative supply -Vee.

so output should saturate at some value near Vcc when hi, and some value near -Vee when low. for your problem, it will be hi only when input is between the two reference voltages.

also it doesn't matter if the input singal is negative or positive as long as it doesn't cross the reference voltages.

so i think your answer is correct. also never mind about that short-circuit protection thing, i assumed a general purpose Op-Amp not a comparator.

Thanks a lot Sir !