Ok, pin 6 and 7 are outputs. 6 is labeled OC and 7 is labeled OD. So what does this lead me to?

 

When I said pin 6 and 7. I was talking about LM324 not the counter.
But pin 5 is a counter pin. Do you know what this pin do?

 

Ah ha that was dumb of me, sorry.
The 6 pin is a inverting input of 2nd comparator
Pin 7 it says is a output of second comparator

 

In this circuit you have only one comparator. First op amp(324) work as a voltage follower. And second op amp work as a comparator. He compares the output voltage from R/2R network with voltage set by 100K pot connect to comparator inverting input.
And if R/R2 network voltage is greater than set by 100K pot the comparator output is set in HIGH state. If R/R2 network voltage is lower than set by 100K pot the comparator output is set in LOW state.
So you need to find out how pin 5 of a 191 counter affects the operation of the counter.

 

OK, thanks. I will try and figure this out. I just have to do a few more of these knowledge check questions and I can move on to the next set of lessons. Thank you for the help.

 

Pin 5 as down/up pin. If I'm reading all this right the answer would be, It "counts upward". Am i correct?

 

Do you understand how this R/2R network work ?
As for the counter. The counter will "count up" until comparator non-inverting input voltage (pin 5) is lower than voltage set by 100K pot. And after R/2R voltage is greater than voltage set by 100K pot the counter "count down".

 

I'm not following. The answer to the question is what? How does the R/2R network work?

 

I'm not following. The answer to the question is what? How does the R/2R network work?

The answer is "The circuit counts upward" until counter reach 0011 and then he will "count down" to reach 0010, and then again count up to 0011 and down to 0010. So for 0.8V at the input,the counter output will be switch between 0010 and 0011 for every clock pulse.

R/R2 network work as a digital to analog converter.

See the example for 0011 at the "input" give as 0.9375mV at "output".
Vout = 5V/16 * 3.

16 - because we have a four-bit counter that counted to three 0011B

 

Thanks, that makes sense. I was on the right track, and had the right answer. Just got a little confused. Thanks again.

 

By the way from which books you have all these questions?

 

Just books issued by my school. I just have one more circuit to analyze, that I will post in just a little bit. Its one more circuit, 2 questions about that circuit. It seems easy to me, but we'll see what you say.

 

Here is the circuit. The question is, identify this circuit? The 2nd question is, If the voltage at e2 is zero this circuit could function as a?

Its the same set of answers to choose from. 1. an adder 2. a subtractor 3. an inverting amp 4. a non-inverting amp

Isn't the first answer a non-inverting amp and question 2 is an inverting amp?

Thank you again for the help.

 

For my this circuit is a differential amplifier and if we ground e2 input - inverting amplifier with DC-offset.

 

Right that is what i thought, its a differential amp, but that is not an option based on the set of answers they gave me. So what do you think is going on here?

 

differential amplifier = subtractor

 

Thanks for that. I'm not sure what answer my specific book wants though. I will go with that. I'll let you know if its correct tomorrow, because come hell or high water I'm done with this "knowledge check" tomorrow. Normally I don't spend this much time on something not being graded, but I really felt I needed to do this one to the fullest. Thanks again.