Hi, i've been trying to do this question for quite some while, but no matter how much i search, i cant even start to attempt it , i'm hoping someone could give me some much needed help

 

We don't your homework for you. You have to show us your attempt at solving the problem and then we can take it from there.

The circuit contains three building blocks.
Can you identify the three blocks and describe the function of each block?

 

I just need to know how to approach it, I assume the flip flops are used as counters, then summing inverting amplifiers i think

 

You can start with figuring out the reference voltage Vref. It is formed by a simple voltage divider.

 

Hello Codfail

Assuming that the Q's of Flip-Flops in the RESET state is 0 (zero volts).
No current flows through any resistor R.

Thus the output of operational amplifier located at the top, will be zero volts.

Those zero volts arrive at the negative input of the operational amplifier, located at the bottom.
On its positive input, we have a voltage level approximately +6 Volts.
These +6 Volts resulting from voltage divider formed by the resistors R2 and R3. This voltage is named: Vref.
Because this Vref level is greater, in the positive input, than the negative input, the output of the operational amplifier will have a positive.

Therefore, when the flip-flips are in their RESET state the LED will be off.
And NO ON as I see in the statement on image that you enclose.

Or not?
I believe !

After the first clock pulse, the flip-flop will change its Q state.
Assuming the voltage level at the Q is +5V, a positive current will flow through the R connected to its Q.
As Rin and Rf are the same value, the gain is 1. So the output of this operational amplifier will be -5V.

What supposed to do, the operational amplifier located at the bottom ??.
when its negative input is -5V and its positive input is +6 V ??

 

Hello Codfail

Assuming that the Q's of Flip-Flops in the RESET state is 0 (zero volts).
No current flows through any resistor R.

Thus the output of operational amplifier located at the top, will be zero volts.

The top op amp is a Summing Amplifier.

The bottom op amp is a Comparator.

So following your example:
We start with all the flip flops outputting 0. At the top op amp: 0+0+0=0 volts. The bottom op amp receives 0 volts and compares it to reference voltage Vref. Vref is some kind of positive voltage (OP has not calculated it yet, but it is greater than 0). So. We compare 0 volts to Vref, 0 volts is less than Vref, this means that output of comparator is driven to the negative rail (-12 volts), so the output is -12 volts. Now look at LED. It has -12 volts at the cathode. It has +12 volts minus voltage across R1 (VR1) at the anode. So the voltage across LED is: 12-VR1-(-12)=24-VR1, basically voltage across LED is positive and large enough to turn LED ON.

That is why LED will stay ON as long as the output of Summing Amplifier is less than Vref.

 

Hello shteii01

Sorry, but I can not believe that.

Vref is positive polarity: +6 V
R2 and R3 form a voltage divider between the +12 V and GND.

Note that the non-inverting input of Op Amp is always positive.
while the inverting input is always negative.
So the comparator output always will have positive voltage.
The LED never lights

Or No??

 

Hi MrCarlos.
You are right.
I was just posting similar message.
The schematics is wrong.
Vref should be negative ( -6V )

 

Hi all

Well if this is the case:
The original question was:
Neglecting propagation delay, how long after the first clock pulse will the LED goes off.

The answer might be:
The LED will continue on forever.
No matter that we have changed the polarity of Vref.

Attached are some pictures to clarify this even be more.
But what the originator of this topic says?

 

It's a trick question. As MrCarlos said, the LED never lights; therefore it can't 'go off' after any clock pulse . But as this is Homework, you need to explain why that is.

 

Hi all

Well if this is the case:
The original question was:
Neglecting propagation delay, how long after the first clock pulse will the LED goes off.

The answer might be:
The LED will continue on forever.
No matter that we have changed the polarity of Vref.

Attached are some pictures to clarify this even be more.
But what the originator of this topic says?

I don't get it that the LED will continue ON forever, when Vref is negative.
Do you mean that only one clock pulse is sent?
The clock input is an 0.5Hz oscillator and the LED will go OFF on the third clock pulse as in your fourth picture.

 

I can not figure out how you guys got Vref to be negative 6 volts.

 

Hello shteii01

If you take a careful look at the attached pictures sure find out.
Remember that all measurements are with respect to ground.

 

I can not figure out how you guys got Vref to be negative 6 volts.

If Vref is positive, the the difference V+ - V- is always positive, the output of the comparator is positive and the LED is never ON.

 

Hello shteii01

If you take a careful look at the attached pictures sure find out.
Remember that all measurements are with respect to ground.

Ah! Thank you. I did not notice that +12 was switched to -12 volts.

 

The schematics is wrong.
Vref should be negative ( -6V )

Is there a source for this question and an associated errata sheet correcting the schematic diagram?

We can not "assume" the schematic is incorrect as the comparator has both -12V and 12V indicated on the comparator.

As drawn, the LED never energizes.

Has the professor given instructions to fix any and all "bad" schematics to ensure the wording of the problem statement over rules the associated diagrams? In the forums, I feel it's proper to "fix" any bad schematic to match the words of the problem, but I don't know how their professor feels about it.

 

Is there a source for this question and an associated errata sheet correcting the schematic diagram?

We can not "assume" the schematic is incorrect as the comparator has both -12V and 12V indicated on the comparator.

As drawn, the LED never energizes.

Has the professor given instructions to fix any and all "bad" schematics to ensure the wording of the problem statement over rules the associated diagrams? In the forums, I feel it's proper to "fix" any bad schematic to match the words of the problem, but I don't know how their professor feels about it.

I'm not sure if the schematic was corrected, i found this question while going through some pass papers for the course.

 

I'm not sure if the schematic was corrected, i found this question while going through some pass papers for the course.

So we are left with it's being a trick question with you explaining why it won't work as drawn, or the professor needs to exercise some due diligence when producing diagrams.

 

Agreed. I wondered why the three resistors were not weighed which would have created a 3-bit DAC.

 

This thread has changed Its perspective.

-From:
finding a simple answer to a question.
Neglecting propagation delay, how long after the first clock pulse, will the LED goes off?
-Until:
Positive criticism for the author of the book or the single sheet attached by the originator of this thread.

The entire system that produced the book or the simply copied sheet.
Either way is a system And . . .

A system without feedback will not work.
There will be NO improvements in products or services.
We continue producing our products or services with the same errors.
(Errors mean opportunities to improve, if we correct the errors We are improving).

It is very clear that there is something to improve in the book or on the page of the book.

-To mention some, not with the intention to point out, just as an example.
As JoeJester said: This seems like a trick question. Or: is there any errata sheet?
He also says a few words concerning the teacher. You can see the last paragraph of post #16.
And Post #18.

While we do not do something (feedback) We will continue with the same opportunities to improve.
But to act, we need at least:
The author of the book,
If possible, its ISBN.
The publishing of the book.
With the intention to point out that they have an opportunity to improve.

In this way we will improve the education system. Improving textbooks.
This is a closed chain very, very long.