Can frequency affect conversion time in a system?

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Increasing the frequency in a system is generally believed to reduce conversion time, as higher sampling rates can lead to quicker data processing. The discussion revolves around a circuit that functions as a digital-to-analog converter, where a counter processes clock pulses to generate binary outputs. The op-amp acts as a comparator, determining when the counter should stop based on the relationship between the analog input signal (Va) and the counter's output (Vx). The circuit's operation hinges on the AND gate allowing signals to pass only when both inputs are active, facilitating the counting process. Ultimately, the system's functionality is dependent on the precise interaction between these components, particularly when Vx equals Va, which halts the counting until a reset occurs.
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I have this system,


http://img696.imageshack.us/img696/488/vau2.jpg

And I'm asked to explain the meaning of increasing the frequency of the system to conversation time. Of course I would imagine the more frequency, the less conversion time.


I presume that is what I found to be "sampling rate". http://en.wikipedia.org/wiki/Sampling_frequency


But I haven't found any formulation to relate frequency of the system to conversion time...is there such one?
 
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and...apparently I'm Evo now...
 
can you describe in words what the circuit does? Presumably, the counter is at zero when you start, and Vx=0...

Oh, and I'm Evo, too now.
 
Kinda weird. According o the question it's a system that converts analog voltage to digital. Though, according to the sketch, it's a D to A system.

I believe that it is a digital to analog converter according to the questions I see.

So going from this standpoint.

we have a counter that counts the clock pulses and present the result usually in binary form. Then you got the converter who converts the binary signal to analog. First a resolution is calculated through the resolution formula, then the binary signal is converted to decimal and using the formula Va = Resolution x The Decimal Value of the Binary signal

We can find out Va.

I think the role of the op-amp is kind of like a switch (not entirely sure). Vcc is the supply voltage. U2 is an AND gate that takes together the inputs from VCC, and the clock pulse to form a combined signal. I think. What does all of that mean... I think I lost myself. Any guidance, please?
 
The key to the circuit is when the counter stops counting. Can you find out when that happens?
 
Va is the analog input signal.
D0-D7 together form the digital output signal.

The counter counts up on a positive pulse, or down on a negative pulse.

Let's assume the counter starts at zero, meaning Vx starts at 0 as mister Quack suggested.
What would then be the output of the op-amp?
 
Va is the analog input signal.
D0-D7 together form the digital output signal.

Are you telling me that anything other than the D to A is basically a mechanism to convert analog to digital?

It appears to me that the counter is in fact the converter.

The counter counts up on a positive pulse, or down on a negative pulse.

Alright.

Let's assume the counter starts at zero, meaning Vx starts at 0 as mister Quack suggested.
What would then be the output of the op-amp?

That's kind of a trick question, because I'm used that in comparators I have the supply voltage-- then I know. But I really don't know what to make of this Vcc that's NOT connected to the comparator, and the fact this weird comparator does not appear to have any supply voltages.

The key to the circuit is when the counter stops counting. Can you find out when that happens?

When V- = V=

or when Vx = Va

Based on the comparator formula.
 
Femme_physics said:
Are you telling me that anything other than the D to A is basically a mechanism to convert analog to digital?

It appears to me that the counter is in fact the converter.
You are right, the counter provides the digital outputs D0-D7

Alright.
Not sure if the counter can count up and down. Given that U2 seems to be a simple logic gate, I'd say it only counts up when it gets a pulse. To make the circuit useful in practice one would have to add a reset switch.

That's kind of a trick question, because I'm used that in comparators I have the supply voltage-- then I know. But I really don't know what to make of this Vcc that's NOT connected to the comparator, and the fact this weird comparator does not appear to have any supply voltages.
U1 works as a comparator. Vcc and ground connections are just not shown.

When V- = V=

or when Vx = Va

Based on the comparator formula.

Exactly.

Now, knowing that, can you again describe in words what the circuit does and how it works?
 
You are right, the counter provides the digital outputs D0-D7

So a converter and counter are the same thing?

U1 works as a comparator. Vcc and ground connections are just not shown.
So how do you explain the Vcc in the drawing?

Exactly.

Now, knowing that, can you again describe in words what the circuit does and how it works?

The clockpulse starts the count, and together with Va the AND gate assures the signal goes out only if both legs receive input. The counter converts the analog signal to binary digits (based on the resolution...number of legs...etc)...and the D to A converter re-converts the digital signal to analog which gets to the minus leg of the comparator. When Vx reaches a high enough value to equal Va the system restarts.
 
  • #10
Femme_physics said:
So a converter and counter are the same thing?
The whole circuit acts as a converter. A counter is just a counter, but in this circuit it provides the output signals.

So how do you explain the Vcc in the drawing?
Don't really know. For a standard OpAmp this should work without the pull up. For an open collector you need the pull up.

The clockpulse starts the count, and together with Va the AND gate assures the signal goes out only if both legs receive input.
Perfect so far.
The counter converts the analog signal to binary digits (based on the resolution...number of legs...etc)
? The counter only receives digital pulses from the AND gate U2...
...and the D to A converter re-converts the digital signal to analog which gets to the minus leg of the comparator. When Vx reaches a high enough value to equal Va the system restarts.
Close, but no cigar. When Vx reaches Va, the comparator switches to 0 and the AND gate U2 blocks any further pulses from the clock. The counter stops.
 
  • #11
Close, but no cigar. When Vx reaches Va, the comparator switches to 0 and the AND gate U2 blocks any further pulses from the clock. The counter stops.

I would say the show stop before the AND gate, in the Op-Amp, but using the AND gate as the show-stopper always is sensible to me.

? The counter only receives digital pulses from the AND gate U2...

The counter RECIEVES digital pulses? Are you sure? The way I see it, it's:

From D to A, to the counter - ANALOG
From the counter to D to A - DIGITAL

I don't see how it changes along the way. Picture this, at Vx you agree that the signal is analog, yes? So...keeps going...reaches the op-amp...op-amp doesn't change it... keeps going... reaches the AND gate with the pulse...still analog...reaches the counter...turns binary...then hits the D to A again and returns to analog!

Voila!

The whole circuit acts as a converter. A counter is just a counter, but in this circuit it provides the output signals.

But you would agree that there are two converters, the counter which acts as a A to D converter, and the D to A converter in the sketch?
 
  • #12
Hi Femme_physics. The counter really is just a binary counter. As input it accepts a binary clock signal to cause it to increment its binary counting. As output we have 8 digital outputs, each carrying a binary 1 or 0. There is no analog level anywhere there.

The D-A converter accepts these 8 binary lines, and produces a corresponding analog output, Vx.

See how you go with that. :smile:
 
  • #13
Femme_physics said:
I don't see how it changes along the way. Picture this, at Vx you agree that the signal is analog, yes? So...keeps going...reaches the op-amp...op-amp doesn't change it.
The op-amp is a comparator, and its output is a digital signal, binary 1 or 0. The resistor to Vcc is symbolic of a dodge that the designer is using to match the comparator's output voltage levels (maybe -Vcc and +Vcc) to what the conventional AND gate expects/requires/demands as input (viz., 0v and +5v).

Everything following the op-amp O/P is digital. :wink:
 
  • #14
Hi Femme_physics. The counter really is just a binary counter. As input it accepts a binary clock signal to cause it to increment its binary counting. As output we have 8 digital outputs, each carrying a binary 1 or 0. There is no analog level anywhere there.

I'm confused. You have a pulse coming in called Va, right? It goes into the AND gate...so it's in fact, the AND gate, who actually converts it to a digital signal. So, is an AND gate starts the process of analog to digital conversion. But an AND gate can only produce 1 or 0. Once the signal goes in the counter, the counter let's out a digital output.

The digital outputs is dependant on its count. It does not depend on Va (from what I understand). Va is only important to know till which number the counter counts to. Because when Vx = Va the system restarts the count.

Did I get it now? And, I still don't get the point of Vcc.

The resistor to Vcc is symbolic of a dodge that the designer is using to match the comparator's output voltage levels (maybe -Vcc and +Vcc) to what the conventional AND gate expects/requires/demands as input (viz., 0v and +5v).

So Vcc is a supply voltage to the AND gate, and has been drawn just for general reference and not for practical usages?
 
  • #15
Femme_physics said:
I'm confused. You have a pulse coming in called Va, right? It goes into the AND gate...so it's in fact, the AND gate, who actually converts it to a digital signal. So, is an AND gate starts the process of analog to digital conversion. But an AND gate can only produce 1 or 0. Once the signal goes in the counter, the counter let's out a digital output.

Va is not coming out of the op-amp.
The op-amp is a voltage comparator that compares Va to Vx.
It's output is either high voltage (digital '1') or low voltage, probably Earth (digital '0').
The digital outputs is dependant on its count. It does not depend on Va (from what I understand). Va is only important to know till which number the counter counts to. Because when Vx = Va the system restarts the count.

Huh? :huh:

Why would the system restart the count if Vx = Va?

It is true that when Vx=Va, the digital output corresponds to the analog value Va.
Did I get it now? And, I still don't get the point of Vcc.

Neither do I.

It doesn't seem to serve any purpose, but it won't hurt either.
It would be more logical if they had drawn Vcc as being connected as a power supply to the op-amp, and if they had drawn Earth as being connected as a (negative) power supply as well.
 
  • #16
Va is not coming out of the op-amp.
The op-amp is a voltage comparator that compares Va to Vx.
It's output is either high voltage (digital '1') or low voltage, probably Earth (digital '0').

Really? So the supply voltages for the op-amps are:

http://img856.imageshack.us/img856/821/vccvccvcc.jpg

Huh? :huh:

Why would the system restart the count if Vx = Va?

Because there's no voltage based on the comparator formula of (V+) - (V-)
 
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  • #17
Femme_physics said:
Really? So the supply voltages for the op-amps are:

http://img856.imageshack.us/img856/821/vccvccvcc.jpg

Yep!
Because there's no voltage based on the comparator formula of (V+) - (V-)

If there is no voltage difference between V+ and V-, the op-amp's output is zero.
That wouldn't reset the counter, would it?
 
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  • #18
Reset...well, no, not reset, but it would stop the system, UNTIL someone or something resets it. It'd be stuck without a reset mechanism.
 
  • #19
Femme_physics said:
Reset...well, no, not reset, but it would stop the system, UNTIL someone or something resets it. It'd be stuck without a reset mechanism.

Well worded! :approve:
 
  • #20
Oohh... then I think I have it figured out... I'll try to put the whole explanation in place tomorrow... thank you very much ILS. You so smart :smile:xoxo
 
  • #21
:blushing:
 
  • #22
:)

I'll have a go now:

The circuit works by getting a digital one coming from the op-amp into the not gate, whereas the clock pulse, provided it's working, also sends a digital 1 signal. Only if those 2 signals are present then the AND gate permits them through into the counter, that starts the count (with a speed determined by the clockpulse) to produce a binary output, and as well to compare the analog value of the binary output (vla a converter) to Va. Provided Va > Vx, the circuit will keep running.
 
  • #23
Very good.
 
  • #24
Thank you!
 
  • #25
Femme_physics said:
The circuit works by getting a digital one coming from the op-amp into the not gate
All correct, except the analog comparator cannot be termed a NOT gate.
 
  • #26
I meant an AND gate. U2 is an AND gate, no?
 
  • #27
Femme_physics said:
I meant an AND gate. U2 is an AND gate, no?
Yes. And U1 is an analog comparator (possibly, but not necessarily, using an op-amp).
 
  • #28
There's actually another confusion that just hit me.

I'm asked to calculate Va, if the counter output is (01101110)2

But Va is just a value that allows the counter to start working. It does not determine the output. Therefor this question is a bit inane...


Unless the question means "what is the minimal value of Va when the counter produces this signal?"
 
  • #29
Va is not just a value to get the counter working.

Va is the analog input signal.
When the counting stops, the digital output matches the analog input signal.

Yes, with a higher Va, the counter will produce the signal for a very short time, but then it will continue counting to the actual value of Va.
 
  • #30
So Va basically defines the MAX value the counter can count to?
 
  • #31
Yep.
 
  • #32
But how can I find the resolution based on Va?

My idea was that Va max defines our Vref, but if Vref isn't defined to us we can't tell the resolution and thus can't solve for Va if given a digital output
 
  • #33
What is Vref?
I don't see it in your problem statement.
Or did you mean Vx?
 
  • #34
Doesn't every D/A has a Vref? How can it function without a analogue input?
 
  • #35
Femme_physics said:
Doesn't every D/A has a Vref? How can it function without a analogue input?

I'm getting confused here.

In your problem statement there is no Vref.

And yes, a DAC would typically have a Vref connected.
Should I assume you mean the voltage that the maximum digital input maps to?

Either way, such a Vref is not really an input, since it is fixed and usually not shown.
 
  • #36
I like Serena said:
I'm getting confused here.

In your problem statement there is no Vref.

And yes, a DAC would typically have a Vref connected.
Should I assume you mean the voltage that the maximum digital input maps to?

Either way, such a Vref is not really an input, since it is fixed and usually not shown.

So to find the resolution of the converter I just use Va Max?
 
  • #37
Femme_physics said:
So to find the resolution of the converter I just use Va Max?

I'm still confused.

Which resolution?
What would Va Max be?
 
  • #38
Suppose Va max is 5. The binary output of the converter in question (the one at the original message) is 01001101. I am asked to find Va in that case. You're asking me "what resolution", but You can't solve it without one because you have the formula Va = Resolution x Output in base 10


Hmm... Do you follow me?
 
  • #39
I'm asked to calculate Va, if the counter output is (01101110)2
In the absence of a specified Vx max (or Va max) you could express your answer in terms of "Vx max", the maximum analog output available from the DAC.

So, if 11111111 → Vx max,
then 01101110 → k⋅Vx max
Find k.

You should be able to express resolution in terms of Vx max

I'm not sure how punctilious one should be in answering this question. But for the comparator to stop the action, Vx necessarily must always be able to exceed Va, so this implies Va max < Vx max
 
  • #40
Femme_physics said:
Suppose Va max is 5. The binary output of the converter in question (the one at the original message) is 01001101. I am asked to find Va in that case. You're asking me "what resolution", but You can't solve it without one because you have the formula Va = Resolution x Output in base 10Hmm... Do you follow me?

Do you have a problem statement?

To find the Va that corresponds to (01001101)2 you would need the maximum Vx output your DAC can give.
 
  • #41
Hi ILS, NasO, I'll clarify the question...


"The following is a main scheme of an A/D converter with 12 bits. Max Va is 5 [V].

In the A/D converter's output, the binary digit 010111110110 was measured. Calculate, what's Vin that was converted to this binary. "

Answer:

IMO I just do this


http://img834.imageshack.us/img834/1905/vinconversion.jpg

Whereas in the formula for resolution

Res = Vmax - Vmin / 2n = 5 / 212 = 0.0012207

Therefor I consider Va max = the Vmax of the converter and use this resolution to reach the answer.
 
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  • #42
Looks right to me!
 
  • #43
Thanks NasO2 :)
 

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