Voltage to Frequency Converter

Hi Everyone

I need to make a VOLTAGE TO FREQUENCY CONVERTER. I am using AD652. I have read datasheet and followed a "SINGLE SUPPLY OPERATION". But i am unable to acheive output. Plz help me to design a working circuit. Please try to make a circuit on single supply. Input voltage will be positive.

Hi,

your picture only shows input and putput.

For sure you need supply and other connections.

Show us your complete circuit according datasheet Fig. 21.

And the informationls like desired frequency range, supply voltage, input voltage range...

Klaus

KlausST said:

Hi,

your picture only shows input and putput.

For sure you need supply and other connections.

Show us your complete circuit according datasheet Fig. 21.

And the informationls like desired frequency range, supply voltage, input voltage range...

Klaus

Click to expand...

Dear Klaus

my input range is 0 to 3.3V and output frequency could be 100Khz (Ful-scale). But right now i am just getting familiar with the IC. i dont know what component i should select so that this ic give me output frequency at any input. Below is circuit i have made so far by the help of Figure 21 (single supply Operation) and Figure 14 (Gain and offset calibration).

Hi,

Did you connect all AGND with DGND?

Klaus?

KlausST said:

Hi,

Did you connect all AGND with DGND?

Klaus?

Click to expand...

yes all ground are connected at their respective locations

Hi,

yes all ground are connected at their respective locations

Click to expand...

This is not what I meant.

Did you connect AGND with DGND?

Klaus

No I did,nt connect AGND with DGND

Hi,

Is there a reason not to connect them?
If not, then connect them and see what happens.

Klaus

Hi,

datasheet says:

In Figure 21, the comparator reference is used as a derived ground; the input voltage is referred to this point as well as to the op amp common mode (Pin 6 is tied to Pin 16).

Click to expand...

In your circuit pin6 is not tied to pin16.

Do this.

Klaus

On the AD652 datasheet, check Figure 21. Single-Supply Mode Positive Voltage Input

If you compare them carefully, you will see the connections in your schematic that are incorrect.

Also, from the same page, there is this note:
Since the input signal source must drive 0.5 mA of full-scale signal current into Pin 7, it must also draw the exact same current from the input reference potential. This current is therefore provided by the 5 V reference.

Your circuit, as wired, won't do that.

Hello ,

KlausST.. I followed the instruction as u mentioned that is pin6 connection with pin16. i did one more thing that i saperated input source. It means ground of input source is not connected with Analogue ground/Digital ground. However, Analogue and Digital Ground is connected together.

Circuit is giving Output frequency. AT 10V = 500 Khz. But waveform is noisy...

Now Another thing is .. As Schemitt Trigger mentioned, that the way Circuit is wired, wont work..

Now what am i supposed to do?

should i make a input current source and modify this circuit.

Here is the current circuit diagram

Hi,

to get rid of the noise your input circuit supply needs to be refereced to GND.

Here a schematic. I tried to keep it simple. you may add a filtering capacitor across R2.
Edit: This circuit is wrong. See posts below

Connect the shown GND to your system GND.

How it works.
It is a differential amplifier.
The differential input is GND (left of R3) and V_In (left of R1).
R1 = R3
R2 = R4
Gain = R2/R1 = 10k/10k = 1

Pin6 is the "output reference". The circuit adds this voltage to the (amplified) difference input voltage.

Input voltage range is now 0..7V... this can be adjusted with gain setting (R1, R2, R3, R4)

Klaus

Hi,

I just recognized there is a mistake in my schematic.

The two input nodes should be interchanged:
Like here:


Sorry.

Klaus

KlausST said:

Hi,

I just recognized there is a mistake in my schematic.

The two input nodes should be interchanged:
Like here:
View attachment 128162

Sorry.

Klaus

Click to expand...

Hi Klaus

Thanka alot for your efforts. well i just simulated this circuit on proteus. I am not best at electronics just below an ordinary. Kindly tell me why did you suggest this circuit for input. why do my circuit is not suitable?


Another thing is; I read datasheet and came to know that input range of amplifier is (Max= Vs-5 ; Min = -Vs+5). therefore, in my case, Circuit was provided Vs=12V. Input range was (Max = 12-5= 7 ; Min=0+5 = 5) that is 5v to 7V.

Now i have converted my circuit on Dual Supply (Vs=+12, -Vs=-12). Input range has extended from -7v to +7v. But i am just giving 0 to 7v. I have observed output from 100mV to onwards. At 100mV; output Frequency is 5KHz. AT 200mV = 10KHz and so on..

Now the problem is; that output waveform is not stable like input-clock. To watch its motion, i have uploaded its video. Kindly have a look.

**broken link removed**

Here is the new circuit with dual supply.

Hi,

Kindly tell me why did you suggest this circuit for input. why do my circuit is not suitable?

Click to expand...

I already gave the answer:

to get rid of the noise your input circuit supply needs to be refereced to GND.

Click to expand...

Your amplifier circuit was not referenced to GND.

****
Video:
It seems the frequency is stable. But how do you generate the pulses of the 2nd channel?

Klaus

Hi,

Frequency on channel-1 is clock-frequency given to AD652 (Pin-10) provided by function generator.

and frequency on Channel-2 is output frequency of AD652 on Pin-11.

Hi,

I was confused, because I expected an output signal of an usual VCO.
So I reread the datasheet.

Your device is a SVCO.
It has an input clock and it divides the clock according the input voltage.
But the output voltage is synchronous to the input clock.

Imagine you have an input clock of exactely 100kHz and you count the pulses of the output signal.
Now let's say input signal if 20% of full scale range.

One expects 20% of 100kHz = 20kHz at the output. 20 pulses per ms.
If you look at the scope picture you see the input clock and the output clock.
You see that every 5 input clocks there is a single output pulse.
This 1:5 ratio equals to 20% input signal.

Now expect 25% input signal.
One expects 25% of 100kHz = 25kHz at the output. 25 pulses per ms.
If you look at the scope picture you see the input clock and the output clock.
You see that every 4 input clocks there is a single output pulse.
This 1:4 ratio equals to 25% input signal.

But what if the input signal is inbetween 20% and 25%?
Let'say 23%.
One expects 23% of 100kHz = 23kHz at the output. 23 pulses per ms.
If you look at the scope picture you see the input clock and the output clock.
You see that sometime the output ratio is 1:4 and sometimes 1:5.
This is similar to what you see in your scope picture.

But your scope setup shows the "integrated" result of multiple full screen samples.

Please set your scope timing to see more pulses, so that you just be able to recognize the input pulses.
And now set it to 'single shot mode'.
Start a measurement.

With a 23% input signal you see 1:4 and 1:5 clock ratios. But if you count the pulses for 1ms = 100 input pulses you will see 23 (+/-1) output pulses.

If so, then your SVCO works like it should work.

Klaus