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  #1  
Old 04-20-2005, 05:55 AM
blue6x blue6x is offline
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good day to everyone again. i would like to ask if someone could explain the logic behind this frequency to voltage conversion, for tachometer application with analog output. my input is signal frequency and i need an output in voltage. The circuit is using the IC NJM4151 (attached file).
The input is fed in pin 7 and the ouput is at pin 1.
Just would like to ask what is the formula that would invert a frequency to a voltage using this IC or perhaps other IC that has similar application.......
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Old 04-20-2005, 03:49 PM
David Bridgen David Bridgen is offline
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Quote:
Originally posted by blue6x@Apr 20 2005, 06:55 AM
good day to everyone again. i would like to ask if someone could explain the logic behind this frequency to voltage conversion, for tachometer application with analog output. my input is signal frequency and i need an output in voltage. The circuit is using the IC NJM4151 (attached file).
The input is fed in pin 7 and the ouput is at pin 1.
Just would like to ask what is the formula that would invert a frequency to a voltage using this IC or perhaps other IC that has similar application.......
The data sheet in your attachment clearly explains the operation of the device in both modes. I can't impove on that.

Figure 5 shows the formula which relates input frequency to output voltage.
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Old 04-20-2005, 05:02 PM
David Bridgen David Bridgen is offline
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Quote:
Originally posted by blue6x@Apr 20 2005, 06:55 AM
good day to everyone again. i would like to ask if someone could explain the logic behind this frequency to voltage conversion, for tachometer application with analog output. my input is signal frequency and i need an output in voltage. The circuit is using the IC NJM4151 (attached file).
The input is fed in pin 7 and the ouput is at pin 1.
Just would like to ask what is the formula that would invert a frequency to a voltage using this IC or perhaps other IC that has similar application.......
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  #4  
Old 04-21-2005, 03:34 AM
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hgmjr hgmjr is offline
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David has correctly directed you to the section of the datasheet where you will find the equation that allows you to compute the voltage output from the device for a given input frequency when the device is used as an F-to-V converter.

If that is all you needed, then you can ignore the explanation that follows.

I reviewed the datasheet that you supplied in your post and what follows is my understanding of the way the device operates when used as an F-to-V converter.

The device is made up of 3 basic sections

- a voltage comparator
- a one-shot
- a constant current source

The input frequency signal feeds the comparator whose output is used to trigger the one-shot. The pulse from the one-shot is used to enable the constant current source for a period of time governed by the width of the pulse generated by the one-shot. The constant current source is used to charge the capacitor/resistor combination that is connected to pin 1.

The voltage that is developed across the capacitor Cb is a function of four things.

- The width of the pulse from the oneshot
- The magnitude of the current being supplied by the constant current source
- The value of the resistor Rb
- The incoming Frequency signal

The maximum voltage across Cb can be computed by multiplying Rb times the magnitude of the current from the constant current source. In figure 5. we have 100000 ohms times 138.7 microamps which comes to 13.87 volts. A reasonable value since the device is being powered by 15 volts.

Basically the device works on the duty cycle principle. You should be able to take the ratio of the one-shot pulse width "T" as determined by 1.1 times Ro times Co and divide it by the period of the incoming frequency to get a ratio that you then multiply by 13.87 volts to arrive at the voltage that will result on the capacitor/resistor connected to pin 1.

Don't forget to observe the warning that the input signal must be in the form of a pulse that is shorter than the pulse width of the one-shot. It is for this reason that the series capacitor and the resistor to ground at the input to the comparator was included in the design of figure 5. It differentiates the incoming square wave signal to satisfy this input pulse requirement.

If you elect to use this device you will need to consider the use of a buffer amplifier between the output at pin 1 and the next stage of your circuit to prevent unwanted loading of the signal at pin 1. The impedance at pin 1 is quite high so it will not tolerate any significant loading without upsetting the accuracy of the F-to-V converter's output.

I hope this helps answer your question.

hgmjr
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  #5  
Old 04-21-2005, 11:53 AM
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hgmjr hgmjr is offline
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blue6x,

Maybe an example would be helpful to clarify my explanation.

Take the F-to-V circuit in figure 5 of the datasheet you posted and apply a 1KHz squarewave to the input.

In this circuit, T=1.1*Ro*Co so the pulsewidth of the one-shot is 75 microseconds. The period of the 1KHz squarewave is 1 millisecond. If my explanation holds true then the output voltage should be:

Vo = (0.000075*13.87)/(0.001). Solving for Vo, the output voltage at pin 1 should then be 1.04 volt.

If the input frequency is 10KHz:

Vo = (0.000075*13.87)/(0.0001). Solving for Vo, the outpupt voltage at pin 1 should be 10.4 volts.

You can see from the second example that as frequency increases the voltage at pin 1 will increase. Keep in mind that if the frequency at the input continues to increase there will come a point at which the output will reach the maximum voltage of 13.87 volts at which point the output will no longer change with increases in the input frequency. This maximum frequency can be determined by taking the inverse of the one-shot pulsewidth. In the case of the circuit in figure 5, Fmax = 1/(0.000075) or 13.3 KHz. In the circuit of figure 5, input frequencies within the operating range of the device that are greater than 13.3 KHz will produce a constant output of 13.87 volts at pin 1.

Let me know if anything that I have described is unclear.

hgmjr
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Old 04-22-2005, 11:07 AM
blue6x blue6x is offline
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thanks, you have clearly explained what i am looking for. I actually used that formula, but i cannot relate the exact output voltage. but you have made it explained very clear. i now know what happens inside, and what opeartion takes place inside. Thanks guys. Also david, thanks.
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  #7  
Old 04-22-2005, 11:19 AM
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hgmjr hgmjr is offline
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I am glad we were able to de-mystify the F-to-V converter for you. My understanding of the device was also enhanced by the exercise.

By the way, you can begin with the formula presented in figure 5. of the datasheet and derive the duty-cycle version that I described. All that is needed is a couple of substitutions and the duty-cycle expression falls right out.

Good Luck.

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  #8  
Old 09-13-2008, 08:08 AM
enur enur is offline
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hi to all (especially mr. david, blue6x, and hgmjr),

I have to face "that". I mean, I have to use "frequency to voltage" (circuit or IC, I don't know where is the best) for my final project. the problem is that the frequency input is high frequency (min 5 MHz). for beginner student, I am still confuse what should I do. would you all guys give some advice to me?
thanks before
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Old 09-14-2008, 04:36 PM
muni muni is offline
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The circuit is using the IC NJM4151 (attached file).

sir i'm a new member to this forum. i've been asking many of the members aboutt how to attach a file into the threads. sadly i could get any reply. please i request you to guide me in this regard. my email is psmuni@gmail.com


thanking you sir
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  #10  
Old 09-14-2008, 04:50 PM
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hgmjr hgmjr is offline
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Quote:
Originally Posted by muni View Post

sir i'm a new member to this forum. i've been asking many of the members aboutt how to attach a file into the threads. sadly i could get any reply. please i request you to guide me in this regard. my email is psmuni@gmail.com


thanking you sir
Greeting muni,

You can go to the FAQ (Frequently Asked Questions) link in the menu line at the top of the forum page and then read the section on posting in the forum.

hgmjr
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