Display Signal Frequency

Discussion in 'The Projects Forum' started by onlyvinod56, Mar 26, 2009.

  1. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
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    HI
    Iam using LM2907 (14 pin) to display the AC signal frequency ranging (10Hz ~ 100Hz). Iam going to use a digital voltmeter 3-1/2 digit (0~200V).
    Can anybody give me the circuit element details for 10Hz / V?
    Please its urgent. Iam designing AC~AC converter where i need to display both input and output frequencies.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Onlyvinod56,
    You are "hijacking" the topic, which causes confusion and is not acceptable on these forums.

    But to answer your question, the LM2907/LM2917 will not be accurate enough for your purpose. You should look at using a microcontroller with a crystal oscillator to count zero crossings, and display the results on an LCD or similar display.
     
  3. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    Hi sgt,
    I dont want to use mc. I want to eliminate the programming schemes in my project. The same project is carrying out by our friends. they are using 89c51 mc, PLC, and even 'labview'.

    I am the one who z trying with the ICs. I had seen in google, that the speed can be displayedon a digital voltmeter where we need some calibration. The DVM is based on ICL7107.

    Coming to the F~V converter. in the page 11 of the datasheet
    http://www.national.com/ds/LM/LM2907.pdf
    the input of the IC is connected to 110v, 60Hz AC. In similar way i will connect 3v (10HZ - 100Hz). The elements mentioned in that schematic are R1 = 111k (i took 110K)
    C1 = 0.01 - 0.1micro farad (i took 0.1micro)
    Vcc = 15V
    The theoritical outputs i got are listed here.
    1.65v - 10Hz
    3.3V - 20Hz
    4.95 - 30Hz
    6.6V - 40Hz
    8.25 - 50Hz
    9.9V - 60hz
    11.55V - 70Hz
    13.2V - 80Hz
    14.85v - 90Hz
    16.5V - 100Hz

    I dont have the function generator right now. I have to design this in my college. I want to clear my problem here, so that i can go for the design.

    Will i get the same in practical case? any modifications required here?
    I can use multiplier / gain which is suitable for my DVM.

    Dont say to use mc please.
     
  4. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Original thread (which DID use mc) was here: http://forum.allaboutcircuits.com/showthread.php?p=12953

    Use a voltage divider. Under similar constraint, I would choose a trim pot and adjust for the required 0.1V per Hz.

    I hope the instructor doesn't require accuracy from this thing...
     
  5. jj_alukkas

    Well-Known Member

    Jan 8, 2009
    751
    5
    ICL 7107 is not fast enough for this purpose.. It takes its own time to display an output.. SO never expect accuracy here.. The other ideas look fine...
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Don't want to, or are not allowed to?
    Oh, it's so much easier trying to track down random noise glitches! :)
    Lucky them!

    Yes, I saw that thread.

    Expect a bright flash, a loud "Kaboom!" and smoke from the IC. Its' maximum input is +-28v (for an LM2907-8). You should use a step-down transformer to about 6.3v or less, and connect one side of the secondary to ground.

    I read R = 100k, and C=0.01uF in the application circuit on page 8 (upper diagram). With those values, Vout = 67Hz/V. Accuracy will largely depend on the precision of the components, but also on manufacturing tolerances, operating temperature, and other uncontrolled elements (like noise on the power supply, etc).

    [eta]Now that I've re-read the datasheet and done some more research, there are some changes that need to be made.
    VO = VCC x fIN x C1 x R1 (times K, the gain, which is typically 1).
    Vcc = 15
    fIN = 10Hz to 100 Hz
    C1 you have chosen to be 0.1uF (100nF)
    Now the problem of the moment is R1, and output saturation level. The output will saturate at Vcc/2, which is 7.5v - so we must stay at or below that. For simplicity, let's choose 5v for the maximum output of the LM2907.
    This leaves R1 to be 33.333...k Ohms.
    5v = 15v x 100Hz x 0.1uF x 33.333...k Ohms
    So, at this point, the output of the LM2907 will be 1v per 20Hz.

    An easier option would be to use a 33nF (0.033uF) capacitor with a 100k resistor; this would eliminate the necessity to recalculate the rest of the formulas in the spreadsheet. If you can't get a 33nF 1% tolerance cap, try to get a value close to it. You'll need to compensate by changing R1.

    In order to increase the output to the desired scale (1v/10Hz), you will need a non-inverting opamp with a gain of 2.
    [/eta]
    Of course, adding an opamp will add it's own errors into the equation (like offset voltage, etc) - the better the opamp you select, the fewer additional problems you will have.

    [eta]You'll need to change C1/R1 as above to keep the output below Vcc/2.[/eta]

    You have some work to do then.

    That is an unknown. I do not have one of these ICs to experiment with, or even a SPICE model to build a simulation with.
    [eta] See the changes in R1/C1 I added above. [/eta]
    Already mentioned. Use a single-supply opamp with a low offset voltage, low noise, and inputs/outputs that can reach ground.

    Keep in mind that the input offset voltage will also be multiplied by the gain of the circuit. An LM324 is an inexpensive quad single-supply opamp, but it has an offset of 2mV or more; this will become 13.4mV or so after amplification. An LT1006 or OP-07 single precision opamp would be a much better choice.

    OK. Instead, I'll say to use a Uc. :D
    While there certainly is a great deal of value in learning linear circuits, this particular application would be far easier to accomplish, far more accurate, and have far fewer components if a single microcontroller had been used instead.
     
    Last edited: Mar 27, 2009
  7. SgtWookie

    Expert

    Jul 17, 2007
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    I have just made some corrections/additions to my reply.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    OK, here you go:

    [​IMG]

    I've eliminated the need for an external opamp. Rather than leaving the internal opamp configured as a voltage follower, I changed the configuration to a noninverting amplifier with a gain range of 1 to 2.5, adjustable by the 5k pot on top. This gives you an output range of 20Hz/1v to 8Hz/1v, which is right in the range of what you need. When the 5k pot has a resistance the same as the 3.3k resistor, your gain should be 2. There are no provisions for compensating for the input offset voltage or for temperature compensation.
     
  9. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
    362
    1
    Thankyou very much sgt..
    my intension is not to hijack the topic. I have read the rules of this forum in which i found

    "1. Before posting a question.

    Use the forum search facility to see if a similar question has been asked before. Chances are your question is similar to an earlier one.
    If your question has not been asked previously, then create a new thread. Please do not hijack another user's thread."

    Thats y i was hanging around this thread. Sorry for this.
    anyways i got ur suggestion. I am going to test that circuit.
    thanks for ur help sgt..
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    I see. Yes, that particular wording could lead to some confusion as how to proceed. Basically, if the thread you found in a search is exactly the same as your questions, go ahead and ask for further clarification in that thread if necessary.

    However, if there are differences (like a different application, etc), you should create your own topic. Reference the thread you'd found during your search that was so similar to your own question by posting a link to it; this may help people help you faster.

    Good. I suggest that before you install the 5k pot, set it to the same resistance that the 3.3k resistor measures. For best results, the 5k pot should be a 10-turn or 21-turn trimpot. If you cannot find such a pot, then you may find it easier to adjust by using a 3k fixed resistor and a 500 Ohm pot in series.

    Measure the output voltage with 0Hz input to find out what your "zero point" is (amplified input offset) using a precision DMM. The specifications for the internal opamp's offset call for 3mV typical with 10mV maximum, so your "zero point" with a gain of 2 will be somewhere in the range of 6mV to 20mV. You could build a resistive divider using a 10k resistor, a 10 Ohm resistor and a 10nF cap from the junction to ground to give you a 15mV reference output. Adjust resistances to match your IC's zero point output.

    Then feed it 100Hz from a precision signal generator, and set the pot so that the output reads 10v, measuring from the reference output to the IC's output. The voltage divider reference output should negate the opamps' input offset voltage.

    Then plot the linearity at 10Hz intervals.
    It would be a good idea to take separate measurements and plot them using your 7107 DMM project.

    Leave the project operating for an hour, and then plot the linearity again.
    This will give you an idea of how much it may drift over time and temperature.

    Let us know your results.
     
  11. onlyvinod56

    Thread Starter Active Member

    Oct 14, 2008
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    So, with 0Hz, Vout will be something around 6-20mV

    Where should i connect this 10K divider...

    Are these the timing elements R(pin 3) & C(pin 2)


    At last..... Iam using LM2907 (14 pin version).
    Iam giving the pin details here. Just chek it out.
    (8 pin) (14 pin)
    1 1
    2 2
    3 3,4 (short)
    4 5
    5 8
    6 9
    7 10
    8 11,12 (short) grounded
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    That is what the specifications indicate if the internal opamp is required to multiply the input voltage by two.
    External to the circuit. It's just a reference offset from ground.

    If you don't understand that, perhaps someone else will explain it to you.

    They have to do with the internal charge pump.

    That's nice.

    Figure out how to adapt what I posted to the IC that you have on hand.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    You know that the 14-pin version's input specification is different from the 8-pin version, right?

    The 8-pin version can go +/-28v. The 14-pin version is 0 to 28v.

    You'll need to either reference the ground side of the transformer to halfway between Vcc, or use an isolating capacitor and voltage divider to keep the AC input signal within limits.
     
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