TC9400 Design Questions

Discussion in 'The Projects Forum' started by robby991, Jul 26, 2011.

  1. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Hello, I am building a frequency to voltage converter using Microchip's TC9400 and I dont' quite understand it's operation. I would like to use this in a single supply configuration of 10V, and I attached a schematic which I will be building.

    Pin 11 is the threshold detector. In the datasheet it says that for single supply, the threshold voltage to turn off the comparator is (VDD-Vss)/2 +/-400mV. With a supply of 10V this gives 4.6V to 5.4V. Now if I have a 0-2V input pulse, I would think that the pin 11 input voltage would be going from 5V to 7V, due to the diode and the 1k resistive divider giving 5V on the anode. If so, then the comparator would always be in the off state and never turn on. Can someone please clarify this for me, because I thought it needs to turn on and off which produces a voltage proporational to this frequency in Cint.

    If anyone can calrify to me how this part of the circuit operates I would appreciate it. Thank you.
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    I think to understand the function one needs to absorb the following excerpts from the datasheet:

    "Each zero crossing at the threshold detector’s input causes a precise amount of charge (q = CREF ∞ VREF) to be dispensed into the op amp’s summing junction."

    Means for me detection is made at zero crossing (where zero means the input threshold voltage level of pin 11).

    Then, in section 7.2:

    "If the frequency source being measured is unipolar, such as TTL or
    CMOS operating from a +5V source, then an AC coupled level shifter should be used. One such circuit is shown in Figure 7-1(a).The level shifter circuit in Figure 7-1(b) can be used in single supply F/V applications. The resistor divider ensures that the input threshold will track the supply voltages. The diode clamp prevents the input from going far enough in the negative direction to turn on the start-up comparator. "

    So, the diode is to prevent the voltage at pin to go more negative then 2.5V below the threshold.
    I understand that the internal threshold level is always Vdd-Vss/2. So in dual symmetrical supplies it will be zero volt. In single supplies it will be higher (half of the power supply voltage).

    In order for the average input AC-voltage to be at the threshold level I have to introduce the DC-offset (datasheet diagram 2x10k +1M resistors).
     
  3. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Thank you for the response, however I have read the description many of times and I am still confused about the threshold level portion.

    I understand the diode is used for level shifting to prevent the voltage at pin to go more negative then 2.5V below the threshold, however I thought this circuit works on the principle of the input frequency turning on and off the threshold detector which switches on and off the switch connected to the 12 pf cap (attached internal block diagram, dual supply version). To do this it must go 2.5V below threshold.

    If the input is always in the threshold voltage range, how does it go low to turn on a off the switch? I think I am missing a fundmental part of this.
     
  4. praondevou

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    Jul 9, 2011
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    Actually not, if it goes below 2.5V of the threshold level it activates the self-start comparator which is not desirable during normal operation. One thing doesn't have to do with the other. Threshold level MUST stay above -2.5V, in your case 5V (half of power supply) - 2.5V = 2.5V.

    I got your point you're asking about HOW the threshold switching is done referring to the circled components (see attachment). This is not explained in the datasheet. The only thing they say is that the threshold detector changes it's output on each zero crossing thus activating/deactivating the switch at the 12pF cap. This can be confirmed by the timing diagram. (see attachment.

    The statement in the datasheet "The nominal threshold of the detector is halfway between the power supplies, or (VDD + VSS)/2 ±400mV" is not being explained in detail in the internal drawing of the IC.

    edit: just saw an error in the circuit you posted, I think it's a 3us delay, not 3ms
     
    Last edited: Jul 26, 2011
  5. praondevou

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    Jul 9, 2011
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    Just saw this on another website and made a drawing on how I think the threshold comparator is made (in principle):

    "The input frequency is applied to the pin11 (non inverting input of the internal comparator). In order to trip the comparator the amplitude of the input frequency must be greater than +/-200mV.Below this level the circuit will not work at any situation.

    Whenever the input signal to the pin 11 of IC1 crosses zero to the negative direction the output of the internal comparator goes low. The 3uS delay circuit enables the Cref charge/discharge circuit after 3uS and this connects the Cref to the reference voltage and this charges the integrating capacitor Cint a specific amount of voltage. In the single supply operation the reference voltage is the potential difference between pin 2 and 7 of the TC9400. Each time the input frequency wave form crosses zero towards positive direction, the output of the internal comparator goes high and this disables the Cref charge/discharge circuit which creates a short circuit across the Cref leads. The voltage across the integrating capacitor Cint is retained because the only discharge path available is the 1M resistor Rint which is a too high and the voltage across Cint is the output voltage. Resistor Rbiasis used to set the bias current of the IC."
     
  6. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    I am having trouble understanding why the input voltage must always be above threshold level, because if this were the case then it would be independent of the frequency of the signal. One can then just apply a DC voltage signal above the threshold value.

    I guess my real question is how the frequency alters the output voltage if it is always above the input switching circuit.
     
  7. praondevou

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    I guess I posted at the same time like you did. :) Have a look at my previous post.
    It must NOT be above threshold level. the input AC signal must be AT threshold level. AND the input AC signal must be ABOVE (2.5V) startup comparator level.
     
  8. robby991

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    Dec 17, 2007
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    So is the start-up comparator not in the datasheet?
     
  9. praondevou

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    Jul 9, 2011
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    not in the last picture you posted.It's in the V/F converter section though.

    I drew another diagram for better understanding
    edit: this is for your particular case only
     
    Last edited: Jul 26, 2011
    robby991 likes this.
  10. robby991

    Thread Starter Active Member

    Dec 17, 2007
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    Got it. I was confused because the datasheet was talking about 2 switching voltages but only showed one component in the diagram (threshold detector). I didn't know the comparator in the v/f config. was connected to the same input line. Thanks!!!
     
  11. windoze killa

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    Feb 23, 2006
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    Hi all. Just a quick question related to this topic. With regards to the input signal, the I/P to the comparator triggers on a zero crossing. Somewhere I read that this occurs on the negative going edge. If this is the case this implies that ANY mark to space ratio is acceptable so long as the pulse width is within acceptable times.

    This also seems to be the same way an LM2917 operates (the one we are having so much trouble with). I have been thinking of swapping it for a TC9400 to see if it can handle our I/P. The input is a 10μS pulse that varies from ≈2K to 10K.

    Could someone please clarify the I/P signal operation.

    PS. Datasheet is written by engineers for engineers.
     
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