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I am looking at the test circuits on the data sheet for the AN6884 VU meter IC, and I cannot understand the connection to pin 8 in the following diagram: enter image description here Pin 8 accepts a positive input voltage. The potentiometer is clear, but the 2.2uf capacitor is not. It appears to be reversed. If I "correct" the orientation, the circuit does not work at all. That's clear, because the input in my test circuit is DC.

This circuit will light the LEDs as the input voltage on pin 8 increases relative to pin 7. As displayed, the LEDs go from fully off to fully on over roughly the entire turn range of the potentiometer. If I remove the 2.2u capacitor, the circuit works, but the range of the potentiometer is greatly reduced. That is to say that the LEDs all come on at a much smaller turn of the potentiometer. Can someone explain what the reversed capacitor is doing here?

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3 Answers 3

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Your description doesn't agree with that in the datasheet. Pin 7 is the output of the internal amplifier, and pin 8 is its input, designed to accept a low-level AC input (57 mV for a 0-dB indication).

It sounds like you're trying to drive this circuit with a signal that includes a considerable DC bias, which explains why all of the LEDs light up right away when you remove (short out) the capacitor.

However, if that's the case, I can't explain why reversing the capacitor doesn't work, unless the actual polarity of the capacitor is backwards from what you think it is. Have you tried a non-polarized capacitor (e.g., ceramic)?

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  • \$\begingroup\$ Right, I get that pin 7 is the output of the amp, but it's part of the negative feedback loop on the amp. Placing it at a higher voltage (within the datasheet's max specified V(7-5)) will both lower the amplification and raise the lower rail of the comparators in the block diagram. To be clear, removing the capacitor doesn't light up the LEDs all at once. It just shortens the range of turn over which the potentiometer works. There's still a (much shorter) range over which a varying number of LEDs come on. \$\endgroup\$
    – Fadecomic
    Nov 21, 2014 at 15:05
  • \$\begingroup\$ I'm not following you. The "lower rail" is the output of the internal V_ref generator. How does doing anything to pin 7 or pin 8 affect that? \$\endgroup\$
    – Dave Tweed
    Nov 21, 2014 at 15:09
  • \$\begingroup\$ I'm not reading it that way. Vref is the high side of the 5 LED comparators in the datasheet block diagram. The LED pins connect to the negative side of the LEDs, and the pins go low when the output of the internal amp goes above Vref (dropped by voltage dividing resistors at each comparator). Setting 7 higher than ground biases the low side of the comparators, allowing them to turn on with a lower output from the amp. \$\endgroup\$
    – Fadecomic
    Nov 21, 2014 at 15:16
  • \$\begingroup\$ Ah. By "low side", you're referring to the inverting inputs of the comparators. Very confusing terminology, since the inverting input can have a higher voltage than the noninverting input. Indeed, it must in order for any of the LEDs to turn on! Most people use "low" to refer to either a lower voltage, or a lower position on the diagram. \$\endgroup\$
    – Dave Tweed
    Nov 21, 2014 at 15:40
  • \$\begingroup\$ True. Mea culpa. I know the correct names, but I conversely find "inverting" and "noninverting" confusing when dealing with comparators, since those terms make more sense when dealing with op-amps. I will be certain to use them here from now on. \$\endgroup\$
    – Fadecomic
    Nov 21, 2014 at 15:47
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Per AN6884'd datasheet,

enter image description here

Pin 8 is the signal input pin, and it's the internal Op amp's positive input. The bias current is specified as -1 µA min., 0 µA max.. That means there is a small current flow out from the pin. So it determines the DC bias of the block capacitor's polarity.

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  • \$\begingroup\$ This makes sense, but why not a diode? Passing Vin through a reverse capacitor seems to be potentially more prone to failure. \$\endgroup\$
    – Fadecomic
    Nov 21, 2014 at 15:20
  • \$\begingroup\$ Capacitor with polarity can be used ac couple or blocking. It's the DC bias determine the polarity. When diode is reverse biased, the input of the Op will lose connection with the input signal, and the voltage will be nearly 0V (take into the bias current, R = 10k ohm, should be 10mV). \$\endgroup\$
    – diverger
    Nov 21, 2014 at 15:30
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you only need to add a 4148 diode forward biased aiming at pin 8 from you wiper on the pot. also add a 1uf from pin 8 to ground. the original cap is in correctly. your just missing the am demodulator the circuit is just like in a diode detector but larger caps.

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    \$\begingroup\$ Welcome to EE.SE. The diode will mess up the response of the chip due to its forward voltage drop. There is also no discharge path with your arrangement. You might want to edit to explain your answer a bit more. You should fix the capitalisation of sentences and electrical units ('F' for farad) to improve legibility and credibility. \$\endgroup\$
    – Transistor
    Aug 3, 2017 at 6:30

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