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Pin 8 accepts a positive input voltage. Vref is the high side of the 5 LED comparators in the datasheet block diagram.

(Datasheet) AN pdf – 5-Dot LED Driver Circuit (1-page)

I will be certain to use them here from now on. Most people use “low” to refer to either a lower voltage, or a lower position on the diagram. 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.

Fadecomic 2 6. If I “correct” the orientation, the circuit does not work at all. Very confusing terminology, since the inverting input can have a higher voltage than the noninverting input. So it determines the DC bias of the block capacitor’s polarity.

That’s clear, because the input in my test circuit is DC. Your description doesn’t agree with that in the datasheet.


If I remove the 2. By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and cookie policyand that your continued use of the website is subject to these policies. The diode will mess up the response of the chip due to its forward voltage drop. I’m not reading it that way. This circuit will light the LEDs as the input voltage on pin 8 increases relative to pin 7.

It just shortens the range of turn over which the potentiometer works. 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.

How does doing anything to pin 7 or pin 8 affect that? However, if that’s the case, I can’t explain why reversing the capacitor doesn’t datadheet, unless the actual polarity of the capacitor is backwards from what you think it is.

Per AN’d datasheet. That means there is a small current flow out from the pin.

AN Datasheet(PDF) – Panasonic Semiconductor

Reverse biased capacitor on IC input pin Ask Question. Setting 7 higher than ground biases the low side of the comparators, allowing them to turn on with a lower output from the amp. I am looking at the test circuits on the data sheet daasheet the AN VU meter IC, and I cannot understand the connection to pin 8 in the following diagram: That is to say that the LEDs all come ann6884 at a much smaller turn of the potentiometer.


To be clear, an688 the capacitor doesn’t light up the LEDs all at once. Pin 8 is the signal input pin, and it’s the internal Op amp’s positive input.

It appears to be reversed. As displayed, the LEDs go from fully off to fully on over roughly the entire turn range of the potentiometer. Capacitor with polarity can be used ac couple or blocking.

AN6884 Datasheet

Sign up using Email and Password. There is also no discharge path with your arrangement. 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. Indeed, it must in order for any of the LEDs to turn on!

Email Required, but never shown. You might want to edit to explain your answer a bit more.