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Can someone explain what the reversed capacitor is doing here? Sign up using Email and Password.
Passing Vin through a reverse capacitor seems to be potentially more prone to failure. As displayed, the LEDs go from fully off to ratasheet on over roughly the entire turn range of the potentiometer.
I know the correct names, but I conversely find “inverting” and “noninverting” confusing when dealing with datashheet, since those terms make more sense when dealing with op-amps.
How does doing anything to pin 7 or pin 8 affect that? It appears to be reversed.
Pin 8 accepts a positive input voltage. Per AN’d datasheet.
Your description doesn’t agree with that in the datasheet. Very confusing terminology, since the inverting input can have a higher voltage than the noninverting datashret. Pin 7 is the output of the internal amplifier, and pin 8 ratasheet its input, designed to accept a low-level AC input 57 mV for a 0-dB indication.
By “low side”, you’re referring to the inverting inputs of the comparators. Post as a guest Name. This makes sense, but why not a diode?
The LED pins connect to the negative side of the Datasheey, and the pins go low when the output of the internal amp goes above Vref dropped by voltage dividing resistors at each vatasheet. So it determines the DC bias of the block capacitor’s polarity. I’m not reading it that way.
AN 데이터시트(PDF) – Panasonic Semiconductor
It’s the DC bias determine the datasjeet. It just shortens the range of turn over which the potentiometer works. Pin 8 is the signal input pin, and it’s the internal Op amp’s positive input.
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 will be certain to use them here from now on. If I remove the 2. 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. Indeed, it must in order for any of the LEDs to turn on! There’s still a much shorter range over which a varying number of LEDs come on. You might want to edit to explain your answer a bit more. This circuit will light the LEDs as the input voltage on pin 8 increases relative to pin 7.
Email Required, but never shown. To be clear, removing the capacitor doesn’t light up the LEDs all at once.
Most people dataaheet “low” to refer to either a lower voltage, or a lower position on the diagram. Vref is the datashret side of the 5 LED comparators in the datasheet block diagram.
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.
That’s clear, because the input in my test circuit is DC.
I’m not following you. Capacitor with polarity can be used ac couple or blocking. Per AN’d datasheetPin 8 is the signal input pin, and it’s the internal Op amp’s positive input. The diode will mess up the response of the chip due to its forward voltage drop. Placing it at a higher voltage within the datasheet’s max specified V will both lower the amplification and raise the lower rail of the comparators in the block diagram.
If I “correct” the orientation, the circuit does not work at all.
(PDF) AN Datasheet PDF Download – 5-Dot LED Driver Circuit
Right, I get that pin 7 is the output of the amp, but it’s part of the negative feedback loop on the amp. That means there is a small current flow out from the pin. I am looking at the test circuits on the data sheet for the AN VU meter IC, and I cannot understand the connection to pin 8 in the following diagram: Have you tried a non-polarized capacitor e. Reverse biased capacitor on IC input pin Ask Question.
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