![]() Each node is then connected to transistors in the same way as above. In a chain of 8 Schottky diodes, pull down voltages at each intermediate node with a resistor that will also function as a current limiter (allow something close to the test current from the datasheet to flow when diode is forward biased). 3.3V can support about 8 Schottky drops (~0.4V) if transistor threshold voltage is less than or equal to 0.4V (otherwise, 7 drops). ![]() To redo calculations with other standard values or specs, the formula is: R i = R TOTAL x (1 - V i / 3.3V) -, derived from voltage divider formula.įinal circuit will look something like this:Īnother method of achieving this effect is to use diode drops. Note: standard 5% resistor values in brackets are not simply the closest match, but are calculated iteratively. This reveals resistor values of: (use voltage divider iteratively) R1 = 624.2 (620) If one aims for a maximum current of ~1mA, total R2R resistance should be near 3.3kΩ. Sound amplitude is traditionally (and biologically!) logarithmic, so the voltage at which each should turn on is linear on a logarithmic scale. When a node exceeds the knee voltage, the FET turns on and drives an LED. A FET can be selected such that its knee voltage or turn-on voltage is something manageable, like 0.5V to 1V, its source grounded and gate connected to the various R2R nodes. The output from the aforementioned edge detector is fed through something similar to an R2R ladder (not necessarily with equal value resistors), which provides a number of nodes with voltages from the input to ground. I'd say you can get away with the bar graph part for under $1 per channel if you shop around.Īn alternative to dedicated drivers and LED arrays is to make your own, borrowing the idea behind R2R ladders and flash ADCs. If you go with 8 elements in your graph, and you need 12*8 ~= 100 LEDs, you can do that for $0.042 cents apiece with these green indicator LEDs, or about $0.34 per channel. Just use some bulk 1206 LEDs for the graph and arrange them on your PCB instead of paying for the prepackaged bar graph. ![]() ![]() Assuming that resistors are basically free, you need a diode, a transistor, decoupling caps (also essentially free), a 1uF peak storage capacitor, and an LED graph. If you don't need the LED current control that this chip provides, you can probably do it cheaper with a classic quad comparator like the LM339 or LM2901 (you don't need anything fancy), which will run you about $0.30 in quantities of 25 (you need 24 for twelve 8-channel graphs). With respect to the cost, the LM3914 that you were looking at is just a cascade of comparators with one input connected through a resistor network to a fixed voltage, and the other to the peak detector input. The 1uF capacitor and 470k resistor gave a good decay rate for watching the audio signals that I was tracking. The Vce of the transistor and Vf of the diode should cancel out. This is a lot cheaper than the opamp circuits you'll find elsewhere. I passed the filtered signals through the following peak detector: I went with the LM3914 and a prepackaged LED bar graph because cost wasn't the issue development time was, but you can take the opposite route. I designed 7 separate opamp bandpass filters (using the Bessel filter topology to minimize distortion). I did a seven-channel version of this project a few months ago.
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