way to generate 0 to 100% duty cycle, fixed freq, at 30khz

page 12, feed the triangle output into one input of a comparitor, and a control V to the other input

martin

Can't be done! At 0 and 100%, frequency is zero...or undefined, depending on your definition. Pick realistic minimum and maximum numbers and go from there. And before somebody jumps all over me, 0 and 100% are easy. it's the .0000001% that's hard. You can do interesting things with two oscillators and a PLL, but even that gets noisy in the limit. Microprocessors are cheaper than 555's if you pick manageable min/max duty factor.

And what's the control input? It's all about the details. mike

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one pot, one capacitor, 2 diodes, and any of the 6 gates in a HEF40106. (leftover gates can be used for oscillators, one shots, VCO's, even inverters..never understood the use for 555's :-)

Most intensely used part in my junkbox for all "use a 555 / PIC" applications.

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JF

You can sort of do it with a 555, a pot and some passives, but you won't quite get all the way to 0% or 100% duty cycle, just very close.

Set up a 555 in the usual astable circuit, using a potentiometer instead of fixed resistors: connect the pot ends to pins 6 and 8, wiper to pin 7. Connect pin 6 to pin 2, and run a timing cap from there to ground. Install a diode between pins 6 and 7, with the cathode at pin 6. This bypasses one of the resistors during rise-time (output high) and enables you to achieve arbitrarily low duty cycle. I think the frequency works out as F = 1 / (.693 RC), where R is the total pot resistance.

I forgot, you also need to add a fixed resistor at the "hot" end of the pot so you don't short the power supply through pin 7 when you adjust the pot all the way to the end. If you use a 1 k fixed resistor for example, make the pot much bigger, like 100 k.

If you really want 0% and 100% (which, as someone pointed out, is DC), then I'd go with the triangle (or a sawtooth, if you want the leading edge to maintain alignment) and comparator trick.

Have Fun! Rich

Of course, there's still some hysteresis at the ends, depending on how you count it. It's, hey I have an example, it's just like my class D tube amp's transfer curve:

Reason being, the comparator doesn't switch all the way (it has some hysteresis), and has finite rise and fall time, so the narrowest pulses (in the limit as duty cycle approaches 0% and 100%) look sort of Gaussian. If you define a pulse as being "on" only past a certain voltage, then there is a region at either end of the duty cycle range where the signal is not DC (solid "on" or solid "off"), but isn't switching either. And these deadbands are what cause the step in the above plot.

If, on the other hand, you count duty cycle as the continuous averaged voltage relative to full amplitude output, then you can get a continuous range, but the ends may not be very linear.

Tim

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curve:

Some comparitors (eg: LT1016) are fairly linear in the switching area and don't produce hysteresis. The output just won't go all the way. This may be more like the desired result for the extreme cases.