Post by thread:[PIC] RC oscillator calculation

Hi all, I've got a PIC project with a speed control that is a pot as the R in RC mode One option I would like is to be able to set a mid-point frequency such that moving the pot fully counter-clockwise will be the same ratio as slow as fully clockwise is as fast The equation would be x * (1/y) < x > x * y For example, in the range 2 to 32, 8 (=x) is the 'midpoint' and y = 4 2 is 8 * 1/4 and 32 is 8 * 4 How would I (can I) resolve the equation to find 'x', the midpoint, if I have just the range, which in this case is 500kHz to 4100kHz ? I can guess at say 1400kHz. 1400/500 = 2.8. 2.8 * 1400 = 3920, which is close but there must be a better way TI

{Quote hidden}> -----Original Message----- > From: spam_OUTpiclist-bouncesTakeThisOuTmit.edu [.....piclist-bouncesKILLspam@spam@mit.edu] On Behalf > Of IVP > Sent: 13 June 2012 11:44 > To: Microcontroller discussion list - Public. > Subject: [PIC] RC oscillator calculation > > Hi all, > > I've got a PIC project with a speed control that is a pot as the R in RC mode > > One option I would like is to be able to set a mid-point frequency such that > moving the pot fully counter-clockwise will be the same ratio as slow as fully > clockwise is as fast > > The equation would be x * (1/y) < x > x * y > > For example, in the range 2 to 32, 8 (=x) is the 'midpoint' and y = 4 > > 2 is 8 * 1/4 and 32 is 8 * 4 > > How would I (can I) resolve the equation to find 'x', the midpoint, if I have > just the range, which in this case is 500kHz to 4100kHz ? > > I can guess at say 1400kHz. 1400/500 = 2.8. 2.8 * 1400 = 3920, which is close > but there must be a better way > SQRT(500kHz * 4100kHz) = 1431.78kHz Cheers Mike ======================================================================= This e-mail is intended for the person it is addressed to only. The information contained in it may be confidential and/or protected by law. If you are not the intended recipient of this message, you must not make any use of this information, or copy or show it to any person. Please contact us immediately to tell us that you have received this e-mail, and return the original to us. Any use, forwarding, printing or copying of this message is strictly prohibited. No part of this message can be considered a request for goods or services. =======================================================================

> The equation would be x * (1/y) < x > x * y This doesn't look like an equation to me, but I think I get what you want. You are looking for a geometric progression with a known value at each end of the scale and the geometric mean in the center. > How would I (can I) resolve the equation to find 'x', the midpoint, > if I have just the range, which in this case is 500kHz to 4100kHz ? > I can guess at say 1400kHz. 1400/500 = 2.8. 2.8 * 1400 = 3920, > which is close but there must be a better way There sure is. Look at the following to compute the middle value: SQRT(500*4100) = 1431.8 Now lets check that: 1431.8 / 500 = 2.8636 4100 / 1431.8 = 2.8636 So, we have found the correct 'middle' value. Assume that the value we read from the pot (lets call it X) ranges from 0 to 1, then we want: 0 to map to 500, 0.5 to map to 1431.8, and 1 to map to 4100 The ugly function: 500 * 2.8636^(X*2) gives you what you want. Lets check it: 500 * 2.8636^0 = 500 * 1 = 500 500 * 2.8636^(0.5*2) = 500 * 2.8636 = 1431.8 500 * 2.8636^(1*2) = 500 * 8.2000 = 4100.1 So, aside from a little rounding error we get the right answer. -- Bob Ammerman RAm Systems