http://forums.leaflabs.com/topic.php?id=476 A place to share, learn, and grow... en-US Fri, 22 Jan 2016 00:18:38 +0000 http://bbpress.org/?v=1.0.2 q http://forums.leaflabs.com/search.php http://forums.leaflabs.com/topic.php?id=476#post-2661 Tue, 16 Nov 2010 20:59:33 +0000 Silntknight 2661@http://forums.leaflabs.com/ <p>The very same project. The details of my project seem to be scattered everywhere. Maybe one day, when I get it all together, I'll also create a new thread to discuss it.</p> <p>Also, I added a new level of complexity to my project. To avoid the hassles of stroke timing, I just decided to let the Maple control the valve timing also. I'm modifying fuel injectors to be fast-acting solenoid valve operators. This creates some VERY interesting possibilities. </p> http://forums.leaflabs.com/topic.php?id=476#post-2632 Tue, 16 Nov 2010 15:32:02 +0000 gbulmer 2632@http://forums.leaflabs.com/ <blockquote><p>optoisolator? you doing some MIDI work? </p></blockquote> <p>I believe the project is the one described in several other threads.<br /> It connects to an internal combustion engine, and controls the spark!<br /> I'd use an optoisolator too :-) </p> http://forums.leaflabs.com/topic.php?id=476#post-2625 Tue, 16 Nov 2010 14:50:28 +0000 poslathian 2625@http://forums.leaflabs.com/ <p>optoisolator? you doing some MIDI work? </p> http://forums.leaflabs.com/topic.php?id=476#post-2594 Mon, 15 Nov 2010 21:56:22 +0000 Silntknight 2594@http://forums.leaflabs.com/ <p>Gbulmer, the equation I used is the equation in the program, not the divider ratio. The 4096 is the maximum value of any analog input. Basically, my equation works off the principle that the voltage in is "x." My voltage divider has R1 as 1 and R2 as 2 (I guess kilohms). The divider is, thus, a 5 to 3.3 divider.</p> <p>I think when I said "voltage" in that quote, I meant temperature. </p> http://forums.leaflabs.com/topic.php?id=476#post-2587 Mon, 15 Nov 2010 19:12:31 +0000 gbulmer 2587@http://forums.leaflabs.com/ <blockquote><p>Now, if I scale this down using a voltage divider (5V to 3V), does this change the 10mV/K to 6.6mV/K?</p></blockquote> <p>To be accurate, a voltage divider could scale by that ratio, it depends on the selection of values. It also depends on the characteristics of the device; poslathians warning about impedence needs to be considered.</p> <blockquote><p>If it doesn't change, I can get the voltage by K = (330/4096)*x </p></blockquote> <p>I don't think that is the ideal ratio.</p> <p>say the temperature is about 20C, about 292K<br /> 292K * 10mV = 2.92V * (330/4096) = 0.235V<br /> Even if the LM335 was at its maxium of 150C (about 422 K) the scale you quote of 10mV/K<br /> would only be a voltage drop of<br /> 422 * 10mv/K = 4.22V * (330/4096) = 0.34V<br /> That is wasting more almost 90% of the ADC's dynamic range (or the top 3+ bits)</p> <p>I think a factor of 1/10 has crept in.</p> <p>Also, there is no resitor standard that includes the value of 4096, 3 sig. fig. is all you get AFAIK.</p> <p>I think it needs something closer to 3.3/5.0, not 0.33/5.0</p> <p>1% resistors are only a few cents, so you can use a wider range of values than E24 standard values to get closer to the computation you want to do. You need to calibrate it anyway, to ensure their are no mistakes that have crept in. </p> http://forums.leaflabs.com/topic.php?id=476#post-2574 Mon, 15 Nov 2010 16:54:51 +0000 Silntknight 2574@http://forums.leaflabs.com/ <p>I'm not sure I have enough space in my project to add an opamp. Recently, I had to add two 16-pin DIPs because I needed an optoisolator and a MOSFET gate driver. As long as every voltage is affected equally, I can simply change my program to compensate for the readings. If a 5V output becomes 3.3V but a 4V output becomes 2.5 instead of 2.6, then I'll have a real problem. Even then, there should be a mathematical workaround. From what you posted, I inferred that the equation that gives me the output based on the output voltage is then scaled by k as well. Thanks, that was really helpful. </p> http://forums.leaflabs.com/topic.php?id=476#post-2548 Mon, 15 Nov 2010 15:17:40 +0000 poslathian 2548@http://forums.leaflabs.com/ <p>Voltage dividers will give you a linear divisor, so your output voltage becomes k*Vin, where k is (R2/(R1+R2)). However, be careful that the added impedance of your voltage divider doesn't effect your sensor output. You may want to buffer the output with an opamp follower before dividing it down. Of course, in that case, you might as well just divide the voltage down with the opamp, give it a gain of 3.3/5 by adjusting the feedback and input resistors. </p> http://forums.leaflabs.com/topic.php?id=476#post-2523 Sun, 14 Nov 2010 19:36:35 +0000 Silntknight 2523@http://forums.leaflabs.com/ <p>Hello again, this time I have a LM335A temperature sensor. It usually runs a good 5V, but I need to scale the output voltage down to 3.3V. The problem is that the output range is limited from 2.33V to 3.73V, because it is calibrated directly in Kelvin and the output is 10mV/K. Now, if I scale this down using a voltage divider (5V to 3V), does this change the 10mV/K to 6.6mV/K? If it doesn't change, I can get the voltage by K = (330/4096)*x. Otherwise, it becomes a different equation (which I haven't bothered to derive).</p> <p>On the same note, would this then affect all such devices whose outputs are based on a linear scale (ex. my pressure sensor)?</p> <p>EDIT: I derived the second equation and it is K = (500/4096)*x. What makes this interesting is that it is the same equation as if I didn't have a voltage divider at all. This may be due to the fact that I divided by the same ratio that I multiplied by earlier. </p>