Timeline for Voltage to PWM Circuit, need to understand frequency
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2019 at 1:41 | vote | accept | Diznaster | ||
| Jan 25, 2019 at 5:53 | comment | added | Diznaster | I put a divider on only the 0-10v signal (not tied to 5v+). It got me closer. It didn't get quite what was calculated and I'm thinking that is because other resistors are also dividing. I did some testing with different values and re-calculating the divider. I'm fairly confident I have something that works now and I understand how to adjust the resistance values. So I don't think I need anymore help on this. I'll report back in a few days, I need to tidy up the mods and also do some other stuff with the kids. | |
| Jan 25, 2019 at 2:48 | comment | added | Diznaster | Unfortunately not with a scope. It is 1.719v to the IN- pin of U2 from the Tresh pin of the 555. This is with the circuit configured back to the original format using R9,R10,R1,R3,R4. C2 was changed from 1nF to 2.2nF to get frequency from about 50kHz to 23kHz. With the new divider it seemed to somewhat work, but the 0-10v input would not drop below around 2v at IN+ on U2 even with it completely disconnected. So I was thinking it was because the divider was now tied to 5V+ and it wasn't before. I'll think on it more, any other ideas let me know I really appreciate the help | |
| Jan 24, 2019 at 14:10 | comment | added | G36 | @Diznaster Can you measure via scope the voltage at U2 inverter input? And what did you use as a C2 capacitor? | |
| Jan 24, 2019 at 5:00 | comment | added | Diznaster | I modified the circuit as described, but it didn't quite work out. Now the duty cycle is only adjustable from 50-100%. At 10V=99%, 9V=96%, 8V=90%, 7V=84%, 6V=79%, 5V=74%, 4V=68%, 3V=63%, 2V=56%, 0V=47%. Before it was close to 0-100% and each 1V+ was a 10% increase in duty cycle. My guess is that it has something to do with R2 connecting to 5V+ in the new simplification theory. Before that the complicated (to me) divider/adjustment circuit was not tied to 5V+.The 0-10V is an independent source. Or maybe I just need to tweak some resistance values, if so which should I adjust/how much? | |
| Jan 23, 2019 at 23:28 | comment | added | Diznaster | Awesome, I was thinking that was all a means to make a voltage divider. Your edit helps me to understand (at least I think I do). It brings the 0-10v into the range of approximately 1.6v-3.3v. Then the comparator can adjust the duty cycle based on when the rising sawtooth crosses the switching threshold. | |
| Jan 23, 2019 at 20:11 | history | edited | G36 | CC BY-SA 4.0 | added 500 characters in body |
| Jan 23, 2019 at 20:07 | comment | added | G36 | @Diznaster I updated my answer. | |
| Jan 23, 2019 at 20:06 | history | edited | G36 | CC BY-SA 4.0 | added 500 characters in body |
| Jan 23, 2019 at 5:31 | comment | added | Diznaster | Hopefully my last question! I'm trying to simplify all the resistors and jumper leading to U2+ (R9,R10,R1,R3,R4). Suppose I don't need the option to jumper to a 0-5V input and only using 0-10V input. Also would like to eliminate adjustability with R1. From testing I know that for best function R1 is set at 8k Ohm between jumper above R3, and 94k Ohm to ground below R3. I'm trying to rationalize all the parallel resistance. I'm guessing it has a lot to do with options and adjustment. I'd like to understand how to make it simpler with less components. | |
| Jan 21, 2019 at 5:05 | comment | added | Diznaster | Yes, Thanks. I realized the D1 was an LED, Duh! | |
| Jan 19, 2019 at 16:59 | comment | added | G36 | @Diznaster D1 is an ordinary green LED. Not needed in the simulation. D2, D3 is the 1N4148 type or any "signal diode" you have. As for the jumper. Simply connect via wire the upper part of the R10 resistor together with 0...5V input. Do you understand this? | |
| Jan 19, 2019 at 3:18 | comment | added | Diznaster | I removed C2, measured it and replaced it with close to what was calculated for desired frequency. It worked as expected and I could just quit, but I still want to learn a few things and try to simulate. The 0-5v/0-10v part of the circuit with the jumper is confusing to me. Could you show a version with only 0-10v. Also I will need values for diodes 1-3, if you can offer any baseline guess on those. I'm honestly reading a lot, using datasheets, and learning, not just asking for answers handed to me. I like to see the answer and figure out how you got to it. | |
| Jan 14, 2019 at 15:21 | comment | added | G36 | @Diznaster You will never know for sure until you change the C2 and measured the output frequency. | |
| Jan 12, 2019 at 0:54 | comment | added | Diznaster | Sorry It's been a while, other stuff... I measured the frequency at 55kHz (I also got a response of "around 50kHz" from seller). So if I understand correctly C2 is probably about 1.5nF. I'm going to remove it and measure it. If confirmed then a replacement of 3.3nF should get me around the desired 25kHz? | |
| Nov 7, 2018 at 16:44 | comment | added | G36 | @Diznaster V+ is the internal voltage provided by AMS1117. And this voltage will be equal to around 5V it should be within this region 4.9V...5.1V. | |
| Nov 6, 2018 at 1:12 | comment | added | Diznaster | Thanks for all the great info so far. Follow up with request of application info. The VCC will be 5V, but the VAnalog signal will be 0-10V (with jumper in 10v position). The 0-10V analog signal is coming from an aquarium controller. I can program it to change signal voltage (0-10V) based on tank temperature, pH, etc. The result of the PWM should be 5V modulated at Fout. This will signal a good 4-pin PC type fan for ventilation. It runs at 12V but uses a 5V PWM for speed. Advice was based on 5V supply, does that mean VCC, or signal? I didn't mention VCC (sorry), does this info change formula? | |
| Nov 3, 2018 at 17:05 | history | edited | G36 | CC BY-SA 4.0 | added 127 characters in body |
| Nov 3, 2018 at 16:47 | history | edited | G36 | CC BY-SA 4.0 | added 495 characters in body |
| Nov 3, 2018 at 16:15 | history | answered | G36 | CC BY-SA 4.0 |