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| Portable Low Frequency square wave generator circuit - Solved - page 3 |
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| soldar:
imo, if you read the OP you will read that is exactly what he is trying to do. |
| Zero999:
--- Quote from: spec on January 31, 2019, 03:33:05 pm ---One thing that surprises me is that no one has commented on the circuits effect on global warming. After all there is a whole 1W2 being dissipated in the output stage.:-DD --- End quote --- Good point. It's being powered from a small battery, so 1.2W of idle power consumption is a killer. --- Quote from: spec on January 31, 2019, 03:17:26 pm ---As I have said before to all you armchair 'experts': Let us see your complete circuit, embodying all these wonderful ideas that you have. --- End quote --- We tend to avoid posting complete circuits here, as it discourages people from thinking. It's often more helpful to post hints and little circuits which the original poster can put together, that way they gain a deeper understanding of how it works, rather than some soldering practice. --- Quote from: spec on January 31, 2019, 03:40:43 pm ---What on earth is all this mark to space ratio stuff all about? --- End quote --- What on earth is all this dual supply rail stuff about? ;) Only one power supply is needed. The trick is to float it. I did hint on this earlier, but I forgot this is the beginners section so should have posted a schematic. 50% duty cycle is achieved by replacing the usual pull-up resistor on the discharge pin with a transistor driven from another 555 (Q1) which inverts the signal, so it's pulled all the way up to +V when charging. If a bridged output isn't needed, the other 555 could be replaced with an NPN BJT, with its base connected to the 555's output via a suitable current limiting resistor. Of course in real life, a NE556 dual timer could be used to save space. One thing to note is the simulation says it outputs +/-9V, but this doesn't represent reality, since the model doesn't mimic the NE555's output stage very accurately. The output voltage will be a little less, much closer to 6V, especially when loaded. It may be necessary to power the circuit from two 9V batteries and a regulator, as the battery voltage will fall, as it discharges. Heck, I'd consider swapping the 9V battery for some AA cells, which have a higher energy density. |
| soldar:
--- Quote from: Zero999 on February 01, 2019, 11:13:34 am --- We tend to avoid posting complete circuits here, as it discourages people from thinking. It's often more helpful to post hints and little circuits which the original poster can put together, that way they gain a deeper understanding of how it works, rather than some soldering practice. --- End quote --- Let me expand on this. This is not a free service where people work for you for free and provide you with turnkey solutions. We are here to help and learn from each other. If you want a turnkey solution you can go hire some engineer and pay them the going rate. Furthermore, many requests are poorly defined, poorly presented and poorly explained or even contradictory. Often getting further clarification from the OP is a slow and painful process, somewhat like pulling teeth. We could be here presenting solutions for days only to have the OP disappear (often happens) or reject different solutions on different grounds (to complicated, too expensive, not fulfilling needs which were not mentioned before, etc). So we take what we are given and kick ideas around and learn from others' ideas and viewpoints. That is the way I see it. |
| Zero999:
I think the OP has gone, but who cares? This is interesting. Has anyone actually tried building the 50% duty cycle 7555 circuit before? I thought I'd breadboard it. I aimed for 1.53kHz, and tried the following values for R and C: 1k & 470nF, 10k & 47nF and 100k & 4.7nF. Firstly I tried using a 5V supply, which I increased to 15V. The results were interesting. Even the 100k and 4.7nF combination didn't give exactly 50% duty cycle. Schematic. R = 1k, C = 470nF, V = 5V 56.3% duty R = 1k, C = 470nF, V = 15V 51.7% duty R = 10k, C = 47nF, V = 15V 48.4% duty R = 100k, C = 4.7nF, V = 15V 48.2% duty Conclusion: if the duty cycle is critical, then don't use the 7555! |
| Zero999:
I thought that wasn't very good, so I tried a different brand IC which gave better results. The ICs used in my previous test were Philips brand. The ones I used for this test were Intersil. Due to time constraints, I didn't bother repeating all of the tests, just the 1k & 470nF and 100k & 4.7nF with a supply voltage of 15V. R = 1k, C = 470nF 53.3% duty R = 100k, C = 4.7nF 49.7% duty |
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