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Hillridge posted:What's on the board? Ah, right, there are no ICs on the board, just a bunch of passive stuff. The mirror command in the board layout just swaps top/bottom layers. It doesn't mirror about an axis.
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Jun 27, 2009 00:29
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I don't get what you're trying to do? maybe some pictures would help? is PINSWAP is what you need... From the Manual Cadsoft posted:Pinswap and Gateswap The PINSWAP command will swap the pins of non-polarised components (resistors/inductors/some capacitors), i.e. This is the same as physically putting the device in "backwards" Or for things like logic gates, will swap input pins since they are equivalent (you'd do this to make the board easier to lay out). GATESWAP is for things like dual op-amps etc. where you have multiple copies of the same thing in one package.
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Jun 27, 2009 01:32
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scuz posted:For those that cared when I asked, I would up getting a Weller WES51 and it sorts rules compared to my old econoiron. However something else has reared its ugly head... I don't have very steady hands either. Usually it's because of too much coffee or nicotine, although I feel like I "phantom type" just a little bit when I'm trying to keep my hands steady. It doesn't impede my soldering all that much, once I got used to it. the trick for me is to realize that you might touch the wrong thing with your iron by accident while you're positioning it, and to come in at the joint from an angle where it won't matter. Once I've touched the iron onto the soon-to-be solder joint, I usually leave it there to heat the joint and then dab the solder into place. That way it's easier to keep the iron steady because it has a point of contact already. You have to be quick with this method though - don't heat a pin for more than a couple of seconds tops. If you manage to make a solder bridge between two pins, don't worry about it - just "drag" it away with the iron or use a solder braid (depending on how big of a bridge we're talking about). One last tip - most solder jobs don't actually need that more than a dab of solder. If you overdo it, shaky hands will make you much more likely to bridge a connection by mistake. Poopernickel fucked around with this message at 07:52 on Jun 27, 2009 |
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Jun 27, 2009 07:49
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If you need to get good at soldering just join a research group that requires 5000 0402 sized surface mount components replaced. My hands are fairly shaky but after that summer I was able to solder a flea's dick to ant's oval office. It's a zen experience. No lie, I've killed a fly in midair by swiping it with the tip of a hot soldering iron.
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Jun 28, 2009 06:04
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Hopefully a quick question here: I'm working on converting a chest freezer to a kegerator. That's all well and good. I had my temperature controller wired up and controlling the thing for several months now. I finally got around to actually installing the thing so that all the wires run through the walls of the chest, instead of just hanging about. However, in the process of doing so, I had to lengthen the wire on my temperature probe (it's a K-type thermocouple.) Now it's reading about 40-50 degrees higher than I expect (so far...it may yet climb.) I know for certain that the probe isn't that hot: I touched it with my finger and it feels like it's closer to, say, 40-50 degrees. What gives? Did I gently caress it up by adding the extra wire? Do I just need to get a probe designed with a longer wire in mind? If I don't solve this, my beer will freeze (controller will think it's 90+ in the chest freezer, even though it's 30 or colder)
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Jun 29, 2009 00:21
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Bad Munki posted:
Ding ding ding If you're going to extend the wire you need to use the same type as the thermocouple. If you don't you've essentially created a second thermocouple in the circuit where the dissimilar metals of the wires meet.
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Jun 29, 2009 01:52
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All right, thanks. It's no biggie, since those things are pretty cheap and the one I had couldn't have worked my purpose anyhow, being too short and all. Here's a marginally-related random question: what sort of response would occur if you hooked a few sensors up in parallel? This is purely hypothetical, I'm just curious. 
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Jun 29, 2009 02:53
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You'll get the average of the temperatures (with some caveats) http://www.allaboutcircuits.com/vol_1/chpt_9/5.html this is a pretty good overview of different TC configurations.
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Jun 29, 2009 03:29
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That was highly interesting. Thanks again!
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Jun 29, 2009 03:55
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Hopefully you guys can help me out! I'm trying to come up with a simple circuit and it's been awhile and I want to bounce my ideas off of you guys. I'm trying to drive an LED that is built into a switch assembly and has a resistor of unknown value inside. The LED is driven at full brightness when 12v is applied. I want to be able to light this LED at full brightness using a source voltage of 3v. Normally, I would just use a relay for this application because I'm more familiar with them, but I feel a transistor is more appropriate as the LED could be flashing several times per second in certain situations. If I understand correctly, I want to connect the anode side of the LED assembly to +12v, the cathode side of the assembly to the collector, the emitter to ground, and the base to my 3v source. Am I on the right track here? What model transistor would suit me best? I essentially want to be sure that I'm saturating the base to get full brightness, but I don't want to burn anything out of have heat issues either. Something I could pick up at the local radio shack for less than a buck would be ideal. Thanks for the help!
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Jun 29, 2009 14:33
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SynMoo posted:Hopefully you guys can help me out! I'm trying to come up with a simple circuit and it's been awhile and I want to bounce my ideas off of you guys. Yeah, you can use a cheap 2N2222 for this. The Rat Shack should have them. Don't forget to use a resistor on the base. Somewhere around 1k should do the trick.
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Jun 29, 2009 16:17
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Mill Town posted:Yeah, you can use a cheap 2N2222 for this. The Rat Shack should have them. Awesome, thanks! I believe the 3v output already has a resistor as it is meant to drive an LED, should I still slap a 1k on there anyway?
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Jun 29, 2009 17:38
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My Dad's HDD died without proper backups, but I've managed to narrow it down to a bad motor controller IC (the scorch mark made it pretty easy). Anyway, I've managed to find a replacement IC from one of those chinese companies that comes up when you google a hard to find part, and they're willing to sell small quantities at a somewhat reasonable price. How do I give them money? Are they likely to accept credit card? Last time I did something like this, all I had to do was get a contact and a price and send it off to purchasing, but now I have to figure it out myself. I'll try asking them myself but I was wondering if anyone has any experience with this kind of thing?
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Jun 29, 2009 18:03
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SynMoo posted:Awesome, thanks! Yeah, the resistor for an LED on 3V will be pretty tiny. If you find your transistor doesn't fully conduct, lower the resistor value.
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Jun 29, 2009 19:14
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Does anyone know of a fixed 3.3 volt regulator that has a very low quiescent current and dropout voltage that takes an input of at least 12 volts? I'd like something equivalent to this 5-volt LP2954IT that I'm already using and very happy with, but a new microcontroller I'm looking at is 3.3 volts only. I guess I suck with Digikey and Google because I haven't had any luck finding anything that's any good.
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Jun 30, 2009 02:49
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BattleMaster posted:Does anyone know of a fixed 3.3 volt regulator that has a very low quiescent current and dropout voltage that takes an input of at least 12 volts? I'd like something equivalent to this 5-volt LP2954IT that I'm already using and very happy with, but a new microcontroller I'm looking at is 3.3 volts only. I guess I suck with Digikey and Google because I haven't had any luck finding anything that's any good. What package? How much current? How low a q-current is "very low" ? Linear Tech makes decent regulators and their website it pretty good at filtering through their catalog. Look into the LT1129 (50uA) or the LT1521(12 uA). And both have even lower power sutdown modes. Why do you care significantly about the q-current, if you're linear regulating 12->3.3v? It seems like rearranging deck chairs on the titanic...
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Jun 30, 2009 07:30
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I'm also confused as to why you would care about the q-current when you're only going to be ~27% efficient at 12V. Also why do you care about the drop out voltage if you're going 12V->3.3? Regardless, I've had a lot of luck with Micrel's products (both switchers and linears), most of which are available from Digikey.
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Jun 30, 2009 18:10
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I forgot to mention that the project is battery powered so I want to get as much life out of it as possible. The project only requires 10 milliamps at the most, and a lot of that is only when reading or writing to the serial EEPROM. The rest of the time it would be in sleep mode with a real-time clock running and drawing less than a milliamp. The LT1129 in the 5-lead TO-220 package looks fine to me, unfortunately I'm not good enough to use surface mount parts yet so it has to be through hole. Okay so let's see where I went wrong: 1. I figured having as low a dropout voltage as possible would let it operate longer before the battery voltage drops too for the regulator to output its specified voltage. Isn't that how that works? 2. I thought that the quiescent current is how much current it draws just by idling. A lot of regulators I've seen have a higher quiescent current by themselves than the rest of my project requires! I don't really want to halve the battery life just because of a crappy regulator. 3. I hooked up my multimeter to the 5 volt version of the project and the current draw seems to be exactly what I would expect adding up the quiescent current of the LP2954 plus the current draw of the rest of the active components. Am I missing something here? What's so special about regulating from 12 to 3.3 volts that makes me asking for a part with a low quiescent current so strange? Edit: I just noticed that Microchip sells some regulators that look like they fit the bill even better, the 3.3 volt version of this sucker in a TO-92 looks pretty perfect. I didn't know Microchip made regulators, I've been using PICs for a while so this was right under my nose the whole time. BattleMaster fucked around with this message at 19:15 on Jun 30, 2009 |
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Jun 30, 2009 19:05
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BattleMaster posted:I forgot to mention that the project is battery powered so I want to get as much life out of it as possible. The project only requires 10 milliamps at the most, and a lot of that is only when reading or writing to the serial EEPROM. The rest of the time it would be in sleep mode with a real-time clock running and drawing less than a milliamp. The LT1129 in the 5-lead TO-220 package looks fine to me, unfortunately I'm not good enough to use surface mount parts yet so it has to be through hole. BattleMaster posted:2. I thought that the quiescent current is how much current it draws just by idling. A lot of regulators I've seen have a higher quiescent current by themselves than the rest of my project requires! I don't really want to halve the battery life just because of a crappy regulator. BattleMaster posted:3. I hooked up my multimeter to the 5 volt version of the project and the current draw seems to be exactly what I would expect adding up the quiescent current of the LP2954 plus the current draw of the rest of the active components. Am I missing something here? What's so special about regulating from 12 to 3.3 volts that makes me asking for a part with a low quiescent current so strange? That drop from 12v -> 3.3v is what kills your power usage, not the Q-current. You will end up dissipating most of your power in the regulator, regardless of the q-current, with this set up. Example for a 10mA load: Power used in the load: P = I*V = 10mA*3.3v = 0.033W Power used in the LDO: P = I*V = 10mA*(12v - 3.3v) = 0.087W The power used in the LDO is already more than 2x the power that is actually being used by the micro. And that's assuming a Q-current of 0.
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Jun 30, 2009 19:33
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So when I hook my multimeter up in series with the battery, how come it doesn't show me that the circuit is drawing twice as much current as the datasheets tell me?
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Jun 30, 2009 19:37
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BattleMaster posted:So when I hook my multimeter up in series with the battery, how come it doesn't show it as drawing twice as much current as it should? It shouldn't? I didn't say that the current draw would be >2x, I said that the power usage is >2x
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Jun 30, 2009 19:38
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SnoPuppy posted:Technically yes, but by the time a 12v battery gets to a voltage anywhere near there, it's practically dead. All true, but it's worth noting that he said the design will be in standby mode most of the time. So if the standby current is, say, 100uA instead of 10mA, then all of a sudden the quiescent current matters a lot. Although it's true most of the power will still be dissipated in the LDO, "most" of 200uA is twice as much as "most" of 100uA.
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Jun 30, 2009 19:39
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So what does that mean for battery life? I am using a 1.3 amp hour battery, I thought that means that it can provide 1.3 amps for an hour or half that for 2, and so on. If it doesn't draw more current does it still adversely affect battery life? Edit: I guess what I mean to say is I don't understand what power dissipation is. Don't be too hard on me, this is my first design where power consumption actually matters 
BattleMaster fucked around with this message at 19:45 on Jun 30, 2009 |
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Jun 30, 2009 19:41
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BattleMaster posted:So what does that mean for battery life? I am using a 1.3 amp hour battery, I thought that means that it can provide 1.3 amps for an hour or half that for 2, and so on. If it doesn't draw more current does it still adversely affect battery life? Yes, that's correct. Preserving quiescent current will lengthen your battery life. I think most people are assuming your power consumption to be 10mA all the time, in which case the amount of different 100uA of quiescent draw makes would be negligible. If your static current draw _is_ closer to the range of 10mA, don't even worry about quiescent current with your linear regulator because it won't be a significant difference. At that range, you're much better off using a switching regulator circuit instead.
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Jun 30, 2009 19:45
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BattleMaster posted:So what does that mean for battery life? I am using a 1.3 amp hour battery, I thought that means that it can provide 1.3 amps for an hour or half that for 2, and so on. If it doesn't draw more current does it still adversely affect battery life? This is true. And if it works for you, then don't fix what isn't broken. However, you could get significantly more battery life by using a reasonably efficient switcher. The total power your battery can provide is 15.6W (12v*1.3A). The total power your micro uses is .033W (3.3v*10mA). But, you waste 0.087W in the LDO. So your total power usage is 0.12W. With 0.12W, you can run for 15.6/0.12 = 130 hours. If you used a switcher that was even 75% efficient, you would only use 0.044 watts. This would let you run for 15.6/0.044 = 354 hours. Now, if you're using the shutdown mode on the micro, your hour use will be much better, but the switcher should almost always win.
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Jun 30, 2009 19:52
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BattleMaster posted:Edit: I guess what I mean to say is I don't understand what power dissipation is. Don't be too hard on me, this is my first design where power consumption actually matters It's cool - I didn't mean to come off as harsh. Just remember that the power used by anything is equal to the voltage across it times the current through it. That's why the LDO is very inefficient - the voltage across it is quite big, in comparison to the voltage across the micro. And poopernickle is right - the Q-current can make a big difference when your part is in shutdown most of the time. If your micro is only active 10% of the time, its average current draw is 10mA*10% = 1mA. Now a 200uA Q-current is something like 20% of the current.
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Jun 30, 2009 19:58
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Thanks for the help guys, I've learned quite a bit from this. I thought regulators were a ton more simple than that, but looks like I may need to look into switching regulators for future designs. It shouldn't be too much of a big deal if I use a linear regulator for this project, though. What I am working on is a battery-powered sensor platform that takes scheduled readings from sensors and stores the data in an SPI EEPROM. The microcontroller is a PIC18F26J50, which has some peripherals that make it great for my design such as the hardware real time clock, an SPI module with DMA, and a USB interface. That's a step up from the PIC18F4550 in my old design which requires a software RTC which wakes the thing up every second to do time/date calculations, and lacks DMA on the SPI module so reads/writes to the EEPROM are much more software intensive and therefore require the thing to stay awake longer. It should spend no more than a few seconds every day outside of sleep mode, and even less than that writing to the EEPROM which which accounts for the bulk of the current draw. The rest of the time it should be drawing no more than 20 microamps for the microcontroller, and another couple of microamps for the EEPROM. I'm still working on selecting sensors - I would like to move away from analog sensors plus the on-chip ADC if possible. I'm probably going to end up going with I2C or SPI sensors and using the second SSP module on board the PIC for those to leave the one with DMA dedicated to the EEPROM. SnoPuppy posted:It's cool - I didn't mean to come off as harsh. Nah, it was very enlightening
BattleMaster fucked around with this message at 20:14 on Jun 30, 2009 |
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Jun 30, 2009 20:11
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SnoPuppy posted:This is true. And if it works for you, then don't fix what isn't broken. I edited the quote above. It may be a little nit-picky, but I think it'll help make things less confusing in the long run. You're explanation is spot on though. In general, I use linear regulators when: - I don't care about power consumption and I want a quick and easy solution - I don't need a lot of current on that supply - I need a very "clean" supply (switchers are more noisy than linears) - I'm doing a small voltage drop, like 3.3V to 1.8V If it's a big drop, or lots of current, a switcher is the way to go.
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Jun 30, 2009 21:40
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For your project, I agree with you - Stick with an LDO. Switching regulators can have higher quiescent draw than an LDO because there are typically a few discrete components in there. You can get big gains in efficiency using a switcher when you've got a big load (uhuhuhuhuhuh load), but the amount of power saved is pretty negligible when your nominal current draw is on the order of 100uA. At those levels, skip the complexity and hassle of a switching regulator and just use an LDO with a low quiescent draw. Poopernickel fucked around with this message at 22:54 on Jun 30, 2009 |
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Jun 30, 2009 22:49
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SnoPuppy posted:This is true. And if it works for you, then don't fix what isn't broken.
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Jul 1, 2009 03:46
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Hey my new internship has reminded me why I want to work in consulting instead of being dedicated to a company. It's so that when someone comes to me asking to work on something completely absurd and futile, like wireless power transmission, I can laugh in their face instead of having to waste my time ahahahfkdjhdgkjfdgfdgdfgf.
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Jul 1, 2009 03:49
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ANIME AKBAR posted:Hey my new internship has reminded me why I want to work in consulting instead of being dedicated to a company. It's so that when someone comes to me asking to work on something completely absurd and futile, like wireless power transmission, I can laugh in their face instead of having to waste my time ahahahfkdjhdgkjfdgfdgdfgf. Is there a story behind that?
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Jul 1, 2009 03:50
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BattleMaster posted:Is there a story behind that? supervisor basically hands me a paper by some MIT physics professor about inductive coupling for magnetic energy transfer and says "hey try to make this work, only make it more efficient." I tell him his demands are physically impossible and he basically says "yeah, well try anyways." edit: oh yeah and he's a chemical engineer.
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Jul 1, 2009 04:01
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lol. I was toying with doing a wireless power transfer project for my final year project, but I quickly found sense. you have to have the RF design reaaaaaally good or you end up with a really crappy transformer.
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Jul 1, 2009 07:34
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I just finished a high school electronics course and thought I'd design a project. Its going to be a D&D dice type of thing, so that you can turn on one switch and have a d8, another to have a d4, and so on. I prototyped it and it worked great, but only for the even numbers. On the odd numbers, it would count 1,2,3,0,1,2,3,4,1,2,3,0,1,2,3,4 etc. I'm using a 4017 Walking Ring counter IC and a 4516 up/down 4-bit counter. The output of the walking ring hits its own reset, the reset of the 4-bit counter, and the preload of the 4-bit counter, depending on which switch is on. I had no idea why it would screw up the count only for odd numbers, and neither did my teacher, so I decided to leave it as it was, since you never really use odd numbers in D&D anyway. But I would still like to know, just because I like to know why when something doesn't work how I expect it to. I have a schematic that I could post if someone thinks it might be helpful, but its drawn on graph paper with no planning, and looks pretty bad. Also I have no idea what a "real" schematic should look like, I kinda wrote stuff down wherever. Unfortunately I broke some of my parts, so I can't really do much testing right now, and its not really a big deal anyway, so don't go spending tons of time on this, but if you happen to know whats going on, I would appreciate it
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Jul 1, 2009 08:36
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SynMoo posted:Hopefully you guys can help me out! I'm trying to come up with a simple circuit and it's been awhile and I want to bounce my ideas off of you guys. So while this circuit works perfectly on the bench, it doesn't in the car because it turns out that the switching circuit for the original LED has a switched ground and a constant +3v instead of the other way around as I had initially thought. It appears, at least initially that it is the ground side has the resistor on it, giving me the low current flow to light my 12v LED assembly properly. I'm starting to wonder if it would just be easier to open the assembly and remove the resistor across the LED and use the direct output as opposed to this circuit.
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Jul 1, 2009 15:42
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ANIME AKBAR posted:there are no switchers that are reasonably efficient at very low power levels. I've wrestled with this problem and the bottom line is that for load currents under 1ma, switchers are useless. I recently had a design where I wanted to power a microcontroller and RF transceiver at 2.0V from a 12V lead acid battery, and they would be asleep 95% of the time (current was about 50uA). For this application, there was no way a switcher could be more efficient than a regulator. I wish linear tech or someone would address the problem, but it's really not as easy as it sounds. Aren't there some pretty small/low-power switched capacitor supplies?
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Jul 3, 2009 02:20
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SecretFire posted:Aren't there some pretty small/low-power switched capacitor supplies? yeah, but the efficient ones are limited to voltage doublers/inverters with no regulation. There are some with regulation, but they never get very efficient, and I've never seen one that could deal with high input voltages, so they'd be useless for applications where linear regulators are inefficient (like stepping from 12V to 2V or whatever).
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Jul 3, 2009 16:38
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BattleMaster posted:So what does that mean for battery life? I am using a 1.3 amp hour battery, I thought that means that it can provide 1.3 amps for an hour or half that for 2, and so on. If it doesn't draw more current does it still adversely affect battery life? Is there any reason you are married to a 12V battery? You can get a ~3.7V li-ion with as much capacity and what I would assume would be a much smaller footprint - it would make a LDO more efficient at any rate.
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Jul 5, 2009 21:28
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Delta-Wye posted:Is there any reason you are married to a 12V battery? You can get a ~3.7V li-ion with as much capacity and what I would assume would be a much smaller footprint - it would make a LDO more efficient at any rate. Because a friend of mine gets them free from the company he works for. They use these 12 volt batteries in medical equipment and throw them out if they go even slightly out of their very rigid specifications. Edit: I'll consider a new battery if space becomes a problem but as it is I'd rather just use more of these 12 volt batteries if I need the design to last longer. BattleMaster fucked around with this message at 22:38 on Jul 5, 2009 |
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Jul 5, 2009 22:30
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