Jump to content
Electronics-Lab.com Community

chylld

Members
  • Posts

    13
  • Joined

  • Last visited

    Never

Everything posted by chylld

  1. it's mainly a safety thing. i guess if i had a switched powerboard mounted onto the side of my workbench it would be ok, but i wanted to avoid the possibility of being able to turn the router off at the power point and not at the switch on the router itself. i've heard horror stories of the router coming on unexpectedly as soon as it got plugged in, and i didn't want to take such a risk with a tool that can take a joint or two off your fingers before you realise it! that said, a powerboard with individual switches for each outlet would be ok, the router would always go into one that's always on, and the vacuum cleaner would get switched on/off from there. thanks for the simple suggestion :)
  2. Heyas, I need to build a power-sensing circuit for my dust collection system in my workshop. i want my 700W vacuum cleaner to turn on automatically when my 1200W router fires up, as it is quickly becoming a tiring hassle to manually turn the vacuum cleaner on and off each time. I'm aware of all the safety precautions needed to work with mains electricity and would like assistance with designing such a circuit so that i can build it myself. Thanks in advance! Jon.
  3. so it looks like i'll need a 3650uF cap - it's all starting to sound a lot more sensible, thanks! by 0-30V transformer i meant the transformer thing itself (in general), not the 0-30V portion of the secondary coil! sorry i didn't make that clear. i'm stupid but believe me i'm not that stupid :) so that pretty much makes things happy on the 12v side of my project, now back to the motors - assuming i can rectify some ac source (be it the same transformer or another one), is it ok give them unsmoothed output? or would it be better to give them smoothed output? the cap i'll need for each will be roughly (5 x 1)/(2 x 100) = 25mF - !?
  4. ahh ok didn't know that. if i take off 0-15V AC, that'll give me 17V minimum (10% ripple) which is ok for the 7812, right? for that i'll need a 7300uF cap... also, assuming it's ok to use the 0-30V 2A transformer i pointed out earlier - did i mention how HUGE that thing is?? it's like 8cm x 7cm x 7cm! why is it so big? is there a 'smaller' way to make a power supply? (cheaply)?
  5. hmmm i thought that taking off from the 12v ac part of the transformer would be ok? i read on this page: so 12v ac in = 12-1.4 (rectifier diodes) = 10.6, 10.6*1.4 (sin peak) = 14.8, and smoothing to within 10% of this peak voltage would result in a minimum of 13.3v which should be enough for the 7812? i do understand that feeding less than 12v to the 7812 is pointless :) one other thing that's been bothering me is the size of the smoothing capacitor, as you pointed out it's way too small. so i used the formula (from the same page) Smoothing capacitor for 10% ripple, C = (5 x Io)/(Vs x f), where Io = output current, Vs = supply voltage and f = frequency. so C = (5 x 1.4)/(14.8 x 50) = 9500uF! isn't that quite big? or am i doing something wrong there? overall i'm getting much closer to a final solution though :P
  6. sorry for the delayed reply audioguru, my internet conked out and only came back this evening. before it went out though i did as much reading up on transformers as i was able to, and in the downtime i think i managed to figure out how they work, or rather, how i can use them. like... if i get it properly, the secondary coil in many transformers is divided up, each giving a portion of the total voltage. an amp-rated unit (as opposed to a VA-rated unit, like the 60VA unit i mentioned earlier) delivers its amp rating across secondary coil voltage range, e.g. 6v @ 1A for the 0-6v portion, or 12v @ 1A for the 0-12v portion. and if the 0-6v portion is wound adjacent (as opposed to on top of) the 6-12v portion, then 1A can be taken from each of the parts simultaneously, since the current limit is mainly based on heat? also, if i take off from the 0v and 6v 'take-offs' from the secondary coil on the transformer, what AC voltage will i actually be reading there? from your calculations (and what my mechatronic engineering friend tells me) it'll be root(2) times that =~ 1.414*6v = 8.484v; and then minus 1.4v for each of the 2 diodes that the current passes through. however the explanation on this page (towards the end) suggests that taking off from the 0-12v portion will give 12v-1.4v=10.6v AC, i.e. no root(2) peak. what gives? regarding my 9V/300mA requirement - it has a couple of main duties: 1) power a timer circuit (2x 555 IC's, 22mA draw) which trips a relay (+35mA) 2) power 3 fans which draw ~90mA @ 9V i didn't mention the relay before so the peak draw of this circuit will be around 327mA. i also decided to bump it up from 9v to 12v, point 1) above accounts for this and point 2) will be achieved through the use of 4 in-series 1n4004 diodes. the thing is, there's another device that will be incorporated into this project that i bought commercially, it runs off an unregulated 12V 1A ac adapter which i figure i can replace with my own unit. so, i found this transformer for less than AUD10: what i plan to do is: - take off 0-12V, rectify and smooth that (470uF) and give the smoothed output to the device originally running off its own ac adapter. the peak sustainable current for the 0-12V bit is (12*2)/(12*sqrt(2)) = 1414mA, 1000mA of which can go to this thing, the rest can go to the next, which is... - feed the smoothed input also to a 7812 (12v 1A) regulator (with 0.1uF capacitors either side) and have that power the 12v circuit above (which in turn powers the fans). this bit draws around 327mA peak as calculated above. - take off 15-17.5, 17.5-20, 27.5-30 (i.e. 3x 2.5v each) and rectify and smooth those and dump that straight onto the motors. i figure it'll give sqrt(2)*2.5-1.4 =~ 2.1v which is ideal because i think 3v will be a bit too much for what i want it to do (also the specified voltage range of the motor is 1.5-3v). i'm taking your advice here and not bothering to regulate the power going to the motors, it's overkill and quite silly actually! so... how am i doing? :)
  7. ahhh ok... that makes a lot more sense. i managed to find another type of transformer, it's more expensive but it looks like it'll do the job: so using your maths, if i take power from the 9v "bit" the peak voltage will be 12.73v and max continuous current 60/that = 4.7A, which is more than the 3A i need. i think regulating this voltage down to 3v is a bit of an ask for the regulator, but the thing is the motors will only be used for a few seconds at a time, and usually one at a time, under normal conditions which is somewhere under 1A (and over 300mA). so the regulator won't be working continuously which should help it from burning up? also, can i power my 9v 300mA load from that same transformer? should i take the power off in parallel just after the transformer (and fuse)? or after the smoothing capacitor?
  8. ahh ok, so when a regulator drops the voltage, it loses it in heat, and the current stays the same? is the current drawn at this stage equal to the current drawn from the mains? i'm pretty sure drawing [email protected] means less than 3A at 230V AC! :) i take it that once the inefficiency of the power supply is taken into account, the power(W) drawn at the 3V DC end will be roughly equal to the power drawn at the 230V AC end? unfortunately the closest thing i can find to a 5V/3A adapter is a 5V/2.5A adapter, which won't be able to sustain all 3 motors at the same time, so it looks like i'll have to build one myself. as seems to be common amongst first-time power supply builders, i'm having trouble picking the right transformer. with this one: does this mean that i can pull 6.3V at 2*15/6.3=4.7A? i keep reading that transformers are rated on VA, so is the VA for the above transformer 15*2 = 30VA? also, would i be able to take 6.3V off it in one circuit and then 9.5V off it in another circuit at the same time? (i have 300mA worth of other things that i wish to run at 9V) i know i'm asking a lot but i've honestly tried all day to better understand these things, i'm slowly coming to grips with it so please bare with me for now! :)
  9. heyas, i have 3 motors that draw a maximum (under load/stall) of [email protected] is it possible to run all of these simultaneously off a 12V 1A ac adapter? i figured that the motors will be drawing 9W total max which is under the 12W the ac adapter can deliver, but i'm not sure if it works this way. i plan to drop the voltage to the motors using some kind of 1A-rated regulator arrangement. if it's not possible to draw [email protected] off a 12V/1A ac adapter, how do i best go about this? thanks in advance :) edit: topic changed from "does an ac adapter's max current rating increase with decreased voltage?" to reflect the latest content of this thread
  10. heyas, as you can probably tell from the subject title i'm quite foreign to electronics but here's my problem: i need to design a circuit that has inputs +9V (or anything really), Gnd, A, B, X, Y. A, B, X and Y are normally not connected. power is provided at +9V and after 3-5 seconds, A connects to B and X connects to Y for a short time (~0.5 secs) and then they diconnect. the +9V stays on since it gets applied at the beginning. fyi, the A-B and X-Y pairs correspond to the power switches on 2 devices that i'm attempting to turn on a short time after voltage appears on the +9V input. i'd like to avoid using relays if at all possible, due to cost. is what i'm trying to do possible, and if so could someone please point me in the right direction? Thanks in advance Jonathan.
×
  • Create New...