using 1.5mA to switch power from battery - Page 1
Forum section: electronics / beginners

author: nsled
author_rank: Contributor
timestamp: January 10, 2016, 03:10:56 pm
content: Hi guys,I have a project I'm working on that requires me to convert 1.5mA into approximate max 500mA, probably less.I'll attach a pic of what I mean, I have a circuit running off 3 AA NiMH batteries but the output current is far to low. How can I use this current to switch power directly from the battery?What I've tried is to use a 2n2222 transistor, I wired the 1.5mA into the base and the battery directly to the collector, earth to the emitter. Doing this just made it "always on", whether the 1.5mA was there or not. What should I be doing?

author: Zero999
author_rank: Super Contributor
timestamp: January 10, 2016, 03:34:06 pm
content: The 2N2222 doesn't have enough gain to enable you to switch 500mA with a base current of 1.5mA.You could use a BJT with a much higher gain, a logic level MOSFET or two 2N2222s cascaded.

author: nsled
author_rank: Contributor
timestamp: January 10, 2016, 03:44:00 pm
content: Thankyou Hero999 for the speedy reply, I'll give that a try tomorrow and report back.Out of curiosity what would it look like when using the MOSFET you mentioned?

author: Ian.M
author_rank: Super Contributor
timestamp: January 10, 2016, 03:54:02 pm
content: All bipolar solutions will fail to deliver 3.6V out from a 3.6V supply because of the Vce_sat of the switching transistor. If you need the full voltage out, you will need to use low gate threshold, low Rds_on MOSFETs. There are specialised high side switch chips that do this, but the low voltage ones are mostly in ultra-miniature 'flip-chip' packages that are extremely experimenter unfriendly.

author: Audioguru
author_rank: Super Contributor
timestamp: January 10, 2016, 03:57:41 pm
content: Since your 2N2222 with not enough base current "stayed turned on" then either you connected it backwards or you blew it up.

author: Audioguru
author_rank: Super Contributor
timestamp: January 10, 2016, 04:04:02 pm
content: I forgot to say that three Ni-MH AA battery cells will not produce 3.6V for long. After a few hours the voltage will drop to only 3V or less when a very sensitive logic-level Mosfet will not turn on well.

author: nsled
author_rank: Contributor
timestamp: January 10, 2016, 06:20:41 pm
content: Sorry I'm an idiot, I had the meter pin in the wrong hole. Its actually 75mA not 1.5mA, but the problem is the same, can this be switched with a single 2N2222? What is the schematic?

author: Zero999
author_rank: Super Contributor
timestamp: January 10, 2016, 06:24:25 pm
content: Quote from: Audioguru on January 10, 2016, 03:57:41 pmHe won't have blown it up by connecting it backwards. The reverse voltage rating of the 2N2222 is 5V and the total supply voltage is a maximum of 4.8V, when the batteries are fresh off the charger. It's more likely he's short circuited it and burnt it out.Quote from: nsled on January 10, 2016, 06:20:41 pmHow did you measure the current? You didn't just set the meter to current mode and connected to the output of the board did you? If so, then you measured the short circuit current, not the maximum output current. If you didn't do it for long, then all will be fine but if you short circuit it for too long, then it will overheat and be damaged.Quote from: nsled on January 10, 2016, 03:44:00 pmYou can do it with a single BJT if the gain is high enough. For example, the ZTX689B is specified to drop just 75mV, when the collector current is 500mA and the base current is 1/200 of that: 2.5mA. In your case the base current is just 1.5mA so it will be slightly worse than that but it should still be usable.http://www.rapidonline.com/pdf/81-0228.pdfAs far as MOSFETs go. You need something which will be able to pass 500mA with a gate voltage of 3V and as lower drain source voltage, as possible. Fortunately, as your current requirement is only 500mA, it doesn't have to be anything special. The old IRL540 is capable of doing what you want, even without a heat sink.http://www.vishay.com/docs/91300/91300.pdfWhat's the circuit on the control board? Do you have a schematic/datasheet?

author: Audioguru
author_rank: Super Contributor
timestamp: January 10, 2016, 06:44:46 pm
content: If the first post showed a schematic with the base resistor value then we would have quickly spotted the measurement error of 75mA showing only 1.5mA.

author: IanB
author_rank: Super Contributor
timestamp: January 10, 2016, 07:28:54 pm
content: Quote from: Audioguru on January 10, 2016, 04:04:02 pmNot really. NiMH cells have a relatively flat discharge curve and they will hold 1.2 V for most of the discharge period, dropping to 1.1 V only near the end. This is especially true of modern good cells like Eneloops.See here, for example:http://www.fenixtactical.com/manuals/sanyo-eneloop-aa-manual.pdf

author: exe
author_rank: Unknown
timestamp: January 10, 2016, 08:23:41 pm
content: Consider using small SSR -- solid state relays. I'm not sure 1.5mA is enough, but 3mA+ should do the job, see datasheets. The only disadvantage I see they are not particularly fast (going above some hundreds of switches per second can be a problem). Honestly, I never tried them in practice, but I'm going to try them soon. They are also not dirt cheap, unfortunately (like $1-3 or so). But for a single application it's okay.

author: nsled
author_rank: Contributor
timestamp: January 12, 2016, 04:16:17 am
content: I've sure made a mess of this thread by the erroneous current measurement, I've attached another pic with more and accurate detail.There's really not much difference between the current draw from the LED driver and the battery but the damn thing wont work, the voltage seems stable as checked by a scope. I haven't connected the speaker yet so the current draw from battery will increase from that shown.What confuses the hell out of me is this, the LED driver is a garden solar light in order to make the MP3 player turn on at night. The short circuit draw is 75mA in the dark and 0mA in the light. But when connected through the transistor as shown it draws a constant 8mA through the LED driver whether it be day or night. How can it be drawing more then the short circuit voltage?? As connected it just stays on day and night.I'm thinking the SSR may be the best option, if my understanding is right its just a LED that triggers a transister? keeping the LED driver driving a LED seems like a good idea. What would be a suitable SSR part?

author: IanB
author_rank: Super Contributor
timestamp: January 12, 2016, 04:44:10 am
content: The way you have shown the transistor wired in your picture is not the same as Hero999 showed you in Reply #1. Electronic circuits are very particular. They will not work unless you wire them up exactly according to the diagram.

author: nsled
author_rank: Contributor
timestamp: January 12, 2016, 04:50:24 am
content: Quote from: IanB on January 12, 2016, 04:44:10 amBut Hero999 based that on my error in thinking it was drawing 1.5mA. I should be able to switch 22mA on with 8mA using one transistor true?

author: IanB
author_rank: Super Contributor
timestamp: January 12, 2016, 04:56:59 am
content: If you took away Tr1 in Hero999's diagram and connected the input directly to the base resistor of Tr2 it would still be different from how you have drawn it.

author: nsled
author_rank: Contributor
timestamp: January 12, 2016, 05:32:16 am
content: It tried that, but the result is the same except the base now draws more current. Same config except load in front of instead of behind the transistor. It just stays on drawing current from the input in the sun when there shouldn't be any current to draw from :/

author: IanB
author_rank: Super Contributor
timestamp: January 12, 2016, 05:48:11 am
content: OK, so maybe more information is needed about the solar LED unit. I presume the LED lights up when the sensor is covered, and it goes out when the sensor is exposed to daylight, correct?Do you have a basic circuit diagram showing the battery, the circuit board and the LED?The LED is probably connected to one of the battery terminals. Which terminal is it, the positive or the negative?When the LED is off, what is the voltage measured across the LED? And when it is on, what is the voltage?This will give more information about how to use the LED to drive a switch.

author: Zero999
author_rank: Super Contributor
timestamp: January 12, 2016, 09:11:50 am
content: It's quite likely the anode of the LED is connected to the positive side of the battery and the cathode is switched.If this is so, you need to use a PNP transistor. If the current gain isn't high enough then you can cascade two PNP transistors, as per my other schematic.

author: GuilTy
author_rank: Contributor
timestamp: January 12, 2016, 11:06:21 am
content: Why don't you use a voltage follower, in order to exploit its current gain.Regards,G

author: nsled
author_rank: Contributor
timestamp: January 12, 2016, 11:22:54 am
content: Yes that's correct, charge during the day and turn on at night. The LED array has been cut off and replaced with a MP3 player.I can show the circuit board, attached and I also created a .gif merging both sides so its easier to see what leads to what here:https://gfycat.com/ElderlyCheeryImperialeagleThere's no LED anymore but if I reattach it its about 3v when on and depending on the light 5mV or so when off.I'm not sure what a voltage follower is or what it does?I went mad on ebay today out of frustration,Bought a reed relayhttp://www.ebay.com.au/itm/191736125597?_trksid=p2057872.m2749.l2649&ssPageName=STRK%3AMEBIDX%3AITmight not work, the coil needs 50mAA solid state relayhttp://www.ebay.com.au/itm/221821497245?_trksid=p2057872.m2749.l2649&ssPageName=STRK%3AMEBIDX%3AITAre these wired the same as a mechanical relay?One of these mosfet module thingshttp://www.ebay.com.au/itm/131644770312?_trksid=p2057872.m2749.l2649&ssPageName=STRK%3AMEBIDX%3AITAnd a bunch of these optocouplershttp://www.ebay.com.au/itm/252224352449?_trksid=p2057872.m2749.l2649&ssPageName=STRK%3AMEBIDX%3AITNo real idea what I'm doing.

author: GuilTy
author_rank: Contributor
timestamp: January 12, 2016, 11:30:52 am
content: This link will take you to a simulation of a voltage follower. There is also an explanation.http://www.falstad.com/circuit/e-amp-follower.htmlI have one question: why did you buy optocuplers?G

author: nsled
author_rank: Contributor
timestamp: January 12, 2016, 11:51:19 am
content: Quote from: GuilTy on January 12, 2016, 11:30:52 amCombined with a transistor it creates a relay does it not? I know the circuit will power a LED the way I want as that's what its designed to do so should trigger an optocoupler. To make more sense of it I want to isolate the signal from the load circuit. At least that was the idea but I freely admit to not knowing much.

author: Zero999
author_rank: Super Contributor
timestamp: January 12, 2016, 02:15:07 pm
content: Quote from: GuilTy on January 12, 2016, 11:06:21 amBecause a BJT voltage follower will drop around 0.6V to 1V and there isn't enough headroom for that.Quote from: GuilTy on January 12, 2016, 11:30:52 amThat will only work, if you can find a reasonably priced op-amp, capable of 500mA output current without dropping more than 200mV in the output stage and can run off 3.6V.Hint: you can't, it's much better to use discrete transistors.

author: GuilTy
author_rank: Contributor
timestamp: January 12, 2016, 02:32:16 pm
content: @Hero999Fair enough!G

author: Audioguru
author_rank: Super Contributor
timestamp: January 12, 2016, 06:32:08 pm
content: Your idea will not work maybe because the LED in many solar garden lights lights up a little when the thing is in the sun.Another problem is that the LED in many solar garden light flickers at the voltage boosting oscillator's frequency so the feed to the base of your transistor when dark is on and off all the time.I have many solar garden lights ($1.00) from Walmart. Many didn't work but were replaced. I measure 1.22V (the battery voltage) across the LED when one is in the sun but the voltage is too low to light the 3V white LED. The 1.22V will turn on your transistor when you need it to turn off.

author: Audioguru
author_rank: Super Contributor
timestamp: January 12, 2016, 06:48:15 pm
content: The pcb from your solar garden light does not show a voltage boosting circuit. Instead it has one transistor used to turn on the LED and the other transistor is used to turn off the first transistor when there is light. But the diode seems to feed the positive solar panel voltage minus a diode voltage drop to the LED when in the sun, so try it with the diode removed. But maybe the diode also charges the 3.6V battery??

author: nsled
author_rank: Contributor
timestamp: January 13, 2016, 08:30:51 am
content: Quote from: Audioguru on January 12, 2016, 06:48:15 pmYes you are right, its running off 3AA's where most lights make use of 1, so no reason to step up voltage. I did check the voltage with a scope and it was smooth. The short circuit voltage in the sun is practically zero with no detectable current flow.But maybe I've been thinking about this all wrong why can't I just remove some of the current limitations from the LED driving circuit? Instead of working around it, change it. Can anyone see a way of doing this? change in a resistor?