Low voltage drop transistor (or other device) - Page 1
Forum section: electronics / beginners

author: permal
author_rank: Regular Contributor
timestamp: December 02, 2018, 06:06:02 pm
content: Hi,Besides transistors, what other devices are available to turn power on/off to other devices?I'm in need of something with a voltage drop lower than 0.5V at ~100mA, 3.3VA relay would have near 0V drop, at the cost of space and increased power draw. Anything integrated available?

author: Benta
author_rank: Super Contributor
timestamp: December 02, 2018, 06:29:42 pm
content: I'd use a DIP or SIP reed relay. Simple and efficient.That being said, 0.1 V drop over a bipolar transistor is normal, on a MOSFET it depends on the on-resistance.

author: tsman
author_rank: Frequent Contributor
timestamp: December 02, 2018, 06:34:22 pm
content: Why don't you want a transistor?Quote from: permal on December 02, 2018, 06:06:02 pmYou can get latching relays which just need a pulse to toggle.

author: janoc
author_rank: Super Contributor
timestamp: December 02, 2018, 06:37:36 pm
content: Quote from: permal on December 02, 2018, 06:06:02 pmEven chips use transistor (whether BJTs or MOSFETs) inside.A low voltage logic level MOSFET could work, there you have Rdson down in tens of milliohms if you pick the right part. So the voltage drop at 100mA would be in millivolts.

author: Zero999
author_rank: Super Contributor
timestamp: December 02, 2018, 06:50:24 pm
content: What voltage does this have to switch? 4V, 400V?Plenty of cheap transistors can do this, see data sheets linked below:https://www.onsemi.com/pub/Collateral/BC337-D.PDFhttps://cdn.sparkfun.com/datasheets/BreakoutBoards/BSS138.pdf

author: permal
author_rank: Regular Contributor
timestamp: December 02, 2018, 07:03:15 pm
content: 3.3V, updated OP.I've never used a Mostfet beforeI'm currently using a NPN transistor, ZXTN25050DFHTA. Changing that to a Mosfet, it'd have to be a P-channel Mosfet, right?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 02, 2018, 07:12:48 pm
content: Are you switching the low side or high side? Or is the load common-GND or common-VCC?If you're using NPN on high side,There's your Problem.NPN ~ NMOS, PNP ~ PMOS.Tim

author: permal
author_rank: Regular Contributor
timestamp: December 02, 2018, 07:30:08 pm
content: Quote from: T3sl4co1l on December 02, 2018, 07:12:48 pmUhm, highside? See image.Why is it a problem to put an NPN transistor on the high side? There's a certain voltage drop over it regardless of what side of the load I put it and the resulting Vloadwould be that much less anyhow? Right? I intentionally put it like this to ensure a common ground between what it controls and other parts.I found thisES answerin which they mention "digital type N-mosfet" which sounds like what I need (It'd still be on the high side due to the existing PCB though), what search terms can I use to find such a (N-channel)mosfet?

author: oPossum
author_rank: Super Contributor
timestamp: December 02, 2018, 07:37:24 pm
content: Quote from: permal on December 02, 2018, 07:30:08 pmThe base voltage would have to exceed the supply voltage to get the lowest possible Vce. That is not a problem with PNP/P-channel because the base/gate is pulled towards ground.You can easily get less than 100 mV Vceat 100 mA with ordinary PNP transistors like PN2907 and 2N4403.

author: permal
author_rank: Regular Contributor
timestamp: December 02, 2018, 07:45:38 pm
content: Quote from: oPossum on December 02, 2018, 07:37:24 pmOh, of course.Thanks for the suggestions, those are however THT mounted, I need a SOT23 to match the existing footprint.CMPDM7002AHCseems like a candidate, but what voltage is required on the gate to open it, I can't see that in the datasheet?

author: electrodacus
author_rank: Super Contributor
timestamp: December 02, 2018, 07:48:37 pm
content: Yes an N-channel like MCH3484 (overkill) will work great and you will have almost no voltage drop.

author: permal
author_rank: Regular Contributor
timestamp: December 02, 2018, 08:19:14 pm
content: Quote from: electrodacus on December 02, 2018, 07:48:37 pmThanks, but sadly the gate and source are switched compared to the SOT-23 so it won't fit the already existing PCB.

author: Zero999
author_rank: Super Contributor
timestamp: December 02, 2018, 08:22:52 pm
content: Quote from: permal on December 02, 2018, 07:03:15 pmThat's an NPN BJT, so would need to be replaced with an N-channel MOSFET.What's wrong with the ZXTN25050DFHTA? It looks perfectly good to me.Quote from: permal on December 02, 2018, 07:45:38 pmThat looks reasonable, also try the IRLML2402TR.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 02, 2018, 08:31:02 pm
content: Yeah, high side. You need a PNP or P-ch. Hack that in there (use pigtails if you have to fudge a footprint), and the enable signal will be inverted (active low). Vce(sat) is typically much lower than Vbe.Tim

author: oPossum
author_rank: Super Contributor
timestamp: December 02, 2018, 08:33:32 pm
content: If you replace the NPN with a N channel MOSFET, you will havegreatervoltage drop. You have to bodge in a P channel device as T3sl4co1l says.

author: permal
author_rank: Regular Contributor
timestamp: December 02, 2018, 08:42:10 pm
content: Quote from: oPossum on December 02, 2018, 08:33:32 pmWait, what? Why greater voltage drop? The on-resistance on for example IRLML2402TR is just 0.25-0.35. At 100mA, that's just 3.5mV, compared to the current ~0.7V.Is it that the 3.3V I can put on the gate isn't enough to open the Mosfet fully? (how do I tell

author: IanB
author_rank: Super Contributor
timestamp: December 02, 2018, 08:48:06 pm
content: Quote from: permal on December 02, 2018, 08:42:10 pmBecause you have to put the transistor on the correct side of the load to obtain good switching properties.For example, an NPN BJT has to go on the low side of the load in order to be a good switch. If you put it on the high side it won't work efficiently.Similarly, a PNP BJT should go on the high side of the load.By similar logic, the same is true of MOSFETs. According to the type of FET you need to put it on the correct side of the load. If you put it on the wrong side it won't switch efficiently.

author: HB9EVI
author_rank: Frequent Contributor
timestamp: December 02, 2018, 08:58:28 pm
content: If you want to switch high side with a N-FET, you need a considerably higher Vgs than the Vdd of 3,3V.Use a P-FET. For 100mA the higher Rdson of P-FETs doesn't really matter; that's an issue if you switch higher currents

author: permal
author_rank: Regular Contributor
timestamp: December 02, 2018, 09:04:27 pm
content: Quote from: IanB on December 02, 2018, 08:48:06 pmOk. So, the rule of thumb is that Ibe/Igdshould never have to pass the load to reach ground for the transistor/mosfet to be an effective switch. Is that correctly understood?Quote from: HB9EVI on December 02, 2018, 08:58:28 pmThat'll invert the control signal though, like T3sl4co1l pointed out. I'll have to think of how that affects the rest of the circuitry.Thanks for all the input so far.

author: HB9EVI
author_rank: Frequent Contributor
timestamp: December 02, 2018, 09:07:43 pm
content: That inverts the control signal, but there are many ways to handle that.otherwise, if you go with the N-FET, you can take a bootstrapped FET driver to solve the problem

author: Zero999
author_rank: Super Contributor
timestamp: December 03, 2018, 12:33:54 am
content: Quote from: permal on December 02, 2018, 09:04:27 pmSorry, I missed the schematic, when I posted previously.The drain.collector current in a MOSFET/BJT is dependant on the potential difference between the source/emitter and gate/base. An NPN/N-channel device starts to turn on then this potential difference is positive and a PNP/P-channel device needs it to be negative to turn on. With the load connected between the source/emitter and 0V and the gatebase tied to +V, the voltage at the source/emitter will rise, thus reducing the potential difference between the source/emitter and gate/base.In short, when the load is connected between the emitter/source and 0V, the transistor will never fully turn on, unless the base/gate voltage rises above the positive supply. Look up emitter follower.If the emitter or source is connected to 0V, with the load between the collector or drain and +V, the fully supply voltage can be applied between the gate and source or base and emitter via a suitable resistor and the transistor can turn fully on.Because a P-channel/PNP device conducts with the gate or base potential difference is negative, it can be used to switch the positive line, but it will reverse the logic.I'm tired now, so ,my wording is probably poor. No doubt someone else can do a better job.