Simple MOSFET question - Page 1
Forum section: electronics / beginners

author: mc
author_rank: Regular Contributor
timestamp: May 20, 2012, 12:50:33 pm
content: Is there any reason why a n-type enhancement MOSFET can't be used to swith the postive side to a load?After much googling, all the examples I can find use n-type to switch the return/ground side, and p-type to switch the supply/power side, and my head now hurts.I've thrown together a quick simulation in LTspice, and it works, but I'm wondering if there's something obvious I've missed.My reason for asking, is I need to disconnect a 2 cell LiPo balancer circuit when the charger is turned of to reduce battery drain (charger/balancer/batteries are all built into the device), and the simplest way I can think of doing it is by using 3 n-types to disconnect the battery cells from the balancer circuit. When outside power is applied and the charger kicks in, power is supplied to the mosfets and the balancer is connected. Or is there some other option I've missed?

author: Psi
author_rank: Super Contributor
timestamp: May 20, 2012, 01:09:11 pm
content: Do you mean putting the load from the source pin to ground (source follower) ?Doing so means you have current gain but no voltage gain and you must overcome the gate threshold voltage.So, for example, if the fet has a gate threshold voltage of 3V then 7V on the gate will result in a max of 4V across your load (source-ground).

author: Zero999
author_rank: Super Contributor
timestamp: May 20, 2012, 01:24:30 pm
content: Quote from: mowicu on May 20, 2012, 12:50:33 pmIt can but the gate voltage needs to be higher than the source, by the threshold voltage, in order to turn it fully on, otherwise the source voltage will float at the gate voltage, minus the gate threshold, at the current being drawn.For example, suppose the MOSFET has a threshold voltage of 4V, at the required load current (which we assume is constant regardless of the voltage) and the power supply voltage is 14V. If the gate is connected to the positive rail, the load voltage will be 10V so there will be 4V lost across the MOSFET. In order to turn the MOSFET fully on, the gate voltage needs to be at least 18V. If you attempted to drive the MOSFET from a 5V, the load voltage would only be 1V.The problem is, if may be inconvenient to add another higher voltage power supply, just to turn on a MOSFET. For AC applications bootstrapping can be used but it requires additional components. Attached is a simple solution using a transistor but the turn off time is very slow.

author: electroguy
author_rank: Regular Contributor
timestamp: May 20, 2012, 01:32:26 pm
content: Quote from: Psi on May 20, 2012, 01:09:11 pm+1Also, your voltage at the gate pin will have to be higher than your input voltage to the drain pin, because when the fet turns on, it's low on resistance means the drain and source pin voltages are very similar (yes i know it depends on the "on resistance" and your load current, etc....).It's ok if you have 2 different voltages available for you to use (ie: dual power supply) but if you only have 1 input voltage, it won't work the best, as you will struggle to get the gate pin voltage to exceed the V-G-S required...

author: electroguy
author_rank: Regular Contributor
timestamp: May 20, 2012, 01:35:16 pm
content: Quote from: Hero999 on May 20, 2012, 01:24:30 pmHero999 described it better than i did. I didn't see his post till after i finished mine. sorry.

author: Zero999
author_rank: Super Contributor
timestamp: May 20, 2012, 01:50:53 pm
content: Whoops I forgot, the example of bootstrapping I posted is really just a voltage doubler used for DC operation. The MOSFET can be slowly turned on/off by applying a 20kHz signal to the base of the BJT.Attached is the bootstrap I was talking about. I'll post the LTSpice file, if you like. The problem is the on time is slow and the load has to pass enough current to keep the capacitor charged.

author: djsb
author_rank: Super Contributor
timestamp: May 20, 2012, 01:52:38 pm
content: How does the highside N Bootstrap circuit work? My initial guess is that the diodes and capacitor form a pump circuit of some kind. And I assume this circuit will only work with an AC signal driving it?David.P.S How does your bootstrap demo circuit work? Yes, Please post the LT spice file. Thanks.

author: Zero999
author_rank: Super Contributor
timestamp: May 20, 2012, 05:07:32 pm
content: Quote from: djsb on May 20, 2012, 01:52:38 pmYes, although a bootstrap wasn't really the correct term, it's just a charge pump voltage doubler.QuoteYes, apply a 20kHz waveform to the base of the transistor and the MOSFET will turn on, once the gate capacitance is charged enough.QuoteThere you go. Try simulating it and looking at the voltages across different components and varying the frequency and duty cycle. You'll see that there's a limit on the maximum on time which depends on the load current.Code:Version 4SHEET 1 880 680WIRE 512 -176 160 -176WIRE 288 -128 -240 -128WIRE 416 -128 288 -128WIRE 160 -112 160 -176WIRE -240 -64 -240 -128WIRE 288 -48 288 -128WIRE 416 -48 416 -128WIRE 160 32 160 -32WIRE 240 32 160 32WIRE -240 64 -240 16WIRE 160 64 160 32WIRE 416 64 416 16WIRE 512 64 512 -176WIRE 512 64 416 64WIRE 416 96 416 64WIRE -176 112 -240 112WIRE -64 112 -96 112WIRE 64 112 0 112WIRE 96 112 64 112WIRE 64 144 64 112WIRE -240 160 -240 112WIRE 160 192 160 160WIRE 288 192 288 48WIRE 288 192 160 192WIRE 416 192 416 160WIRE 416 192 288 192WIRE 64 224 64 208WIRE 160 224 160 192WIRE 160 224 64 224WIRE 160 272 160 224WIRE -240 288 -240 240WIRE 160 384 160 352FLAG -240 64 0FLAG 160 384 0FLAG -240 288 0SYMBOL nmos 240 -48 R0SYMATTR InstName M1SYMATTR Value IRL3915SYMBOL Misc\\EuropeanResistor 144 -128 R0SYMATTR InstName R1SYMATTR Value 100kSYMBOL voltage -240 -80 R0WINDOW 123 0 0 Left 0WINDOW 39 0 0 Left 0SYMATTR InstName V1SYMATTR Value 12VSYMBOL diode 400 -48 R0SYMATTR InstName D1SYMBOL Misc\\EuropeanResistor -80 96 R90WINDOW 0 5 56 VBottom 0WINDOW 3 27 56 VTop 0SYMATTR InstName R2SYMATTR Value 1MegSYMBOL Misc\\EuropeanPolcap 400 96 R0SYMATTR InstName C1SYMATTR Value 100µFSYMBOL Misc\\EuropeanResistor 144 256 R0SYMATTR InstName RLSYMATTR Value 100RSYMBOL voltage -240 144 R0WINDOW 123 0 0 Left 0WINDOW 39 0 0 Left 0SYMATTR InstName V2SYMATTR Value PULSE(0 12V 0 0 0 100m 200m)SYMBOL npn 96 64 R0SYMATTR InstName Q1SYMBOL cap 0 96 R90WINDOW 0 0 32 VBottom 0WINDOW 3 32 32 VTop 0SYMATTR InstName C2SYMATTR Value 100nSYMBOL diode 80 208 R180WINDOW 0 24 72 Left 0WINDOW 3 24 0 Left 0SYMATTR InstName D2TEXT -226 376 Left 0 !.tran 10

author: mc
author_rank: Regular Contributor
timestamp: May 20, 2012, 08:25:31 pm
content: Thanks for the replies guys.I knew there was something I was missing, so time to go and think of other solutions