Help sizing a MOSFET - Page 1
Forum section: electronics / beginners

author: vmallet
author_rank: Regular Contributor
timestamp: April 17, 2019, 12:15:25 am
content: I want to experiment driving a 12v/85ohm/145mA relay coil with a 5v signal using a MOSFET.I have a ON Semi 2N7000 (60V 200mA,https://www.mouser.com/datasheet/2/308/2N7000-D-33553.pdf) and a ON Semi BS170 (60V 500mA,https://www.mouser.com/datasheet/2/308/BS170-1118810.pdf)Would they both work just fine? Would one (or both) not be appropriate? Is one recommended over the other? Should I use another one altogether?I'm hoping I can learn how to pluck the right info from the data sheets.Thanks!

author: David Hess
author_rank: Super Contributor
timestamp: April 17, 2019, 12:24:20 pm
content: The 2N7000 is marginal at that load current with only 5 volts of drive and some units may overheat. The BS170 is a better choice and all parts should work.You can get this by figuring out from the datasheet by what the drain-to-source resistance will be with a gate drive of 5 volts.

author: Zero999
author_rank: Super Contributor
timestamp: April 17, 2019, 12:51:18 pm
content: What's wrong with a plain old BJT? The BC338 should do that, even with a miserable 3mA of base drive, it will have an on voltage of under 1V.http://www.mouser.com/ds/2/149/BC337-193546.pdf

author: exe
author_rank: Unknown
timestamp: April 17, 2019, 01:34:10 pm
content: Just don't forget base resistor for bjt.

author: David Hess
author_rank: Super Contributor
timestamp: April 17, 2019, 02:20:56 pm
content: Quote from: Zero999 on April 17, 2019, 12:51:18 pmNothing is wrong with a BJT but vmallet asked about two specific power MOSFETs. A 2N4401 would be my first choice but the BC337/BC338 would be equally good.

author: vmallet
author_rank: Regular Contributor
timestamp: April 17, 2019, 06:34:56 pm
content: Thank you for your replies.Quote from: David Hess on April 17, 2019, 12:24:20 pmTrying to figure this out, not seeing it yet. On the 2N7000 datasheet I see: Rds(on)@Vgs=4.5v@Id=75mA = 6ohm max, with an Rds(on) otherwise stated at 5ohms on the first page. How do I draw conclusions w.r.t. the drive voltage that would be appropriate?Quote from: Zero999 on April 17, 2019, 12:51:18 pmLike David said I was hoping to get a bit of practical understanding of mosfets. I did use a 2N4401 in my first attempthttps://www.eevblog.com/forum/beginners/adding-a-horn-on-a-delay-where-to-begin/msg2330295/#msg2330295(using the power supply you suggested) and it seemed fine on the bench.When playing with the 2N4401 I also felt it was easy to not drive it hard enough and risk overheating by not being fully saturated. I gave it 20mA on its base and I felt it was a lot of current when a mosfet could do this "for free"*. (*newbie understanding caveat)

author: MarkF
author_rank: Super Contributor
timestamp: April 17, 2019, 09:26:11 pm
content: Quote from: vmallet on April 17, 2019, 06:34:56 pmI believe the two items to look at are:The Continuous Drain Current (ID) = 200mA >vs< for the relay 145mA requirement.The Total Power Dissipation (PD) = 350mW >vs< for the relay of P = I2RDS= (0.145)2* 6 = 126mW requirement.The continuous current may be the driving factor here (i.e. 75% of max).

author: David Hess
author_rank: Super Contributor
timestamp: April 19, 2019, 11:21:31 am
content: Quote from: vmallet on April 17, 2019, 06:34:56 pmThe problem is that Vgs(threshold) can vary significantly between MOSFETs of the same type. For the 2N7000, it can vary between 0.8 and 3.0 volts according to the datasheet so pretend that that 5 volts of drive is actually 2.8 volts because a part has a high Vgs(th). Now the drain current graph shows less than 100 milliamps of output current. But this assumes a worse case on the graph which is probably not possible.So instead look at the static drain-source on-resistance specification at 4.5 volts which is 6 ohms maximum. 6 ohms at 145 milliamps is 0.87 volts and 126 milliwatts. But the junction-to-ambient thermal resistance is 357 C/W so the temperature rises 45 C and the power dissipation limit drops to 244 milliwatts. That is not too bad but a factor of 2 may only be acceptable under ideal conditions.I also would not rely on a 200 milliamp continuous current part to drive 145 milliamps reliably over the long term.Quote from: vmallet on April 17, 2019, 06:34:56 pmA forced beta of 20 means 7.25 milliamps of drive current for a 145 milliamp load which is reasonable. 20 milliamps results in a forced beta of 7.25 which is a little high but will not hurt anything. A forced beta of 10 is common but up to a point, using a higher current transistor allows for a higher forced beta and lower drive current.This shows why a higher current part like the 2N4401 should be used instead of a 200 milliamp 2N3904. At higher currents, current gain drops so while a 2N3904 could switch a 145 milliamp load, it will require excessive base drive to do it because of lower current gain at a load current approaching its maximum.

author: Zero999
author_rank: Super Contributor
timestamp: April 19, 2019, 04:49:34 pm
content: Don't worry about completely saturating the BJT. People make a big deal of it, but it's not necessary, as long as the voltage drop is low enough for the load to work and not result in excessive power dissipation in the BJT, then near saturation is fine.The 2N4401 is specified hFE> 100, when IC= 150 mAdc and VCE= 1.0 VDC, which would be an IB= 1.5mA, giving a power dissipation of just 150mW. I suggested a slightly higher forced beta of about 50, IB= 3mA, just to be sure, especially at lower temperatures. If you're deriving the base current from the 12V supply, using a linear regulator or resistor, then an IBof 20mA would use 240mW, which is more power wasted, than you'd save with a lower on voltage.