Drain Source Voltage question - Page 1
Forum section: electronics / beginners

author: DuncanSteel
author_rank: Contributor
timestamp: February 06, 2017, 09:19:08 pm
content: HiI would like to control (Car, side electrical mirror), with 5V from Atmega32u4 pin & a N-MOSFET for learning purposes. I estimate that this motor requires 12V & up to 1.5A to operate.I have found thisMOSFET& I would like to know if my choice is correct & I understand what Im doing?1) At 5V the mosfet will be fully open & its drain to source resistance will be 0.045ohm2) Max current that can go through drain to source is 2.6A ?3) At 25 degrees this mosfet can dissipate 1W without a heat sink, I will be dissipating 0.1W at 1.5A (are my calculations correct?)4) Drain Source Voltage Vds: 20V - I dont understand what this is. 20V Voltage drop from drain to source? Does this mean that I cant apply just 12V at the Drain & hope that the motor will start up? I understand that there will be some voltage drop, but how much at 1.5A ? im a bit confused.

author: jaromir
author_rank: Unknown
timestamp: February 06, 2017, 09:34:07 pm
content: 1) Yes2) Correct **3) 0,045Ohmx1,5A=0,0675V. 0,0675V*1,5A=0,1W, so it's OK4) That is maximal allowable voltage between D and S to still operate as transistor. When the transistor is off (Vgs=0V), the drain "is seeing" the full supply voltage, in this case 12V*. Once you apply more than 20V between D and S, manufacturer guarantees nothing and anything can happen. From nothing to black hole forming and swallowing universe.Very likely, it will survive few (tens) volts of overloading, then it starts to behave either as conductor or insulator, often in non-reversible way.* 20V MOSFET could too low (with respect to Vds) for automotive environment, where voltage spikes on 12V rail could be way higher. Especially when switching motors or other inductive loads.------** EDIT: it is 0,045Ohm typical, 0,057Ohms maximal. You always have to deal with worst-case scenarios, so take 0,057Ohm figure. The calculation prom point 3 then gives you 0,13W, which is still OK.

author: langwadt
author_rank: Super Contributor
timestamp: February 06, 2017, 09:45:28 pm
content: I'd use something like a vnd7nv04 instead, it is meant for automotive and protected against shorts and overtemperature

author: Benta
author_rank: Super Contributor
timestamp: February 06, 2017, 09:47:49 pm
content: QuoteCorrect. Standard today is 55 V for for protected FETs, 60 V for normal MOSFETs + TVS in an automotive environment.

author: Ian.M
author_rank: Super Contributor
timestamp: February 06, 2017, 10:10:27 pm
content: Another trap for young players is the motor transient startup current, which is the same as the locked rotor stall current and is only limited by the winding DC resistance,Typically, its an order of magnitude more than the running current, so for a motor that draws 1.5A running, you'll probably need a 25A or 30A rated MOSFET.

author: Benta
author_rank: Super Contributor
timestamp: February 06, 2017, 10:27:39 pm
content: Quote from: Ian.M on February 06, 2017, 10:10:27 pmA very good point. You can estimate the starting current by placing an ohmmeter across the motor terminals and rotating the shaft very slowly. You'll probably be surprised at the result (expect 0.5...2 ohms).

author: Ian.M
author_rank: Super Contributor
timestamp: February 07, 2017, 12:35:42 am
content: Quote from: Benta on February 06, 2017, 10:27:39 pmEmphasis onslowly. The aim is to turn the shaft half a turn as slowly as possible, ideally pausing every couple of degrees, waiting for the ohmmeter to settle, noting the minimum reading reached so far then resuming movement. All PM motors will act as generators and to a lesser extent, wound stator motors due to residual magnetism, so if you spin the shaft too fast, it will upset or possibly even damage the ohmmeter.

author: DuncanSteel
author_rank: Contributor
timestamp: February 08, 2017, 03:10:27 pm
content: That was helpful, thank you. There is just 1 more thing.I ordered the partSI2302DDS-T1-GE3,(Datasheet hire)& hooked it up but the motor started spinning even tho I did not apply 5V to the gate (when I apply 5V to the gate motor stops). How comes?Should not the mosfet start conducting when I apply 5V to the gate? How comes it works in reverse?I mean I would like it to conduct only when I apply 5V to the gateEdit, never mind, I ether burned the first mos or it was already broken, good thing I ordered 2 of them.Problem solved.

author: Zero999
author_rank: Super Contributor
timestamp: February 08, 2017, 06:25:19 pm
content: Quote from: langwadt on February 06, 2017, 09:45:28 pmGood but I think he's after something SOT-23.Quote from: DuncanSteel on February 08, 2017, 03:10:27 pmOh no, that's not the right way to do it. You've got it configured as a source follower. The MOSFET turns on when the voltage between the gate and source reaches a certain threshold. As you have it drawn, when 5V is applied to the gate, the MOSFET will turn on, current will flow through the motor which is in series with the source. A voltage will develop across the motor, lifting the source voltage up, thus reducing the potential difference between the gate and source. The circuit will reach an equilibrium, with about 4V across the motor and 1V between the gate and source. If the current is high enough, then the MOSFET will overheat, as it will have 8V across it, at whatever current the motor draws at 4V. Since the MOSFET is in a tiny package, it won't take much power to destroy it.The source should be connected to 0V and the motor between the drain and +12V. Then when 5V is applied to the gate, the full 5V appears across the gate and source, the MOSFET turns on with a tiny voltage drop across it.Have you considered the start up current, when the motor is first turned on? It might still be too high for the MOSFET to handle.QuoteNo, the datasheet specifies 710mW, which is probably a bit optimistic.