MOSFET newbie - Page 1
Forum section: electronics / beginners

author: littlebill
author_rank: Regular Contributor
timestamp: June 28, 2015, 07:37:56 pm
content: ok i boughthttp://www.irf.com/product-info/datasheets/data/irfz44n.pdfi am playing with it and now i have learned i need a higher voltage for gate to turn it on fully.based on the datasheet above, i see the min voltage difference is 2v? is that correct?how do i determine what the voltage difference needs to be to get full amperage?also is there a way drive this when the source voltage is as high as the gate voltage?ex.. 5v source running a 5v relay

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 28, 2015, 07:47:21 pm
content: QuoteNo, that's just the threshold voltage , where the MOSFETs starts barly to conduct. In the same row you also find the associated drain current: 250µAYou have to look further into the datasheet and you will find a ID=f(VDS) graph with the gate soure voltage as parameter.QuoteSee above.QuoteYes, these circuits are called "high side driver"QuoteYou don't need a MOSFET to switchcouple mAfor a relay. What does the relay is intended to switch?P.S.: 5V gate source voltage is rather tight for these particular MOSFET. It works, but only up to about ID=3A (without too much losses)

author: SeanB
author_rank: Super Contributor
timestamp: June 28, 2015, 07:50:24 pm
content: Fig 1 and fig 2 in the datasheet tells you how much gate source voltage you need for it to turn on fully for a particular drain current. 5v will allow you to pass around 5A, and if you take the GS voltage to 6V it will handle the full rated current as a low resistance switch.

author: littlebill
author_rank: Regular Contributor
timestamp: June 28, 2015, 08:15:12 pm
content: thanks so far, how do i determine when the unit will need a heatsink?

author: Zero999
author_rank: Super Contributor
timestamp: June 28, 2015, 08:20:29 pm
content: Quote from: littlebill on June 28, 2015, 08:15:12 pmYou can work this out given the temperature rise is 65°C per Watt of power dissipated (see page 1 of the data sheet) and the maximum junction temperature is 175°C.

author: schwarzrmsu
author_rank: Contributor
timestamp: June 28, 2015, 09:59:01 pm
content: Two major things you need to determine before you decide you need a heat sink:1.) Maximum amount of current being switched through the MOSFET2.) Maximum ambient temperatureThen you do a couple of simple calculations below:Power Dissipated = (Maximum Current)^2 * Rds(on)Temperature of Junction = (Power Dissapatted)*RjaThen make sure Temperature of Junction calculated is less then the maximum allowable junction temperature shown in the datasheet. If not, then you will need to implement a heat sink in order to decrease the Rja figure in your calculations.

author: littlebill
author_rank: Regular Contributor
timestamp: June 28, 2015, 10:54:40 pm
content: thanks for the info guys!

author: suicidaleggroll
author_rank: Super Contributor
timestamp: June 28, 2015, 11:23:10 pm
content: Quote from: littlebill on June 28, 2015, 07:37:56 pmThe best way is to study the Rds(on) vs Vgs or Id vs Vgs figures in the datasheet, but a good rule of thumb is you need 2x the nominal Vgs(th) to turn it on fully. Your device has a min and max Vgs(th) of 2 and 4 respectively. So let's call the nominal 3. 2x nominal = 6v, so you should aim to give that FET a Vgs of at least 6v to get a decent Rds(on).Quote from: littlebill on June 28, 2015, 07:37:56 pmSwitches come in two varieties, high side and low side. A high side switch goes between the + supply and the + input on the device, with the device's - tied directly to the circuit -. A low side switch goes between the device's - and the circuit -, with the device's + tied directly to the supply.MOSFETs come in two varieties as well, Pch and Nch. With a Pch, you need to bring the gate below the source to allow current to flow from the source to the drain. With an Nch, you need to bring the gate above the source to allow current to flow from the drain to the source. Everything is reversed. Each of these has a "natural fit" with the two types of switches mentioned above.Pch MOSFETs work best as a high side switch. Tie the FET source to the supply +, the FET drain to the load +, and the load - to the supply -. Then just bring the gate to the supply + to turn it off, and bring the gate to ground to turn it on.Nch MOSFETs work best as a low side switch. Tie the load + to the supply +, the load - to the drain, the source to the supply -. Then just bring the gate to ground to turn it off, and bring the gate high to turn it on.You bought an Nch FET, so you're best off using it as a low side switch. Nch MOSFETscanbe used for high side switching, but to do that you need a special gate driving to bring the gate even higher than the source (which is already at your supply +) to turn it on. This can be a pain, it's usually easier to just use a Pch if you need high side switching.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 29, 2015, 04:21:02 am
content: You need no heatsink, if the MOSFET is just switching a relay...hence my question what the relay is switching...

author: schwarzrmsu
author_rank: Contributor
timestamp: June 29, 2015, 10:19:53 am
content: Quote from: suicidaleggroll on June 28, 2015, 11:23:10 pmI deal with a lot of high side switching using Nch Power MOSFET's. If you are simply triggering a relay coil I would use a Nch and orient the circuit so the MOSFET is on the lowside with respect to the coil. If you can not do that look at implementing a charge pump to double your gate voltage and set up a simple circuit to turn on and off your gate voltage.