Re: [funwithtransistors] IRF3205 (104 amps??)


May 11, 2013

 


----------------------------

#1333 May 11, 2013

I have a package of 10 IRF 3205 N channel mosfets. They are rated

for 55 volts and 104 Amps @ 10V. In the data sheet it talks about 50

watt current dissipation, and a package current limitation (junction

current) of 75A.



Not being an expert at reading data sheets, none of this adds up to

me............ The TO220 package conductor size is absurdly small to

be talking about 75 amps. 110 amps @ 10V = 1100 watts, far in excess

of the 50 watts mentioned. Package limitation of 75 amps adds up to 750 W.



It's completely confusing to me....... What kind of current can I

really put through these things? I have an application for them where

I had planned to use a number of them in parallel bolted to an aluminum

plate for cooling. Current max should be around 200A @ 12V for a short

interval (occasional) of about 30 seconds. They're cheap (about 80

cents each). I'm not sure how many of them I really need to parallel

for this. Any thoughts? I'm thinking they simply are not a suitable

part for the job.



Howard



----------------------------

#1334 May 11, 2013

Those are all maximum limits. .So you can NOT run the device such that it hits every lint at once. .you have to choose. .For example if you choose to go with 50 volts, clearly you can put 100 amps through it or you'd run into the 50W limit. . So at 50 volts you have a 1A limit but at 0.5 volts you could put more current through it. .Once you select a voltage or a current then you must check the over limits. .Your design can't go past any one of them. .You will have to think about corner cases and start up transients and so on to make sure NONE of those limits are passed even with shorted outputs and so on. . I'd so with a safety margin of at least 50%

You say you need 200A at 12V. .That is 2400W. .But are you building a switch? .If so then likely there is only maximum voltage when the switch is off with no current and when the current is maximum there is little drop across the device. .So I doubt you need 2400w. . .You will have to plot some waveforms.



----------------------------

#1335 May 11, 2013

Hi Howard.

The Dissipation figure is completely different to the switching figure.

You could quite happily be switching a 50v load taking 50 amps without difficulty - as long as the fet switches on and off CLEANLY.

If you biassed the fet so that if had 20v ACROSS IT while 2.5 A were flowing then there would be 50w dissipated by the fet.

Check the RDS-on figure in the datasheet. It is 0.008 ohm. If you switch the fet Hard-on then you can think of it as a 0.008 ohm resistor with 50A going through it (the fet then will only have 0.4v across it) and the dissipation will be 20W (not 2500 as you were thinking)

I agree that 50A is hard to imagine going through a fet like this, Personally I would halve this for my design.

So, parallel as many as you want to stay within this figure, and make sure they are given plenty of Gate drive

Cheers,

Andrew



----------------------------

#1336 May 11, 2013

You can do this graphically, plot current and voltage drop and then plot their product (Watts) the worst case is going to be in the "half-on" case because certainly when it is full off there is zero watts. . To do this you will need to be able to plot the controlling voltage on the gate and know how fast it raises and falls. .What is the frequency? .Is this 60Hz ir a PWM design? .

The very best way is to do a simulation in Spice and then ask Spice to plot power dissipation in watts. .Spice makes it easy



----------------------------

#1337 May 12, 2013

Andrew & Chris:

    I'm still pretty ignorant of how to read a data sheet and whatthe various items listed mean.  Your comments answer somequestions.    I'm not  sure how to read gate saturation from thedata sheet, but it appears from the charts to be somewhere in the10-12 V range.   I will be working entirely with 12 VDC andnothing else and the mosfets will function as switches.   The loadis entirely resistive, and on time will range up to about 30seconds, and will be extremely intermittent.  The mosfets will beswitched using an ordinary push button, which should take the gatevoltage from zero up to 12 volts pretty rapidly if the switch isdecent.   Perhaps a P channel mosfet with a lower saturationvoltage should be used to do the switching in response to the pushbutton to minimize issues with corrosion related resistance in thepush button.



    Thanks for the explanation of power dissipation......... Ididn't understand that..... I think I do now.   It looks to melike an array of 4 or 5 of these should easily handle the load ifset up properly.   It also looks like the bias resistor value isimportant since it appears that my working voltage is aboutsaturation voltage, and I want full saturation as well as rapiddischarge of gate capacitance.



                                                Howard





On 05/11/2013 01:20 PM, andrew jardine wrote:



Hi Howard.

The Dissipation figure is completely different tothe switching figure.

You could quite happily be switching a 50v loadtaking 50 amps without difficulty - as long as thefet switches on and off CLEANLY.

If you biassed the fet so that if had 20v ACROSS ITwhile 2.5 A were flowing then there would be 50wdissipated by the fet.

Check the RDS-on figure in the datasheet. It is 0.008ohm. If you switch the fet Hard-on then you can think ofit as a 0.008 ohm resistor with 50A going through it(the fet then will only have 0.4v across it) and thedissipation will be 20W (not 2500 as you were thinking)

I agree that 50A is hard to imagine going through a fetlike this, Personally I would halve this for my design.

So, parallel as many as you want to stay within this figure,and make sure they are given plenty of Gate drive

Cheers,

Andrew







----------------------------

#1338 May 12, 2013

Why not just use a relay? .

One thing about a push button. .The the contacts will bounds at the millisecond level or faster. .They never open and close as fast as you think. .So when you push it the power mosfets will be switched on and off about 50 times each time the button goes up of down. .If there is anything that looks like a filter the on/off bounding will appear as a slow raising plus. . The best thing is to make de-bounce circuit or design the mosfets to handle the slow raise time of a few milliseconds

a debouncer is just a low pass RC with possibly a comparator, something as simple as a logic gate.

In other words, I think you have to handle the "half on" case but only for a faction of a second.



----------------------------

#1339 May 12, 2013

Greetings...Added to your idea.. why not a relay, or the initialpushbutton, plus a .01 capacitor across the contacts.. The .01 takes care of thecontact bounce..Larry ve3fxq ----- Original Message -----From:Chris AlbertsonTo: funwithtransistors@yahoogroups.comSent: Sunday, May 12, 2013 11:20 AMSubject: Re: [funwithtransistors] IRF3205(104 amps??)

 Why not just use a relay?  

One thing about a push button.  The the contacts will bounds at themillisecond level or faster.  They never open and close as fast as youthink.  So when you push it the power mosfets will be switched on and offabout 50 times each time the button goes up of down.  If there isanything that looks like a filter the on/off bounding will appear as a slowraising plus.   The best thing is to make de-bounce circuit or design themosfets to handle the slow raise time of a few milliseconds

a debouncer is just a low pass RC with possibly a comparator, somethingas simple as a logic gate.

In other words, I think you have to handle the "half on" case but onlyfor a faction of a second.



----------------------------

#1340 May 13, 2013

I was not aware of thebounce................ A relay would be fine so long as it had thecapacity...... except that I lack the space... I have the mosfets,so I thought I'd try them..... Thanks for the tip about thecap.......... I'm just learning this stuff............ These kindsof tips help prevent me from learning the hard way. ;-)... Iprefer to learn from the experience of others!



... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...... ... ... ... ... Howard

On 05/12/2013 09:37 AM, larry allen wrote:



Greetings...Added to your idea.. why not a relay, orthe initial pushbutton, plus a .01 capacitor across thecontacts.. The .01 takes care of the contact bounce..Larry ve3fxq.----- Original Message -----From: Chris AlbertsonTo: funwithtransistors@yahoogroups.comSent: Sunday, May 12, 201311:20 AMSubject: Re:[funwithtransistors] IRF3205 (104 amps??)



.Why not just use a relay? .

One thing about a push button. .The the contacts willbounds at the millisecond level or faster. .They neveropen and close as fast as you think. .So when you push itthe power mosfets will be switched on and off about 50times each time the button goes up of down. .If there isanything that looks like a filter the on/off bounding willappear as a slow raising plus. . The best thing is to makede-bounce circuit or design the mosfets to handle the slowraise time of a few milliseconds

a debouncer is just a low pass RC with possibly acomparator, something as simple as a logic gate.

In other words, I think you have to handle the "halfon" case but only for a faction of a second.



----------------------------

#1341 May 13, 2013

Hi all.I would also put a resistor from the gate to ground (assuming the button connects The gate to 12v)The value will depend upon the capacitor value.This is essential. It is needed to discharge the cap. The gate of the fet is such a high impedance that the gate capacitance itself (a few pf) can keep the gate on for seconds!Try this as an experiment... put a small bulb as the load and with no discharge resistor the bulb stays on for seconds!The danger point is when the bulb/load is starting to dim. Now the fet is entering it's analogue state and is not just a switch. It is now going to seriously dissipate power and get hot.You do not want it in this state for more than milliseconds. Of course you can use this gate capacitance to your advantage....it can be your debounce capacitor, but the resistor is essential.Have fun!Andrew



Sent from Samsung Mobile



----------------------------

#1342 May 13, 2013

I would include full gate protection with a capacitor, 15 volt zener

diode, and the pull down resistor in parallel with the gate close to

the MOSFET if it is only being driven by a mechanical switch. I would

include a series resistor between the zener and parallel

capacitor/pull down resistor if I was worried about high voltage from

the switch as well.



On Mon, 13 May 2013 21:03:42 +0100, andybiker andybiker@...>

wrote:

>Hi all.

>I would also put a resistor from the gate to ground (assuming the button connects.

>The gate to 12v)

>The value will depend upon the capacitor value.

>This is essential. It is needed to discharge the cap. The gate of the fet is such a high impedance that the gate capacitance itself (a few pf) can keep the gate on for seconds!

>Try this as an experiment... put a small bulb as the load and with no discharge resistor the bulb stays on for seconds!

>The danger point is when the bulb/load is starting to dim. Now the fet is entering it's analogue state and is not just a switch. It is now going to seriously dissipate power and get hot.

>You do not want it in this state for more than milliseconds..

>Of course you can use this gate capacitance to your advantage....it can be your debounce capacitor, but the resistor is essential.

>Have fun!

>Andrew

>

>Sent from Samsung Mobile







----------------------------

#1343 May 13, 2013

do you have a schematic diagram of yourproblem?Larry ve3fxq ----- Original Message -----From:DavidTo: funwithtransistors@yahoogroups.comSent: Monday, May 13, 2013 4:47 PMSubject: Re: [funwithtransistors] IRF3205(104 amps??)

 I would include full gate protection with a capacitor, 15 voltzenerdiode, and the pull down resistor in parallel with the gate closetothe MOSFET if it is only being driven by a mechanical switch. Iwouldinclude a series resistor between the zener andparallelcapacitor/pull down resistor if I was worried about high voltagefromthe switch as well.

On Mon, 13 May 2013 21:03:42 +0100,andybiker andybiker@...>wrote:

>Hiall.>I would also put a resistor from the gate to ground (assuming thebutton connects >The gate to 12v)>The value will dependupon the capacitor value.>This is essential. It is needed to dischargethe cap. The gate of the fet is such a high impedance that the gatecapacitance itself (a few pf) can keep the gate on for seconds!>Trythis as an experiment... put a small bulb as the load and with no dischargeresistor the bulb stays on for seconds!>The danger point is when thebulb/load is starting to dim. Now the fet is entering it's analogue state andis not just a switch. It is now going to seriously dissipate power and gethot.>You do not want it in this state for more thanmilliseconds. >Of course you can use this gate capacitance to youradvantage....it can be your debounce capacitor, but the resistor isessential.>Have fun!>Andrew>>Sent from SamsungMobile



----------------------------

#1344 May 13, 2013

Wouldn’t the addition of a capacitor across the contacts of the switch just increases the transition time from open circuit to closed circuit (which I believe is what you wanted to avoid)?  The idea is to switch the FETs quickly from non-conducting to conducting so they don’t have to dissipate a lot of power.   For this application, you might want to use a set-reset latch.  To avoid the FETs altogether, you might want to consider a “contactor” which is a type of relay specifically made for switching high currents.  Jerry F.   From: funwithtransistors@yahoogroups.com [mailto:funwithtransistors@yahoogroups.com] On Behalf Of StoneToolSent: Monday, May 13, 2013 9:57 AMTo: funwithtransistors@yahoogroups.comSubject: Re: [funwithtransistors] IRF3205 (104 amps??)



----------------------------

#1345 May 13, 2013

With all these questions, this is why I would like to seea schematic diagram...Larry ----- Original Message -----From:Gerald FeldmanTo: funwithtransistors@yahoogroups.comSent: Monday, May 13, 2013 5:17 PMSubject: RE: [funwithtransistors] IRF3205(104 amps??)

 Wouldn.t the addition of acapacitor across the contacts of the switch just increases the transition timefrom open circuit to closed circuit (which I believe is what you wanted toavoid)?  The idea is to switch the FETs quickly from non-conducting toconducting so they don.t have to dissipate a lot of power.   Forthis application, you might want to use a set-reset latch.

To avoid the FETs altogether,you might want to consider a .contactor. which is a type of relay specificallymade for switching high currents.

Jerry F. 

From: funwithtransistors@yahoogroups.com[mailto:funwithtransistors@yahoogroups.com] On Behalf OfStoneToolSent: Monday, May 13, 2013 9:57 AMTo:funwithtransistors@yahoogroups.comSubject: Re: [funwithtransistors]IRF3205 (104 amps??)



This Site's Privacy Policy
Google's privacy policies

S
e
n
i
o
r
T
u
b
e
.
o
r
g