Author Topic: Why IRF510 is more hot than IRF3205?  (Read 640 times)

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Offline Chriss

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Why IRF510 is more hot than IRF3205?
« on: October 13, 2017, 06:42:34 AM »
Hi!
I wish just check if my thinking is ok.

I have to power on 7 car bulbs which have 21W/pcs through 7 separated FET's.
The car bulb on 12V is actually using 1.7A current consumption.

So, the main problem is the heat what I have to get cool as possible, because the housing
of the device should be water proof and filled with some sort of silicon.

I made some test's with two FET's.

1. IRF3205
2. IRF510

Each other meat my requirement for one bulb/FET.
When I tested the IRF3205 with Vg=10V and the bulb was attached on the FET 24H the FET
was on ambient temperature, what means for me cool.

When I made the same test with the IRF510, but with a very short time, compared to the first testing(10min),
the FET was around 80C.
What means for me hot.

In both testing method I monitored the the current, which was flow through the bulb,
and it was 1.7A.
For me that means the FET was properly in an ON state.

Question:
Why was the IRF510 so hot with the same bulb and same setup?

If I put 7 IRF510 in the same box and fill it with silicon, I assume it would be
very hot after 10-15H of working, mean in an ON stage.

What I think:
The "problem" with the IRF510 is the Rds(on)=0.54 \$\Omega\$ but on the IRF3205 the Rds(on)=0.008 \$\Omega\$.

Am I right?

Thanks in advance.
My best regards.
 

Offline capt bullshot

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Re: Why IRF510 is more hot than IRF3205?
« Reply #1 on: October 13, 2017, 06:51:24 AM »
Yes
Safety devices hinder evolution
 
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Offline Chriss

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Re: Why IRF510 is more hot than IRF3205?
« Reply #2 on: October 13, 2017, 07:18:38 AM »
Would it be a more worse way to replace the IRF510 with a BD679 power transistor?

I really focusing on to power the mentioned bulb but with lower temp of the powering FET or transistor as possible.
I'm also limited with space in the case where I have to put my design with 7 FET's or transistors.
So it is hard to put heat sinks but it must be filled with some sort of water resistant material like silicon.

I can use the IRF3202 but the price is almost 3x higher in my areal than the IRF510.
This is the reason why I search another component maybe...

Thanks.
 

Online technogeeky

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Re: Why IRF510 is more hot than IRF3205?
« Reply #3 on: October 13, 2017, 08:50:17 AM »
So when I did the calculation for this, I got a result I didn't understand. But I'll just start at the beginning.

First, let's just make a simplifying assumption:

  • the bulb is instantly on at full power(no hysteresis effects)
  • the bulb burns 1.7 amps at 12 volts
  • therefore, the bulb acts like a (12 volt / 1.7 amp = 7.05 ohm) resistor


Then, when the FET is on, then the situation is just two resistors in series Rbulb and Rrds_on.

The power for two resistors in series is:

P = V^2 / (R1 + R2).

So for our two fets, we have:

Pirf510 = (12 V)^2 / (Rbulb + Rrds_on)
Pirf510 = (12 V)^2 / (7.05 ohms + 0.54 ohms)
Pirf510 = 18.97 watts

Pirf3205 = (12 V)^2 / (Rbulb + Rrds_on)
Pirf3205 = (12 V)^2 / (7.05 ohms + 0.008 ohms)
Pirf3205 = 20.40 watts

But at first glance, this seemed wrong. I set out to do this calculation to show that the wasted energy you are burning is an even worse problem than the heating IRF510. But apparently it is using less energy!  :wtf:

I found two ways to show this is the wrong view. First, let's actually compare to the situation where no mosfet is used:

Pbest_case = (12 V)^2 / (Rbulb + 0)
Pbest_case = (12 V)^2 / (7.05 ohms)
Pbest_case = 20.42 watts

A-ha. This begins to make sense. With our 0.008 ohm rds_on mosfet, we're really close to the ideal situation. But what's really happening?

Well, I drew up a stupidly simple simulation.

Here we can see what's going on very easily. The bulb in the IRF540 case is using about 2.5W less power than in the IRF3205 case (which is basically identical to the case with no mosfet). This is an amount of power that's probably imperceptible to you. However, the IRF540 switch is burning 1.26 watts as heat, compared to 0.023 watts for the IRF3205. This is about 55 times more power being wasted!
 
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Offline Chriss

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Re: Why IRF510 is more hot than IRF3205?
« Reply #4 on: October 13, 2017, 11:24:43 AM »
technogeeky
Thanks for this nice explanation.
This is an interesting angle of view.

The actually problem is the high RDS(ON) of the IRF510

When I computed the data for the IRF510 and IRF3205 by the given data from the datasheet,
I used an ambient temperature of 60oC for both FET.

My result end up with the IDSmax which can be used without a heat sink:

IRF510 - IDSmax[email protected]oC
IRF3205 - IDSmax[email protected]oC

The IRF510 can't be used in my design.
[email protected]oC ambient temp. is to low current parameter what the IRF510 can handle at that temp.

The IRF3205 is still in game but it is a bit over specified for my design.

I'm searched forward, for some other FET which could meet my requirement in my design.
I found this one, the IRFD014 which is in a HVMDIP package.

When I made my calculation again I got a result for IDSmax[email protected]oC.
It means for me, the awaiting max ambient temp. could be 60oC and even in that
scenario the bulb will consume near 1.7A which is 0.45A less than the IDSmax@60oC.

Can somebody approve my calculation?
Can this IRFD014 in a HVMDIP package really handle that current?
I'm a little suspicious about that...

Thank's
My best regards.
 

Online xani

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Re: Why IRF510 is more hot than IRF3205?
« Reply #5 on: October 13, 2017, 10:01:02 PM »
IRFD has 0.20 RDSon @10V and according to datasheet can  dissipate 1W on it's own. (there is no detals but I assume that's assisted by fat copper track)

That's 0.34V/[email protected] so IMO you should be fine heat-wise but datasheet states 1.7A as

Can somebody approve my calculation?
Can this IRFD014 in a HVMDIP package really handle that current?
I'm a little suspicious about that...
Maybe read the datasheet then

Continous drain current [email protected], [email protected] so you will be right on the edge of specs. Power-wise you're fine,  0.34V/[email protected]  altho that will really depend on how thermally conductive whatever you fill it with will be. Just remember that cold bulb will take more than 1.7A

IMO it is too small one, very little marigin for error. Just go on digikey/mouser and search for MOSFET with RDSon < 0.1 (for low losses) and continous current >5A (just to be on safe side)
 
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Offline ThomasDK

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Re: Why IRF510 is more hot than IRF3205?
« Reply #6 on: October 14, 2017, 08:01:18 AM »
Look at IRF530 or IRF540. They are "jelly bean" parts, so should be available everywhere.
 
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Offline Chriss

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Re: Why IRF510 is more hot than IRF3205?
« Reply #7 on: October 16, 2017, 05:19:26 AM »
Here is what type of FET I found and I think it would meet my requirement:

IRF540N
- RDS(ON)=0.044Ohm

When I made the calculation about the max current what can be handled
on an ambien temp of 60oC, I got the result about 6.4A.
Which would be more than enough.

I assume most of time the device will operate at an ambien temperature
not more than 25-30oC.

In that scenario the FET should handel an Imax=7.36A without a heat sink.
The max IDS=2.2A - what the bulb will use.

I think the FET will be nice cool.

What you think?

Thanks.
My best regards.
« Last Edit: October 16, 2017, 05:23:20 AM by Chriss »
 

Online Audioguru

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Re: Why IRF510 is more hot than IRF3205?
« Reply #8 on: October 16, 2017, 08:33:02 AM »
The on-resistance of the IRF540 Mosfet is a maximum of 0.044 ohms only when its chip is prevented from heating. The resistance increases as it is heated by current in it. You can re-calculate its resistance and its heating with the bulb current.
 
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