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help reading a datasheet

L

lerameur

Jan 1, 1970
0
Hi,

I am looking at digikey web site the IRFZ44 TO-220AB, there are many
chips available.
I am a bit confuse concerning the resistance Vs Watt coloumn.
I Will take two randomly
the IRFZ44VZ-ND WITH 12mOhm Rds and 92W power @31 amps 10v
and
the IRFZ44R-ND WITH 28mOhm Rds and 150W power @31 amps 10v
I guess the more the resistance the more power...?
does that mean more power it is able to dissipate.
I would suppose I should get the 28mOhm one because there is less
loss because less resistance.. is this the way to think? or does
these data means something else?
Does the IRFZ44VZ handles the current more efficiently?

thanks

K
 
L

lerameur

Jan 1, 1970
0
IR is notorious for spec'ing absurd power dissipations and currents on
their fets. 150 watts in a TO-220 is borderline crazy, and would need
heroic heat sinking. Derate their power specs by at least 2:1.

But 31 amps into a 28 mOhm switch is only 27 watts, OK with a good
heat sink. 12 mOhms is only 12 watts, even better.

John

wouldn't the least the resistance value be the better option in any
cases?
The way I see it, the more resistance Rds, the more power it will
loose through heat sinking.
E = IR, P = IE.
so by doubling R, (P=I * E ) the power onto the heat sink will be 4x
as much.
Therefore I should be using a small Rds value.
as I read above, the power column displays how well the MOsfter can
dissipate heat. if I am wrong tell me.
I intend to use a lot of current, therefore huge heatsinking. I never
bothered with these values, but now they are critical and it affect
design.

K
 
as I read above, the power column displays how well the MOsfter can
dissipate heat. if I am wrong tell me.
- Show quoted text -

You are wrong, the power column tells you what to expect the
transistor to generate in wasted power as heat. A lower number here
is desirable unless other aspects are more important to the overall
design.
 
L

lerameur

Jan 1, 1970
0
If you're switching some given load, I is probably pretty constant, so
power dissipation in the transistor is linear on Rds-on, not squared.

P = I^2 * Rds, where I is probably constant.


But yes, the lower Rds-on, the lower the heat lost in the fet. Make
sure you apply enough gate voltage to turn it on good.


OK, but it's better to buy more silicon (lower resistance fets, even
several in parallel) than to use one and a huge heat sink. Heat sinks
are big and expensive, transistors are small and cheap.

And it's creepy to run currents like 30 amps through one of those
skinny TO-220 leads. More fets is better.

What's your actual load current?

John

HI,

well I am running about 1.7 amps (eventually at lot more) I am using a
IRFZ44 and its heating up like crazy.. and melts.
The gate runs at 22v while the drain and source are at 10v. This last
info is taken from Pspice. I have measured 1.7 in my circuit. I would
think there is something wrong with my circuit, its only 1.7 amps (no
heatsink).

K
 
L

lerameur

Jan 1, 1970
0
If you're switching some given load, I is probably pretty constant, so
power dissipation in the transistor is linear on Rds-on, not squared.

P = I^2 * Rds, where I is probably constant.


But yes, the lower Rds-on, the lower the heat lost in the fet. Make
sure you apply enough gate voltage to turn it on good.


OK, but it's better to buy more silicon (lower resistance fets, even
several in parallel) than to use one and a huge heat sink. Heat sinks
are big and expensive, transistors are small and cheap.

And it's creepy to run currents like 30 amps through one of those
skinny TO-220 leads. More fets is better.

What's your actual load current?

John

HI,

well I am running about 1.7 amps (eventually at lot more) I am using a
IRFZ44 and its heating up like crazy.. and melts.
The gate runs at 22v while the drain and source are at 10v. This last
info is taken from Pspice. I have measured 1.7 in my circuit. I would
think there is something wrong with my circuit, its only 1.7 amps (no
heatsink).

K
 
P

Phil Allison

Jan 1, 1970
0
lerameur


You are wrong, the power column tells you what to expect the
transistor to generate in wasted power as heat.

** Absolute nonsense.

The Pd figure tells you the LIMIT on how much heat can be dissipated by
the device under ideal heatsinking conditions and at a stated case temp.

A lower number here
is desirable unless other aspects are more important to the overall
design.

** The reverse is true.

A higher Pd figure indicates that in a given application and with other
parameters the same, the chip will remain cooler.

In the OP's example, the "R" device has a much lower thermal resistance
from chip to case then the "VZ" device - ie 1.0 degrees C per watt as
against 1.6.

Turns a 90 watt rating into 150.


....... Phil
 

neon

Oct 21, 2006
1,325
Joined
Oct 21, 2006
Messages
1,325
lerameur said:
Hi,

I am looking at digikey web site the IRFZ44 TO-220AB, there are many
chips available.
I am a bit confuse concerning the resistance Vs Watt coloumn.
I Will take two randomly
the IRFZ44VZ-ND WITH 12mOhm Rds and 92W power @31 amps 10v
and
the IRFZ44R-ND WITH 28mOhm Rds and 150W power @31 amps 10v
I guess the more the resistance the more power...?
does that mean more power it is able to dissipate.
I would suppose I should get the 28mOhm one because there is less
loss because less resistance.. is this the way to think? or does
these data means something else?
Does the IRFZ44VZ handles the current more efficiently?

thanks

K
I did not look at the spec data but let me say this power dissipation of a device is stricly related to packaging or how fast can the heat be removed from the device.
 
L

lerameur

Jan 1, 1970
0
lerameurwrote:

Something is wrong.  Under those conditions (zero volts drop
drain to source (since both are at about 10V) there should
be no heat at all.  If you mean that the drain and source
are close to 10 volts, then the heat should be low.  For
instance, if the source is at 9.9 and the drain is at 10
volts, then the device should be producing 0.1V*1.7A=0.17
watts of heat.  you might notice that as a slight rise above
stone cold, but not much more.  No heat sink should be
needed for this condition.

What voltage drop do you measure across the actual MOSFET?

Hi,

I posted a snapshot of my circuit (wellpart of it and the outcome)
I am putting two batteries in series:
http://picasaweb.google.com/Kronmaster/ElectronicMisc/photo#5239244417875908290
you may need to download the jpg to see it better
From the simulation , it gives me the same voltage in drain and
source.

K
 
L

lerameur

Jan 1, 1970
0
lerameurwrote:

Yes, that simulation shows almost zero volts drop across the
device, so should produce almost zero heat.

Do you really have a separate floating battery in your
physical circuit just to drive the gate, like the simulation
shows?

I have two optocoupler driving from a a 6v battery, the other
batteries are 12v.
 
L

lerameur

Jan 1, 1970
0
lerameurwrote:

So... not the schematic shown in the simulator.  Please post
the actual schematic that produces the hot transistor.

I will do this tonight when I get back at home
thanks

k
 
L

lerameur

Jan 1, 1970
0
lerameurwrote:

So... not the schematic shown in the simulator.  Please post
the actual schematic that produces the hot transistor.

John, I implemented the simulator schematic, do you want to see the
full version of that?
I have two circuits, using 3 batteries (A, B and C). from these
circuits I am using mosfets to create a 24v (battery A, B) section
charging into the 12v section(battery C) , or battery, once full I
switch the circuit around 24v is now battery A and C.
I added three screen shots, one odd thing, when i simulate the
separate circuit I get 1.2 amps, but when I add the two circuits, I
get 2.2 amps.
step 2 circuit (step1 is just the same but reverse.:
http://picasaweb.google.com/Kronmaster/ElectronicMisc/photo#5239559975908529826

Full circuit:
http://picasaweb.google.com/Kronmaster/ElectronicMisc/photo#5239559918179568482

I decided to parallel my mosfet last night, I was missing one mosfet
to finish all the pairing. when I turned on the switch that single
mosfet blew up like I never seen before, half my breadboard is now
black...

K
 
L

lerameur

Jan 1, 1970
0
lerameurwrote:

(snip)


On the step 2 circuit I see 12 volt batteries called V1, V10
and V14.  On the full schematic I see V1 labeled A that
looks like V1 on step2, V5 labeled B that looks like V14 on
step2, but V7 labeled C is 6 volts, not like the 12 volt V10
on step 2.

Why is C a 6 volt battery?

Battery C is turned the wrong way round to be in series with
battery A (both positive ends connected).

You have no current limiting resistors in the LED (input)
side of your opto couplers.  If you hook them up this way
they will be destroyed very quickly.  If you will drive
these from something that can hold almost 5 volts while
delivering say, 10 mA, then the resistor in series should be
about 330 ohms.



I'll study these a bit, but they are an awful tangle.  I
would not be surprised if you have trouble following these
as you build the actual circuit.  I think the time you took
to better organize these drawings would pay off in fewer
blasted parts caused by mis-connections.


Pretty obviously, your mosfets formed a short circuit across
a battery.  The weak link acted as the fuse (which you
should add to the circuit).

I can post the same circuit without the voltage and current showing
will that help?

K
 
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