help reading a datasheet

L

lerameur

Guest
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
 
lerameur wrote:
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?
Click to expand...
Each of those specs measures something different. The spec
that is most important to you, depends on the application.

Rds is the resistance of the fully turned on channel at some
specified gate to source voltage (which can be different for
different devices). To a first approximation, the lower the
Rds, the lower the DC resistive heating the device will have
for a given drain current.

The power rating tells you how much heat can be pulled out
of the die into a heat sink that holds the cast at a given
temperature. This is mostly about die area and package
thermal resistance. If your application is a highly
efficient one (that produces little heat in the die) this
spec is not very important to you. If you will use the
device as a load to test power supplies (intentionally dump
lots of power into the die) this this may be the most
important spec. The data sheet writers try to cover any
possible application.

The maximum amps and volts spec also apply to high power
applications, not high efficiency ones, since low loss
applications will not apply such large voltages across the
die while it also passes large current.

For more specifics, please post links to the data sheets, so
we are looking at the same data. The devil is often in the
details.

--
Regards,

John Popelish
 
lerameur wrote:
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
Click to expand...
Think basic physics. E = IR, P = IE.

Higher Rds(on) means that for a given current the device will burn up
more power (I'll leave it up to you to figure out the relationship
between current (I), resistance (R) and power (P) -- you can derive it
from the above two relations). I suspect that those power figures are
just the amount of dissipation you can expect at 31 amps, not any
recommended operating range.

Each device has a certain thermal resistance to its case, and a certain
maximum junction temperature that it can stand. It's your job to figure
out how much power you can dump from the device (which may require that
you design in heat sinks to conduct heat away from the device case; such
design isn't trivial), then use that to figure out what Rds(on) you need
to look for to meet your other requirements.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
 
On Tue, 26 Aug 2008 10:22:01 -0700 (PDT), lerameur
<lerameur@yahoo.com> wrote:

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
Click to expand...
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
 
On Aug 26, 4:11 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 26 Aug 2008 10:22:01 -0700 (PDT), lerameur



leram...@yahoo.com> wrote:
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

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
Click to expand...
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
 
On Tue, 26 Aug 2008 13:27:14 -0700 (PDT), lerameur
<lerameur@yahoo.com> wrote:

On Aug 26, 4:11 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 26 Aug 2008 10:22:01 -0700 (PDT), lerameur



leram...@yahoo.com> wrote:
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

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.
Click to expand...
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.

Therefore I should be using a small Rds value.
Click to expand...
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.


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.
Click to expand...
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
 
lerameur wrote:
On Aug 26, 4:11 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 26 Aug 2008 10:22:01 -0700 (PDT), lerameur



leram...@yahoo.com> wrote:
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
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
All else being equal, yes (with due deference to John Larkin's "low
Click to expand...
aggregate Rds(on) through heatsinking").

But all else is rarely equal. Lower Rds(on) means higher gate
capacitance which is a pain when you're switching fast, or a more
expensive part which is a pain when you're trying to shave pennies out
of a product, or a harder to second-source part, which is a pain when
your vendor decides they want to discontinue the part or sells 100000 of
them to Ford without bothering to tell the factory that they did it, or,
or, or etc.

So I wouldn't run out and buy the World's Lowest Rds(on) without
thinking hard about the other compromises.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
 
On Aug 26, 2:27 pm, lerameur <leram...@yahoo.com> wrote:
as I read above, the power column displays how well the MOsfter can
dissipate heat. if I am wrong tell me.
- Show quoted text -
Click to expand...
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.
 
lerameur wrote:

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).
Click to expand...
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?

--
Regards,

John Popelish
 
On Aug 26, 6:31 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 26 Aug 2008 13:27:14 -0700 (PDT), lerameur



leram...@yahoo.com> wrote:
On Aug 26, 4:11 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 26 Aug 2008 10:22:01 -0700 (PDT), lerameur

leram...@yahoo.com> wrote:
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

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.

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.

Therefore I should be using a small Rds value.

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.

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.

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
Click to expand...
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
 
On Aug 26, 6:31 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 26 Aug 2008 13:27:14 -0700 (PDT), lerameur



leram...@yahoo.com> wrote:
On Aug 26, 4:11 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 26 Aug 2008 10:22:01 -0700 (PDT), lerameur

leram...@yahoo.com> wrote:
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

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.

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.

Therefore I should be using a small Rds value.

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.

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.

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
Click to expand...
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
 
lerameur wrote:

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.
Click to expand...
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?

--
Regards,

John Popelish
 
lerameur wrote:
On Aug 27, 1:31 pm, John Popelish <jpopel...@rica.net> wrote:
lerameurwrote:
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#523924441...
you may need to download the jpg to see it better
From the simulation , it gives me the same voltage in drain and
source.
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.
Click to expand...
So... not the schematic shown in the simulator. Please post
the actual schematic that produces the hot transistor.

--
Regards,

John Popelish
 
On Aug 27, 1:24 am, John Popelish <jpopel...@rica.net> wrote:
lerameurwrote:
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).

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?

--
Regards,

John Popelish
Click to expand...
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
 
On Aug 27, 1:31 pm, John Popelish <jpopel...@rica.net> wrote:
lerameurwrote:
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#523924441...
you may need to download the jpg to see it better
From the simulation , it gives me the same voltage in drain and
source.

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?

--
Regards,

John Popelish
Click to expand...
I have two optocoupler driving from a a 6v battery, the other
batteries are 12v.
 
On Aug 27, 1:43 pm, John Popelish <jpopel...@rica.net> wrote:
lerameurwrote:
On Aug 27, 1:31 pm, John Popelish <jpopel...@rica.net> wrote:
lerameurwrote:
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#523924441....
you may need to download the jpg to see it better
From the simulation , it gives me the same voltage in drain and
source.
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.

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

--
Regards,

John Popelish
Click to expand...
I will do this tonight when I get back at home
thanks

k
 
lerameur wrote:

(snip)
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.
Click to expand...
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 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
Click to expand...
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.

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...
Click to expand...
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).

--
Regards,

John Popelish
 
On Aug 27, 1:43 pm, John Popelish <jpopel...@rica.net> wrote:
lerameurwrote:
On Aug 27, 1:31 pm, John Popelish <jpopel...@rica.net> wrote:
lerameurwrote:
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#523924441....
you may need to download the jpg to see it better
From the simulation , it gives me the same voltage in drain and
source.
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.

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

--
Regards,

John Popelish
Click to expand...
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
 
On Aug 28, 11:25 am, John Popelish <jpopel...@rica.net> wrote:
lerameurwrote:

(snip)

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.

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 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#523955997...

Full circuit:
http://picasaweb.google.com/Kronmaster/ElectronicMisc/photo#523955991...

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.

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...

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).

--
Regards,

John Popelish
Click to expand...
I can post the same circuit without the voltage and current showing
will that help?

K
 
On Aug 28, 1:16 pm, lerameur <leram...@yahoo.com> wrote:
On Aug 28, 11:25 am, John Popelish <jpopel...@rica.net> wrote:



lerameurwrote:

(snip)

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.

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 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#523955997....

Full circuit:
http://picasaweb.google.com/Kronmaster/ElectronicMisc/photo#523955991....

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.

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...

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).

--
Regards,

John Popelish

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

K
Click to expand...
Ok I added it anyway:
http://picasaweb.google.com/Kronmaster/ElectronicMisc/photo#523962020...

K
 
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