Darlington Transistor vs. MOSFET
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Posts: 46,084
The other day I set up an automotive fuel injector driven with an IRL520
MOSFET to capture some scope traces showing how it handled the voltage
spikes. My trusty Tek 2465B scope did a nice job, and I expect to post the
photos and article sometime this weekend on my website.
With the recent discussion of the TIP120 Darlington, I thought it would be
interesting to compare it with the IRL520 MOSFET. I couldn't in good
conscience use the TIP120 with a fuel injector because it is only rated at
60 volts -- considerably less than the raw injector spikes. So for this
project I replaced the injector with a 10 ohm power resistor (the actual
resistance at operating temperature was 9.8 ohms).
One thing to keep in mind about Darlingtons is that, by their very nature,
there is a substantial voltage drop between the collector and emitter. This
causes more heat dissipation in the device, and results in less power to the
load. In some cases it is acceptable, in others it isn't. It may result in
the need for a heat sink, making it more expensive overall than a different
transistor which wouldn't need a heat sink under the same conditions.
The TIP120's characteristics can be seen by looking at the data sheet,
located here:
http://www.onsemi.com/productSummary/0,4317,TIP120,00.html
You might want to download the PDF data sheet and take a look at the graphs
on page 5. Note that the graphs for the NPN TIP120 are on the left side of
the page -- the graphs on the right are for the complimentary PNP devices...
Notice on Figure 8 (left side graph) the DC current gain at various
collector currents is shown. Each curve is for a different operating
temperature. At room temperature, the TIP120 gain is over 3,000 for
collector current from 1 to 4 amps.
The next graph in Figure 9 is even more revealing, showing how much base
current you need for full saturation (minimum collector-emitter voltage
drop) at different collector currents. Take a look at the curve for Ic
(collector current) of 2.0 amps. Notice that as you look across the
horizontal axis, the collector voltage (drop) decreases as the base current
increases. A base current of about 5 mA or so gets the collector voltage
about as low as possible. So a 1k ohm base resistor (or a little lower)
would be perfect at 2.0 amps. But notice that the collector voltage is
still 1.0 volts when the transistor is fully saturated (on). That means
that it will be turning 2 Watts into heat (power = voltage * current), which
heats the transistor and steals the power from your load. Notice that the
6.0 amp curve needs about 15 mA for full saturation, and there is now a 1.6
volt drop in the device - resulting in 9.6 watts turning to heat in the
transistor instead of powering your load. Better buy a good heat sink!
Here is a comparison between a TIP120 and an inexpensive "logic-level"
IRL520N MOSFET. Both can be easily driven by a Stamp pin. Each device was
operated as a "saturated switch" on the ground side of the load. In other
words, the +13.85 VDC power was applied to one side of the load (a 9.8 ohm
power resistor), and the transistors switched the current between the other
side of the load and ground. Ambient temperature was 71 degrees Fahrenheit.
Temperature measurements were taken with a Fluke 80T-IR infrared temperature
probe, and displayed on a Fluke 87 meter.
Using the TIP120, I connected a 20k ohm pot between the 5V control signal
and the base. Varying the resistance between 20k ohms and 500 ohms, there
was almost no variation in the voltage drop across the transistor (collector
voltage), indicating that it was saturating nicely with only a fraction of a
milliamp of base current. I left it toward the lower end of the resistance
range so the transistor was fully on. After letting the temperatures
stabilize for about 15 minutes, here are the readings...
Collector voltage (drop from collector to emitter) = .774 V
Voltage across load resistor = 13.0 V
Current though load and transistor = 1.33 A
Power delivered to load resistor = 17.3 W
Power turned to heat in the transistor = 1.03 W
TIP120 Temperature = 191 deg F
Temperature rise over ambient = 120 deg F
I then dropped in the IRL520N MOSFET (no wiring changes needed) and got the
following results under identical conditions:
Drain voltage (drop from drain to source) = .224 V (equivalent to collector
voltage, above)
Voltage across load resistor = 13.5 V
Current though load and transistor = 1.38 A
Power delivered to load resistor = 18.6 W
Power turned to heat in the transistor = 0.31 W
IRL520N Temperature = 111 deg F
Temperature rise over ambient = 40 deg F
There are some pretty significant differences.... power lost in the
transistor (turned to heat) went from 1.03 W in the TP120 to 0.31 W in the
MOSFET. As a result the MOSFET ran a lot cooler -- a temperature rise of 40
deg F over ambient, vs. a 120 deg F rise in the TIP120.
The load received more power with the IRL520 - about 8% more - since less of
it was wasted as heat in the transistor. Even more dramatic differences can
be obtained by substituting the lower resistance IRL530N or IRL540N.
I will post a link when I get the scope trace photos on my website with the
MOSFET driving the fuel injector. I ran it for several hours and the MOSFET
handled the voltage spikes from the injector nicely without any other spike
suppression components. Its temperature was only a few degrees above
ambient (although the injector got pretty hot, since the fuel flow normally
helps cool it).
Hope you found this interesting...
Randy
www.glitchbuster.com
MOSFET to capture some scope traces showing how it handled the voltage
spikes. My trusty Tek 2465B scope did a nice job, and I expect to post the
photos and article sometime this weekend on my website.
With the recent discussion of the TIP120 Darlington, I thought it would be
interesting to compare it with the IRL520 MOSFET. I couldn't in good
conscience use the TIP120 with a fuel injector because it is only rated at
60 volts -- considerably less than the raw injector spikes. So for this
project I replaced the injector with a 10 ohm power resistor (the actual
resistance at operating temperature was 9.8 ohms).
One thing to keep in mind about Darlingtons is that, by their very nature,
there is a substantial voltage drop between the collector and emitter. This
causes more heat dissipation in the device, and results in less power to the
load. In some cases it is acceptable, in others it isn't. It may result in
the need for a heat sink, making it more expensive overall than a different
transistor which wouldn't need a heat sink under the same conditions.
The TIP120's characteristics can be seen by looking at the data sheet,
located here:
http://www.onsemi.com/productSummary/0,4317,TIP120,00.html
You might want to download the PDF data sheet and take a look at the graphs
on page 5. Note that the graphs for the NPN TIP120 are on the left side of
the page -- the graphs on the right are for the complimentary PNP devices...
Notice on Figure 8 (left side graph) the DC current gain at various
collector currents is shown. Each curve is for a different operating
temperature. At room temperature, the TIP120 gain is over 3,000 for
collector current from 1 to 4 amps.
The next graph in Figure 9 is even more revealing, showing how much base
current you need for full saturation (minimum collector-emitter voltage
drop) at different collector currents. Take a look at the curve for Ic
(collector current) of 2.0 amps. Notice that as you look across the
horizontal axis, the collector voltage (drop) decreases as the base current
increases. A base current of about 5 mA or so gets the collector voltage
about as low as possible. So a 1k ohm base resistor (or a little lower)
would be perfect at 2.0 amps. But notice that the collector voltage is
still 1.0 volts when the transistor is fully saturated (on). That means
that it will be turning 2 Watts into heat (power = voltage * current), which
heats the transistor and steals the power from your load. Notice that the
6.0 amp curve needs about 15 mA for full saturation, and there is now a 1.6
volt drop in the device - resulting in 9.6 watts turning to heat in the
transistor instead of powering your load. Better buy a good heat sink!
Here is a comparison between a TIP120 and an inexpensive "logic-level"
IRL520N MOSFET. Both can be easily driven by a Stamp pin. Each device was
operated as a "saturated switch" on the ground side of the load. In other
words, the +13.85 VDC power was applied to one side of the load (a 9.8 ohm
power resistor), and the transistors switched the current between the other
side of the load and ground. Ambient temperature was 71 degrees Fahrenheit.
Temperature measurements were taken with a Fluke 80T-IR infrared temperature
probe, and displayed on a Fluke 87 meter.
Using the TIP120, I connected a 20k ohm pot between the 5V control signal
and the base. Varying the resistance between 20k ohms and 500 ohms, there
was almost no variation in the voltage drop across the transistor (collector
voltage), indicating that it was saturating nicely with only a fraction of a
milliamp of base current. I left it toward the lower end of the resistance
range so the transistor was fully on. After letting the temperatures
stabilize for about 15 minutes, here are the readings...
Collector voltage (drop from collector to emitter) = .774 V
Voltage across load resistor = 13.0 V
Current though load and transistor = 1.33 A
Power delivered to load resistor = 17.3 W
Power turned to heat in the transistor = 1.03 W
TIP120 Temperature = 191 deg F
Temperature rise over ambient = 120 deg F
I then dropped in the IRL520N MOSFET (no wiring changes needed) and got the
following results under identical conditions:
Drain voltage (drop from drain to source) = .224 V (equivalent to collector
voltage, above)
Voltage across load resistor = 13.5 V
Current though load and transistor = 1.38 A
Power delivered to load resistor = 18.6 W
Power turned to heat in the transistor = 0.31 W
IRL520N Temperature = 111 deg F
Temperature rise over ambient = 40 deg F
There are some pretty significant differences.... power lost in the
transistor (turned to heat) went from 1.03 W in the TP120 to 0.31 W in the
MOSFET. As a result the MOSFET ran a lot cooler -- a temperature rise of 40
deg F over ambient, vs. a 120 deg F rise in the TIP120.
The load received more power with the IRL520 - about 8% more - since less of
it was wasted as heat in the transistor. Even more dramatic differences can
be obtained by substituting the lower resistance IRL530N or IRL540N.
I will post a link when I get the scope trace photos on my website with the
MOSFET driving the fuel injector. I ran it for several hours and the MOSFET
handled the voltage spikes from the injector nicely without any other spike
suppression components. Its temperature was only a few degrees above
ambient (although the injector got pretty hot, since the fuel flow normally
helps cool it).
Hope you found this interesting...
Randy
www.glitchbuster.com
Comments
Good stuff, well researched and written. Thanks for sharing, I found it very
interesting, and I will probably never make a FIS. But then, never say
never!
Jonathan
www.madlabs.info
Original Message
From: "Randy Jones" <randyjones@w...>
To: <basicstamps@yahoogroups.com>
Sent: Friday, February 21, 2003 10:24 PM
Subject: [noparse][[/noparse]basicstamps] Darlington Transistor vs. MOSFET
> The other day I set up an automotive fuel injector driven with an IRL520
> MOSFET to capture some scope traces showing how it handled the voltage
> spikes. My trusty Tek 2465B scope did a nice job, and I expect to post
the
> photos and article sometime this weekend on my website.
>
> With the recent discussion of the TIP120 Darlington, I thought it would be
> interesting to compare it with the IRL520 MOSFET. I couldn't in good
> conscience use the TIP120 with a fuel injector because it is only rated at
> 60 volts -- considerably less than the raw injector spikes. So for this
> project I replaced the injector with a 10 ohm power resistor (the actual
> resistance at operating temperature was 9.8 ohms).
>
> One thing to keep in mind about Darlingtons is that, by their very nature,
> there is a substantial voltage drop between the collector and emitter.
This
> causes more heat dissipation in the device, and results in less power to
the
> load. In some cases it is acceptable, in others it isn't. It may result
in
> the need for a heat sink, making it more expensive overall than a
different
> transistor which wouldn't need a heat sink under the same conditions.
>
> The TIP120's characteristics can be seen by looking at the data sheet,
> located here:
>
> http://www.onsemi.com/productSummary/0,4317,TIP120,00.html
>
> You might want to download the PDF data sheet and take a look at the
graphs
> on page 5. Note that the graphs for the NPN TIP120 are on the left side
of
> the page -- the graphs on the right are for the complimentary PNP
devices...
>
> Notice on Figure 8 (left side graph) the DC current gain at various
> collector currents is shown. Each curve is for a different operating
> temperature. At room temperature, the TIP120 gain is over 3,000 for
> collector current from 1 to 4 amps.
>
> The next graph in Figure 9 is even more revealing, showing how much base
> current you need for full saturation (minimum collector-emitter voltage
> drop) at different collector currents. Take a look at the curve for Ic
> (collector current) of 2.0 amps. Notice that as you look across the
> horizontal axis, the collector voltage (drop) decreases as the base
current
> increases. A base current of about 5 mA or so gets the collector voltage
> about as low as possible. So a 1k ohm base resistor (or a little lower)
> would be perfect at 2.0 amps. But notice that the collector voltage is
> still 1.0 volts when the transistor is fully saturated (on). That means
> that it will be turning 2 Watts into heat (power = voltage * current),
which
> heats the transistor and steals the power from your load. Notice that the
> 6.0 amp curve needs about 15 mA for full saturation, and there is now a
1.6
> volt drop in the device - resulting in 9.6 watts turning to heat in the
> transistor instead of powering your load. Better buy a good heat sink!
>
> Here is a comparison between a TIP120 and an inexpensive "logic-level"
> IRL520N MOSFET. Both can be easily driven by a Stamp pin. Each device
was
> operated as a "saturated switch" on the ground side of the load. In other
> words, the +13.85 VDC power was applied to one side of the load (a 9.8 ohm
> power resistor), and the transistors switched the current between the
other
> side of the load and ground. Ambient temperature was 71 degrees
Fahrenheit.
> Temperature measurements were taken with a Fluke 80T-IR infrared
temperature
> probe, and displayed on a Fluke 87 meter.
>
> Using the TIP120, I connected a 20k ohm pot between the 5V control signal
> and the base. Varying the resistance between 20k ohms and 500 ohms, there
> was almost no variation in the voltage drop across the transistor
(collector
> voltage), indicating that it was saturating nicely with only a fraction of
a
> milliamp of base current. I left it toward the lower end of the
resistance
> range so the transistor was fully on. After letting the temperatures
> stabilize for about 15 minutes, here are the readings...
>
> Collector voltage (drop from collector to emitter) = .774 V
> Voltage across load resistor = 13.0 V
> Current though load and transistor = 1.33 A
> Power delivered to load resistor = 17.3 W
> Power turned to heat in the transistor = 1.03 W
> TIP120 Temperature = 191 deg F
> Temperature rise over ambient = 120 deg F
>
> I then dropped in the IRL520N MOSFET (no wiring changes needed) and got
the
> following results under identical conditions:
>
> Drain voltage (drop from drain to source) = .224 V (equivalent to
collector
> voltage, above)
> Voltage across load resistor = 13.5 V
> Current though load and transistor = 1.38 A
> Power delivered to load resistor = 18.6 W
> Power turned to heat in the transistor = 0.31 W
> IRL520N Temperature = 111 deg F
> Temperature rise over ambient = 40 deg F
>
> There are some pretty significant differences.... power lost in the
> transistor (turned to heat) went from 1.03 W in the TP120 to 0.31 W in the
> MOSFET. As a result the MOSFET ran a lot cooler -- a temperature rise of
40
> deg F over ambient, vs. a 120 deg F rise in the TIP120.
>
> The load received more power with the IRL520 - about 8% more - since less
of
> it was wasted as heat in the transistor. Even more dramatic differences
can
> be obtained by substituting the lower resistance IRL530N or IRL540N.
>
> I will post a link when I get the scope trace photos on my website with
the
> MOSFET driving the fuel injector. I ran it for several hours and the
MOSFET
> handled the voltage spikes from the injector nicely without any other
spike
> suppression components. Its temperature was only a few degrees above
> ambient (although the injector got pretty hot, since the fuel flow
normally
> helps cool it).
>
> Hope you found this interesting...
>
> Randy
> www.glitchbuster.com
>
>
>
>
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the Subject and
Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
Thanks
Bill
--- In basicstamps@yahoogroups.com, "Jonathan Peakall"
<jpeakall@m...> wrote:
> Randy,
>
> Good stuff, well researched and written. Thanks for sharing, I
found it very
> interesting, and I will probably never make a FIS. But then, never
say
> never!
>
> Jonathan
>
> www.madlabs.info
>
>
Original Message
> From: "Randy Jones" <randyjones@w...>
> To: <basicstamps@yahoogroups.com>
> Sent: Friday, February 21, 2003 10:24 PM
> Subject: [noparse][[/noparse]basicstamps] Darlington Transistor vs. MOSFET
>
>
> > The other day I set up an automotive fuel injector driven with
an IRL520
> > MOSFET to capture some scope traces showing how it handled the
voltage
> > spikes. My trusty Tek 2465B scope did a nice job, and I expect
to post
> the
> > photos and article sometime this weekend on my website.
> >
> > With the recent discussion of the TIP120 Darlington, I thought
it would be
> > interesting to compare it with the IRL520 MOSFET. I couldn't in
good
> > conscience use the TIP120 with a fuel injector because it is
only rated at
> > 60 volts -- considerably less than the raw injector spikes. So
for this
> > project I replaced the injector with a 10 ohm power resistor
(the actual
> > resistance at operating temperature was 9.8 ohms).
> >
> > One thing to keep in mind about Darlingtons is that, by their
very nature,
> > there is a substantial voltage drop between the collector and
emitter.
> This
> > causes more heat dissipation in the device, and results in less
power to
> the
> > load. In some cases it is acceptable, in others it isn't. It
may result
> in
> > the need for a heat sink, making it more expensive overall than a
> different
> > transistor which wouldn't need a heat sink under the same
conditions.
> >
> > The TIP120's characteristics can be seen by looking at the data
sheet,
> > located here:
> >
> > http://www.onsemi.com/productSummary/0,4317,TIP120,00.html
> >
> > You might want to download the PDF data sheet and take a look at
the
> graphs
> > on page 5. Note that the graphs for the NPN TIP120 are on the
left side
> of
> > the page -- the graphs on the right are for the complimentary PNP
> devices...
> >
> > Notice on Figure 8 (left side graph) the DC current gain at
various
> > collector currents is shown. Each curve is for a different
operating
> > temperature. At room temperature, the TIP120 gain is over 3,000
for
> > collector current from 1 to 4 amps.
> >
> > The next graph in Figure 9 is even more revealing, showing how
much base
> > current you need for full saturation (minimum collector-emitter
voltage
> > drop) at different collector currents. Take a look at the curve
for Ic
> > (collector current) of 2.0 amps. Notice that as you look across
the
> > horizontal axis, the collector voltage (drop) decreases as the
base
> current
> > increases. A base current of about 5 mA or so gets the
collector voltage
> > about as low as possible. So a 1k ohm base resistor (or a
little lower)
> > would be perfect at 2.0 amps. But notice that the collector
voltage is
> > still 1.0 volts when the transistor is fully saturated (on).
That means
> > that it will be turning 2 Watts into heat (power = voltage *
current),
> which
> > heats the transistor and steals the power from your load.
Notice that the
> > 6.0 amp curve needs about 15 mA for full saturation, and there
is now a
> 1.6
> > volt drop in the device - resulting in 9.6 watts turning to heat
in the
> > transistor instead of powering your load. Better buy a good
heat sink!
> >
> > Here is a comparison between a TIP120 and an inexpensive "logic-
level"
> > IRL520N MOSFET. Both can be easily driven by a Stamp pin. Each
device
> was
> > operated as a "saturated switch" on the ground side of the
load. In other
> > words, the +13.85 VDC power was applied to one side of the load
(a 9.8 ohm
> > power resistor), and the transistors switched the current
between the
> other
> > side of the load and ground. Ambient temperature was 71 degrees
> Fahrenheit.
> > Temperature measurements were taken with a Fluke 80T-IR infrared
> temperature
> > probe, and displayed on a Fluke 87 meter.
> >
> > Using the TIP120, I connected a 20k ohm pot between the 5V
control signal
> > and the base. Varying the resistance between 20k ohms and 500
ohms, there
> > was almost no variation in the voltage drop across the transistor
> (collector
> > voltage), indicating that it was saturating nicely with only a
fraction of
> a
> > milliamp of base current. I left it toward the lower end of the
> resistance
> > range so the transistor was fully on. After letting the
temperatures
> > stabilize for about 15 minutes, here are the readings...
> >
> > Collector voltage (drop from collector to emitter) = .774 V
> > Voltage across load resistor = 13.0 V
> > Current though load and transistor = 1.33 A
> > Power delivered to load resistor = 17.3 W
> > Power turned to heat in the transistor = 1.03 W
> > TIP120 Temperature = 191 deg F
> > Temperature rise over ambient = 120 deg F
> >
> > I then dropped in the IRL520N MOSFET (no wiring changes needed)
and got
> the
> > following results under identical conditions:
> >
> > Drain voltage (drop from drain to source) = .224 V (equivalent to
> collector
> > voltage, above)
> > Voltage across load resistor = 13.5 V
> > Current though load and transistor = 1.38 A
> > Power delivered to load resistor = 18.6 W
> > Power turned to heat in the transistor = 0.31 W
> > IRL520N Temperature = 111 deg F
> > Temperature rise over ambient = 40 deg F
> >
> > There are some pretty significant differences.... power lost in
the
> > transistor (turned to heat) went from 1.03 W in the TP120 to
0.31 W in the
> > MOSFET. As a result the MOSFET ran a lot cooler -- a
temperature rise of
> 40
> > deg F over ambient, vs. a 120 deg F rise in the TIP120.
> >
> > The load received more power with the IRL520 - about 8% more -
since less
> of
> > it was wasted as heat in the transistor. Even more dramatic
differences
> can
> > be obtained by substituting the lower resistance IRL530N or
IRL540N.
> >
> > I will post a link when I get the scope trace photos on my
website with
> the
> > MOSFET driving the fuel injector. I ran it for several hours
and the
> MOSFET
> > handled the voltage spikes from the injector nicely without any
other
> spike
> > suppression components. Its temperature was only a few degrees
above
> > ambient (although the injector got pretty hot, since the fuel
flow
> normally
> > helps cool it).
> >
> > Hope you found this interesting...
> >
> > Randy
> > www.glitchbuster.com
> >
> >
> >
> >
> >
> >
> > To UNSUBSCRIBE, just send mail to:
> > basicstamps-unsubscribe@yahoogroups.com
> > from the same email address that you subscribed. Text in the
Subject and
> Body of the message will be ignored.
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
> >
> >
randyjones@w... writes:
> The next graph in Figure 9 is even more revealing, showing how much base
> current you need for full saturation (minimum collector-emitter voltage
> drop) at different collector currents. Take a look at the curve for Ic
> (collector current) of 2.0 amps.
Hi Randy,
Good information.....
I completely understand everything you wrote, One point though. In response
to the original post regarding the use of a TIP120, that user indicated
he/she needed to switch a 2 amp relay.
I am relatively sure the 2 amps that person was talking about was the switch
side of the relay and not the coil side, and that would be the basis of
his/her concern for the value of the base resistor. As I am sure you are
aware, if a relay requires 2 amps to be acitaved, there is something wrong
with that relay.
None the less, your information on Darlington vs Mosfet is still acurate,
applicaple and thorough!!!!!!!!!
Good work!
Regards,
Ken
[noparse][[/noparse]Non-text portions of this message have been removed]
billak@f... writes:
> Sounds really good. Where can I get these MOSFETs?
>
> Thanks
>
> Bill
>
<A HREF="www.glitchbuster.com">www.glitchbuster.com</A>
[noparse][[/noparse]Non-text portions of this message have been removed]
>Good information.....
>I completely understand everything you wrote, One point though. In response
>to the original post regarding the use of a TIP120, that user indicated
>he/she needed to switch a 2 amp relay.
>
>I am relatively sure the 2 amps that person was talking about was the switch
>side of the relay and not the coil side, and that would be the basis of
>his/her concern for the value of the base resistor. As I am sure you are
>aware, if a relay requires 2 amps to be acitaved, there is something wrong
>with that relay.
You simply can't make that assumption.
Grab a horn relay from an American-made old clunker in the junkyard. The
kind of horn relay I am talking about is in a metal can with 3 terminals:
Batt, Horn, Switch. Measure the coil resistance. Awfully low, isn't
it. Now hook it up to a 12V battery and measure the coil current. Darn -
its over 2 Amps. Even more darn - the relay is getting too hot to hold.
Now try the same thing for the starter solenoid from the same clunker. You
might find a coil current in excess of 10 Amps.
The original poster did not say where the relay was from or even what kind
of relay it was. An awful lot of stuff is built for intermittent use for
cost reasons. One of the ways of saving cost in relays and starter
solenoids is to reduce the size and number of turns of wire used to make
the actuator coils.
dwayne
--
Dwayne Reid <dwayner@p...>
Trinity Electronics Systems Ltd Edmonton, AB, CANADA
(780) 489-3199 voice (780) 487-6397 fax
Celebrating 19 years of Engineering Innovation (1984 - 2003)
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Do NOT send unsolicited commercial email to this email address.
This message neither grants consent to receive unsolicited
commercial email nor is intended to solicit commercial email.
I was trying to simplify my question by using a 2 amp relay. But to be
totally honest its a Push Tubular Solenoid with a resistance of 5.8 ohms.
FYI I'm using it to depress a button that needs about 500 grams of force at
1/8 inch, so this rube goldberg affair. Any advice on a better way then a
Push Tubular Solenoid would be great as the solenoid does take a lot of
power and heats up when cycled a dozen times or so.
I haven't any experiece with muscel wires or electric pistons, could these
be a option ??
Bob Phillips