Mixing
Controlling Solenoid Valves with ULN2003A, Inrush Current, Arduino
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I want to control a total of five solenoid valves with my Arduino. The solenoid walves operate at 12 volts and have a power consumption of 5 watts per solenoid valve, which equals a current of roughly 420 milliamps.
My idea is to use a ULN2003A transistor array to do this. The ULN2003A datasheet says it supports voltages of up to 50 volts and a maximum current of 500 milliamps per output.
My question is: Is the ULN2003A with its 500 milliamps maximum current rating per output sufficient to reliably switch the solenoid valves? I'm asking because solenoid valves are inductive loads and if I'm not mistaken, they are subject to inrush current that far exceeds the normal current.
arduino
asked 5 hours ago
Chris1309
61
New contributor
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
I want to control a total of five solenoid valves with my Arduino. The solenoid walves operate at 12 volts and have a power consumption of 5 watts per solenoid valve, which equals a current of roughly 420 milliamps.
My idea is to use a ULN2003A transistor array to do this. The ULN2003A datasheet says it supports voltages of up to 50 volts and a maximum current of 500 milliamps per output.
My question is: Is the ULN2003A with its 500 milliamps maximum current rating per output sufficient to reliably switch the solenoid valves? I'm asking because solenoid valves are inductive loads and if I'm not mistaken, they are subject to inrush current that far exceeds the normal current.
arduino
asked 5 hours ago
Chris1309
61
New contributor
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to control a total of five solenoid valves with my Arduino. The solenoid walves operate at 12 volts and have a power consumption of 5 watts per solenoid valve, which equals a current of roughly 420 milliamps.
My idea is to use a ULN2003A transistor array to do this. The ULN2003A datasheet says it supports voltages of up to 50 volts and a maximum current of 500 milliamps per output.
My question is: Is the ULN2003A with its 500 milliamps maximum current rating per output sufficient to reliably switch the solenoid valves? I'm asking because solenoid valves are inductive loads and if I'm not mistaken, they are subject to inrush current that far exceeds the normal current.
arduino
asked 5 hours ago
Chris1309
61
New contributor
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to control a total of five solenoid valves with my Arduino. The solenoid walves operate at 12 volts and have a power consumption of 5 watts per solenoid valve, which equals a current of roughly 420 milliamps.
My idea is to use a ULN2003A transistor array to do this. The ULN2003A datasheet says it supports voltages of up to 50 volts and a maximum current of 500 milliamps per output.
My question is: Is the ULN2003A with its 500 milliamps maximum current rating per output sufficient to reliably switch the solenoid valves? I'm asking because solenoid valves are inductive loads and if I'm not mistaken, they are subject to inrush current that far exceeds the normal current.
arduino
arduino
asked 5 hours ago
Chris1309
61
New contributor
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Chris1309
61
New contributor
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Chris1309
61
New contributor
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Chris1309
61
asked 5 hours ago
Chris1309
61
61
New contributor
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Chris1309 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
You need to read the data sheet carefully- the maximum current per output can be closer to 50mA than 500mA, depending on the conditions. See figures 4 and 5 in the datasheet.
Surge is not an issue with DC solenoids, but you do need to connect the COM to the supply voltage.
For such high current I would recommend discrete MOSFETs (logic level) and diodes across the solenoids. For example, AO3400A and 1N40005/M5.
answered 1 hour ago
Spehro Pefhany
196k4139389
add a comment |Â
up vote
1
down vote
The current won't surge as you expect. Because the solenoids are intended for constant application of the rated 12VDC, they are built
with enough internal resistance to limit the current; the rated
current is the largest that (with 12V) they will ever have.
In addition to inductance (which can be subject to nonlinear
effects), the solenoid has internal resistance, by design.
Solenoids for AC application with less internal resistance,
would also limit current according to AC frequency and
their inductance.
As a practical matter, the ULN2003A also has thermal limits and (at the
ground pin) a total-of-all-outputs current that cannot exceed
2.5A; at 420 mA per section, six solenoids will be too much
to drive simultaneously.
answered 2 hours ago
Whit3rd
4,3851019
add a comment |Â
up vote
1
down vote
A few things on solenoids (either relays or valves), since you are using a ULN2003 I'll assume you are going to drive it with a voltage step:
- When apply a voltage step, the current will rise until the electro-magnetic force exceeds static friction.
- At this point the plunger (for a valve) or the armature (for a relay) will start moving and create back-EMF which will reducing the current as movement accelerates.
- Once the movement ends (usually abruptly), back-EMF goes to zero and the current will raise to stall value.
- Usually at this point the voltage is reduced (by PWM) to ~40% of initial value as less force is needed to maintain position as opposed to initiate motion (keep in mind that force is linked to current, not voltage).
Datasheets often specs both values: boost (initiate movement) and hold (maintain position).
A good rule of thumb is that hold is ~40% of boost.
By default, I'd assume the power is given in static (where the valve will be most of the time) so the 420mA would be the hold current then you'd need overhead to be able to activate it properly (~800mA).
answered 48 mins ago
mcv
716
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You need to read the data sheet carefully- the maximum current per output can be closer to 50mA than 500mA, depending on the conditions. See figures 4 and 5 in the datasheet.
Surge is not an issue with DC solenoids, but you do need to connect the COM to the supply voltage.
For such high current I would recommend discrete MOSFETs (logic level) and diodes across the solenoids. For example, AO3400A and 1N40005/M5.
answered 1 hour ago
Spehro Pefhany
196k4139389
add a comment |Â
up vote
2
down vote
You need to read the data sheet carefully- the maximum current per output can be closer to 50mA than 500mA, depending on the conditions. See figures 4 and 5 in the datasheet.
Surge is not an issue with DC solenoids, but you do need to connect the COM to the supply voltage.
For such high current I would recommend discrete MOSFETs (logic level) and diodes across the solenoids. For example, AO3400A and 1N40005/M5.
answered 1 hour ago
Spehro Pefhany
196k4139389
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You need to read the data sheet carefully- the maximum current per output can be closer to 50mA than 500mA, depending on the conditions. See figures 4 and 5 in the datasheet.
Surge is not an issue with DC solenoids, but you do need to connect the COM to the supply voltage.
For such high current I would recommend discrete MOSFETs (logic level) and diodes across the solenoids. For example, AO3400A and 1N40005/M5.
answered 1 hour ago
Spehro Pefhany
196k4139389
You need to read the data sheet carefully- the maximum current per output can be closer to 50mA than 500mA, depending on the conditions. See figures 4 and 5 in the datasheet.
Surge is not an issue with DC solenoids, but you do need to connect the COM to the supply voltage.
For such high current I would recommend discrete MOSFETs (logic level) and diodes across the solenoids. For example, AO3400A and 1N40005/M5.
answered 1 hour ago
Spehro Pefhany
196k4139389
edited 1 hour ago
answered 1 hour ago
Spehro Pefhany
196k4139389
answered 1 hour ago
Spehro Pefhany
196k4139389
answered 1 hour ago
Spehro Pefhany
196k4139389
196k4139389
add a comment |Â
add a comment |Â
up vote
1
down vote
The current won't surge as you expect. Because the solenoids are intended for constant application of the rated 12VDC, they are built
with enough internal resistance to limit the current; the rated
current is the largest that (with 12V) they will ever have.
In addition to inductance (which can be subject to nonlinear
effects), the solenoid has internal resistance, by design.
Solenoids for AC application with less internal resistance,
would also limit current according to AC frequency and
their inductance.
As a practical matter, the ULN2003A also has thermal limits and (at the
ground pin) a total-of-all-outputs current that cannot exceed
2.5A; at 420 mA per section, six solenoids will be too much
to drive simultaneously.
answered 2 hours ago
Whit3rd
4,3851019
add a comment |Â
up vote
1
down vote
The current won't surge as you expect. Because the solenoids are intended for constant application of the rated 12VDC, they are built
with enough internal resistance to limit the current; the rated
current is the largest that (with 12V) they will ever have.
In addition to inductance (which can be subject to nonlinear
effects), the solenoid has internal resistance, by design.
Solenoids for AC application with less internal resistance,
would also limit current according to AC frequency and
their inductance.
As a practical matter, the ULN2003A also has thermal limits and (at the
ground pin) a total-of-all-outputs current that cannot exceed
2.5A; at 420 mA per section, six solenoids will be too much
to drive simultaneously.
answered 2 hours ago
Whit3rd
4,3851019
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The current won't surge as you expect. Because the solenoids are intended for constant application of the rated 12VDC, they are built
with enough internal resistance to limit the current; the rated
current is the largest that (with 12V) they will ever have.
In addition to inductance (which can be subject to nonlinear
effects), the solenoid has internal resistance, by design.
Solenoids for AC application with less internal resistance,
would also limit current according to AC frequency and
their inductance.
As a practical matter, the ULN2003A also has thermal limits and (at the
ground pin) a total-of-all-outputs current that cannot exceed
2.5A; at 420 mA per section, six solenoids will be too much
to drive simultaneously.
answered 2 hours ago
Whit3rd
4,3851019
The current won't surge as you expect. Because the solenoids are intended for constant application of the rated 12VDC, they are built
with enough internal resistance to limit the current; the rated
current is the largest that (with 12V) they will ever have.
In addition to inductance (which can be subject to nonlinear
effects), the solenoid has internal resistance, by design.
Solenoids for AC application with less internal resistance,
would also limit current according to AC frequency and
their inductance.
As a practical matter, the ULN2003A also has thermal limits and (at the
ground pin) a total-of-all-outputs current that cannot exceed
2.5A; at 420 mA per section, six solenoids will be too much
to drive simultaneously.
answered 2 hours ago
Whit3rd
4,3851019
answered 2 hours ago
Whit3rd
4,3851019
answered 2 hours ago
Whit3rd
4,3851019
answered 2 hours ago
Whit3rd
4,3851019
4,3851019
add a comment |Â
add a comment |Â
up vote
1
down vote
A few things on solenoids (either relays or valves), since you are using a ULN2003 I'll assume you are going to drive it with a voltage step:
- When apply a voltage step, the current will rise until the electro-magnetic force exceeds static friction.
- At this point the plunger (for a valve) or the armature (for a relay) will start moving and create back-EMF which will reducing the current as movement accelerates.
- Once the movement ends (usually abruptly), back-EMF goes to zero and the current will raise to stall value.
- Usually at this point the voltage is reduced (by PWM) to ~40% of initial value as less force is needed to maintain position as opposed to initiate motion (keep in mind that force is linked to current, not voltage).
Datasheets often specs both values: boost (initiate movement) and hold (maintain position).
A good rule of thumb is that hold is ~40% of boost.
By default, I'd assume the power is given in static (where the valve will be most of the time) so the 420mA would be the hold current then you'd need overhead to be able to activate it properly (~800mA).
answered 48 mins ago
mcv
716
add a comment |Â
up vote
1
down vote
A few things on solenoids (either relays or valves), since you are using a ULN2003 I'll assume you are going to drive it with a voltage step:
- When apply a voltage step, the current will rise until the electro-magnetic force exceeds static friction.
- At this point the plunger (for a valve) or the armature (for a relay) will start moving and create back-EMF which will reducing the current as movement accelerates.
- Once the movement ends (usually abruptly), back-EMF goes to zero and the current will raise to stall value.
- Usually at this point the voltage is reduced (by PWM) to ~40% of initial value as less force is needed to maintain position as opposed to initiate motion (keep in mind that force is linked to current, not voltage).
Datasheets often specs both values: boost (initiate movement) and hold (maintain position).
A good rule of thumb is that hold is ~40% of boost.
By default, I'd assume the power is given in static (where the valve will be most of the time) so the 420mA would be the hold current then you'd need overhead to be able to activate it properly (~800mA).
answered 48 mins ago
mcv
716
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A few things on solenoids (either relays or valves), since you are using a ULN2003 I'll assume you are going to drive it with a voltage step:
- When apply a voltage step, the current will rise until the electro-magnetic force exceeds static friction.
- At this point the plunger (for a valve) or the armature (for a relay) will start moving and create back-EMF which will reducing the current as movement accelerates.
- Once the movement ends (usually abruptly), back-EMF goes to zero and the current will raise to stall value.
- Usually at this point the voltage is reduced (by PWM) to ~40% of initial value as less force is needed to maintain position as opposed to initiate motion (keep in mind that force is linked to current, not voltage).
Datasheets often specs both values: boost (initiate movement) and hold (maintain position).
A good rule of thumb is that hold is ~40% of boost.
By default, I'd assume the power is given in static (where the valve will be most of the time) so the 420mA would be the hold current then you'd need overhead to be able to activate it properly (~800mA).
answered 48 mins ago
mcv
716
A few things on solenoids (either relays or valves), since you are using a ULN2003 I'll assume you are going to drive it with a voltage step:
- When apply a voltage step, the current will rise until the electro-magnetic force exceeds static friction.
- At this point the plunger (for a valve) or the armature (for a relay) will start moving and create back-EMF which will reducing the current as movement accelerates.
- Once the movement ends (usually abruptly), back-EMF goes to zero and the current will raise to stall value.
- Usually at this point the voltage is reduced (by PWM) to ~40% of initial value as less force is needed to maintain position as opposed to initiate motion (keep in mind that force is linked to current, not voltage).
Datasheets often specs both values: boost (initiate movement) and hold (maintain position).
A good rule of thumb is that hold is ~40% of boost.
By default, I'd assume the power is given in static (where the valve will be most of the time) so the 420mA would be the hold current then you'd need overhead to be able to activate it properly (~800mA).
answered 48 mins ago
mcv
716
edited 3 mins ago
answered 48 mins ago
mcv
716
answered 48 mins ago
mcv
716
answered 48 mins ago
mcv
716
716
add a comment |Â
add a comment |Â
Chris1309 is a new contributor. Be nice, and check out our Code of Conduct.
Chris1309 is a new contributor. Be nice, and check out our Code of Conduct.
Chris1309 is a new contributor. Be nice, and check out our Code of Conduct.
Chris1309 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f400821%2fcontrolling-solenoid-valves-with-uln2003a-inrush-current-arduino%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password