Hi all

i have a 7amp power supply (for a project im doing) powering a wiper motor
off a car. the constant current of the motor is about 1.5amp but obviously
the start up current is rather more than 7 amps and it sometimes causes the
power supply to momentarily shut down. Can i use a capacitor of a
reasonable size to hold enough juice for start up and if so, what size?

cheers

steve

ps changing the power supply is not an easy option....

On 21/04/2011 12:51, Mr Sandman wrote:

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use a
capacitor of a reasonable size to hold enough juice for start up and if
so, what size?

You would need a *very* large capacitor. C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

--
Andy

Mr Sandman wrote:
Hi all

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use a
capacitor of a reasonable size to hold enough juice for start up and if
so, what size?

cheers

steve

ps changing the power supply is not an easy option....
You calculate the approx value from

CV=IT

or C= IT/V

C in farads , I in amps extra current over that supplied by the PSU, T
the duration in seconds for the excess current and V the voltage dip
that you can tolerate.

Say 10 amps, for 100mS and voltage dip of 5 volts.

so 0.2F or 200,000uF

Bob

Andy Wade wrote:
On 21/04/2011 12:51, Mr Sandman wrote:

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use a
capacitor of a reasonable size to hold enough juice for start up and if
so, what size?

You would need a *very* large capacitor. C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

Its not THAT bad. The model car racing boys use supercapacitors to get
an instant startline boost.

Try a few thoushand 'muffs'

Bob Minchin wrote:
Mr Sandman wrote:
Hi all

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use a
capacitor of a reasonable size to hold enough juice for start up and if
so, what size?

cheers

steve

ps changing the power supply is not an easy option....
You calculate the approx value from

CV=IT

or C= IT/V

C in farads , I in amps extra current over that supplied by the PSU, T
the duration in seconds for the excess current and V the voltage dip
that you can tolerate.

Say 10 amps, for 100mS and voltage dip of 5 volts.

so 0.2F or 200,000uF

100ms is pretty long. The MOMENT the armature starts to move, the back
EMF starts to be there.


Bob

On Apr 21, 12:51*pm, "Mr Sandman" wrote:
Hi all

i have a 7amp power supply (for a project im doing) powering a wiper motor
off a car. *the constant current of the motor is about 1.5amp but obviously
the start up current is rather more than 7 amps and it sometimes causes the
power supply to momentarily shut down. *Can i use a capacitor of a
reasonable size to hold enough juice for start up and if so, what size?

cheers

steve

ps changing the power supply is not an easy option....

The normal way for "DCmotors" is to start up with a resistor in series
which is switched out/shorted out when the motor starts up.

"harry" wrote in message
...

On Apr 21, 12:51 pm, "Mr Sandman" wrote:
Hi all

i have a 7amp power supply (for a project im doing) powering a wiper motor
off a car. the constant current of the motor is about 1.5amp but
obviously
the start up current is rather more than 7 amps and it sometimes causes
the
power supply to momentarily shut down. Can i use a capacitor of a
reasonable size to hold enough juice for start up and if so, what size?

cheers

steve

ps changing the power supply is not an easy option....

The normal way for "DCmotors" is to start up with a resistor in series
which is switched out/shorted out when the motor starts up.

Is there an "off the shelf" solution available for next to nowtish?

Steve

On Apr 21, 1:07*pm, Andy Wade wrote:
On 21/04/2011 12:51, Mr Sandman wrote:



i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use a
capacitor of a reasonable size to hold enough juice for start up and if
so, what size?

You would need a *very* large capacitor. *C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. *I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.


NT

"Tabby" wrote in message
...

On Apr 21, 1:07 pm, Andy Wade wrote:
On 21/04/2011 12:51, Mr Sandman wrote:



i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use a
capacitor of a reasonable size to hold enough juice for start up and if
so, what size?

You would need a *very* large capacitor. C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.



I've got an old battery booster than don't work.... maybe thats got some big
caps in it?

steve

On Apr 21, 8:07*pm, "Mr Sandman" wrote:
"Tabby" *wrote in message

...

On Apr 21, 1:07 pm, Andy Wade wrote:



On 21/04/2011 12:51, Mr Sandman wrote:

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use a
capacitor of a reasonable size to hold enough juice for start up and if
so, what size?

You would need a *very* large capacitor. *C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. *I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got some big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.


NT

After serious thinking Mr Sandman wrote :

"harry" wrote in message
...

On Apr 21, 12:51 pm, "Mr Sandman" wrote:
Hi all

i have a 7amp power supply (for a project im doing) powering a wiper motor
off a car. the constant current of the motor is about 1.5amp but obviously
the start up current is rather more than 7 amps and it sometimes causes the
power supply to momentarily shut down. Can i use a capacitor of a
reasonable size to hold enough juice for start up and if so, what size?

cheers

steve

ps changing the power supply is not an easy option....

The normal way for "DCmotors" is to start up with a resistor in series
which is switched out/shorted out when the motor starts up.

Is there an "off the shelf" solution available for next to nowtish?

Steve

Car headlight bulb in series, with a switch to short out the lamp?

--
Regards,
Harry (M1BYT) (L)
http://www.ukradioamateur.co.uk

On Thu, 21 Apr 2011 18:46:56 +0100, "Mr Sandman"
wrote:



"harry" wrote in message
...

On Apr 21, 12:51 pm, "Mr Sandman" wrote:
Hi all

i have a 7amp power supply (for a project im doing) powering a wiper motor
off a car. the constant current of the motor is about 1.5amp but
obviously
the start up current is rather more than 7 amps and it sometimes causes
the
power supply to momentarily shut down. Can i use a capacitor of a
reasonable size to hold enough juice for start up and if so, what size?

cheers

steve

ps changing the power supply is not an easy option....

The normal way for "DCmotors" is to start up with a resistor in series
which is switched out/shorted out when the motor starts up.

Is there an "off the shelf" solution available for next to nowtish?

Steve

If you are into electronics an emitter follower/ darlington pair would
work. Feed the base via a resistor from the supply and use a capacitor
from the base to ground to allow the Voltage to build up to max.

A BC108 and 2n3055 should be the only outlay apart from the odd few
pence for the R/C feed.

An even simpler solution might be an NTC thermistor.

HN

no im referring to those boosters you carry round to start your battery in
the event of not having a charger.

steve

"Tabby" wrote in message
...

On Apr 21, 8:07 pm, "Mr Sandman" wrote:
"Tabby" wrote in message

...

On Apr 21, 1:07 pm, Andy Wade wrote:



On 21/04/2011 12:51, Mr Sandman wrote:

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use
a
capacitor of a reasonable size to hold enough juice for start up and
if
so, what size?

You would need a *very* large capacitor. C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got some
big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.


NT

In message , Mr
Sandman writes
no im referring to those boosters you carry round to start your battery
in the event of not having a charger.

steve

"Tabby" wrote in message
...

On Apr 21, 8:07 pm, "Mr Sandman" wrote:
"Tabby" wrote in message

...

On Apr 21, 1:07 pm, Andy Wade wrote:



On 21/04/2011 12:51, Mr Sandman wrote:

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i
use a
capacitor of a reasonable size to hold enough juice for start up
and if
so, what size?

You would need a *very* large capacitor. C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got
some big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.

Does the motor have to maintain its torque when under load? If it's not
that critical, a low-value series resistor might suffice.

For example, assuming a 12V supply, and that, when stationary, the motor
resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on
current to 7A.

If the motor takes 1.5A when running, its effective resistance will be 8
ohms. This is relatively small compared with the 0.7 ohms, and when
lightly loaded, and the voltage across the motor will be 11V (the
resistor dropping 1V).

If you load the motor, it will slow down, and draw more current. If you
keep on loading it, the resistor will drop more and more voltage, so the
motor torque will begin to suffer. But, depending on your application,
this might be acceptable.

If you think this simplistic approach might be adequate, you could
measure the actual resistance of the motor, and calculate the correct
value for the required resistor.
--
Ian

Mr Sandman wrote:


"harry" wrote in message
...

On Apr 21, 12:51 pm, "Mr Sandman" wrote:
Hi all

i have a 7amp power supply (for a project im doing) powering a wiper
motor
off a car. the constant current of the motor is about 1.5amp but
obviously
the start up current is rather more than 7 amps and it sometimes
causes the
power supply to momentarily shut down. Can i use a capacitor of a
reasonable size to hold enough juice for start up and if so, what size?

cheers

steve

ps changing the power supply is not an easy option....

The normal way for "DCmotors" is to start up with a resistor in series
which is switched out/shorted out when the motor starts up.

Is there an "off the shelf" solution available for next to nowtish?

PTC thermistors work.

high resistance till hot.

Steve

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about 1.5amp
but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can i use
a
capacitor of a reasonable size to hold enough juice for start up and
if
so, what size?

You would need a *very* large capacitor. C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got some
big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.

Does the motor have to maintain its torque when under load? If it's not
that critical, a low-value series resistor might suffice.

For example, assuming a 12V supply, and that, when stationary, the motor
resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on
current to 7A.

If the motor takes 1.5A when running, its effective resistance will be 8
ohms. This is relatively small compared with the 0.7 ohms, and when
lightly loaded, and the voltage across the motor will be 11V (the
resistor dropping 1V).

If you load the motor, it will slow down, and draw more current. If you
keep on loading it, the resistor will drop more and more voltage, so the
motor torque will begin to suffer. But, depending on your application,
this might be acceptable.

If you think this simplistic approach might be adequate, you could
measure the actual resistance of the motor, and calculate the correct
value for the required resistor.

Thanks Ian,

it only stirs water at a very low speed (wiper motor out of a car on slow
speed) if i put load on the motor by holding the output shaft the current
increases but not in my application as i don't load the motor it only draws
1.5amp when running but on start up it is drawing more and tripping the
power supply. The resistance of the motor is 1ohm (the faster speed is
1.5ohm...it'd be nice to have the option to use it too)

What wattage will the resistor need to be to cope with the current going
through it?

steve

In message , Mr
Sandman writes
i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about
1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can
i use a
capacitor of a reasonable size to hold enough juice for start up
if
so, what size?

You would need a *very* large capacitor. C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got
some big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.

Does the motor have to maintain its torque when under load? If it's not
that critical, a low-value series resistor might suffice.

For example, assuming a 12V supply, and that, when stationary, the motor
resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on
current to 7A.

If the motor takes 1.5A when running, its effective resistance will be 8
ohms. This is relatively small compared with the 0.7 ohms, and when
lightly loaded, and the voltage across the motor will be 11V (the
resistor dropping 1V).

If you load the motor, it will slow down, and draw more current. If you
keep on loading it, the resistor will drop more and more voltage, so the
motor torque will begin to suffer. But, depending on your application,
this might be acceptable.

If you think this simplistic approach might be adequate, you could
measure the actual resistance of the motor, and calculate the correct
value for the required resistor.

Thanks Ian,

it only stirs water at a very low speed (wiper motor out of a car on
slow speed) if i put load on the motor by holding the output shaft the
current increases but not in my application as i don't load the motor
it only draws 1.5amp when running but on start up it is drawing more
and tripping the power supply. The resistance of the motor is 1ohm
(the faster speed is 1.5ohm...it'd be nice to have the option to use it too)

What wattage will the resistor need to be to cope with the current
going through it?

Using the 'I squared R' formula, 1 ohm at 1.5A is 2.25W. But you might
be wise to allow for more current. The power rises rapidly (as the
square of the current), so I'd go for 5W or more. Depending one what you
have in you junk box, it might be more convenient to use several lower
wattage resistors in parallel (eg at least five x 4.7 ohms or ten x 10
ohms). This enables you (if you need to) to alter the total value by
adding or removing resistors.
--
Ian

On Fri, 22 Apr 2011 18:16:00 +0100, John Rumm
wrote:

On 22/04/2011 10:24, The Natural Philosopher wrote:
Mr Sandman wrote:


"harry" wrote in message
...

On Apr 21, 12:51 pm, "Mr Sandman" wrote:
Hi all

i have a 7amp power supply (for a project im doing) powering a wiper
motor
off a car. the constant current of the motor is about 1.5amp but
obviously
the start up current is rather more than 7 amps and it sometimes
causes the
power supply to momentarily shut down. Can i use a capacitor of a
reasonable size to hold enough juice for start up and if so, what size?

cheers

steve

ps changing the power supply is not an easy option....

The normal way for "DCmotors" is to start up with a resistor in series
which is switched out/shorted out when the motor starts up.

Is there an "off the shelf" solution available for next to nowtish?

PTC thermistors work.

You could shunt the motor with one...

Series, surely? A shunt would divert current from the motor, but
would present an even greater load for the PSU.


high resistance till hot.

That would be NTC would it not?

That is correct.

The trouble is the thermistor is a bit wasteful as it has to dissipate
power to maintain it's temperature.

I never saw one used to supply a 7A load, although they are probably
available.

The main reservation I'd have is the amount of heat produced and from
past experience the reliability due to the metal bonding not expanding
and contracting in sympathy with the ceramic type medium of the body.

HN

On Apr 23, 11:18*am, Ian Jackson
wrote:
In message , Mr
Sandman writes

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about
1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can
i use *a
capacitor of a reasonable size to hold enough juice for start up
if
so, what size?

You would need a *very* large capacitor. *C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. *I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got
some big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.

Does the motor have to maintain its torque when under load? If it's not
that critical, a low-value series resistor might suffice.

For example, assuming a 12V supply, and that, when stationary, the motor
resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on
current to 7A.

If the motor takes 1.5A when running, its effective resistance will be 8
ohms. This is relatively small compared with the 0.7 ohms, and when
lightly loaded, and the voltage across the motor will be 11V (the
resistor dropping 1V).

If you load the motor, it will slow down, and draw more current. If you
keep on loading it, the resistor will drop more and more voltage, so the
motor torque will begin to suffer. But, depending on your application,
this might be acceptable.

If you think this simplistic approach might be adequate, you could
measure the actual resistance of the motor, and calculate the correct
value for the required resistor.

Thanks Ian,

it only stirs water at a very low speed (wiper motor out of a car on
slow speed) if i put load on the motor by holding the output shaft the
current increases but not in my application as i don't load the motor
it only draws 1.5amp when running but on start up it is drawing more
and tripping the power supply. *The resistance of the motor is 1ohm
(the faster speed is 1.5ohm...it'd be nice to have the option to use it too)

What wattage will the resistor need to be to cope with the current
going through it?

Using the 'I squared R' formula, 1 ohm at 1.5A is 2.25W. But you might
be wise to allow for more current. The power rises rapidly (as the
square of the current), so I'd go for 5W or more. Depending one what you
have in you junk box, it might be more convenient to use several lower
wattage resistors in parallel (eg at least five x 4.7 ohms or ten x 10
ohms). This enables you (if you need to) to alter the total value by
adding or removing resistors.

reducing start current, thus reducing starting torque, is ok for
stirring water.


NT

On Sat, 23 Apr 2011 08:56:59 -0700 (PDT), Tabby
wrote:

On Apr 23, 11:18*am, Ian Jackson
wrote:
In message , Mr
Sandman writes

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about
1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can
i use *a
capacitor of a reasonable size to hold enough juice for start up
if
so, what size?

You would need a *very* large capacitor. *C = I * dt/dv, where I is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which might be very
small. *I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got
some big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.

Does the motor have to maintain its torque when under load? If it's not
that critical, a low-value series resistor might suffice.

For example, assuming a 12V supply, and that, when stationary, the motor
resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on
current to 7A.

If the motor takes 1.5A when running, its effective resistance will be 8
ohms. This is relatively small compared with the 0.7 ohms, and when
lightly loaded, and the voltage across the motor will be 11V (the
resistor dropping 1V).

If you load the motor, it will slow down, and draw more current. If you
keep on loading it, the resistor will drop more and more voltage, so the
motor torque will begin to suffer. But, depending on your application,
this might be acceptable.

If you think this simplistic approach might be adequate, you could
measure the actual resistance of the motor, and calculate the correct
value for the required resistor.

Thanks Ian,

it only stirs water at a very low speed (wiper motor out of a car on
slow speed) if i put load on the motor by holding the output shaft the
current increases but not in my application as i don't load the motor
it only draws 1.5amp when running but on start up it is drawing more
and tripping the power supply. *The resistance of the motor is 1ohm
(the faster speed is 1.5ohm...it'd be nice to have the option to use it too)

What wattage will the resistor need to be to cope with the current
going through it?

Using the 'I squared R' formula, 1 ohm at 1.5A is 2.25W. But you might
be wise to allow for more current. The power rises rapidly (as the
square of the current), so I'd go for 5W or more. Depending one what you
have in you junk box, it might be more convenient to use several lower
wattage resistors in parallel (eg at least five x 4.7 ohms or ten x 10
ohms). This enables you (if you need to) to alter the total value by
adding or removing resistors.

reducing start current, thus reducing starting torque, is ok for
stirring water.


NT
Now that is deep!


HN

In message , H. Neary
writes
On Sat, 23 Apr 2011 08:56:59 -0700 (PDT), Tabby
wrote:

On Apr 23, 11:18*am, Ian Jackson
wrote:
In message , Mr
Sandman writes

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about
1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can
i use *a
capacitor of a reasonable size to hold enough juice for start up
if
so, what size?

You would need a *very* large capacitor. *C = I * dt/dv, where
is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which
be very
small. *I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or
mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got
some big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.

Does the motor have to maintain its torque when under load? If it's not
that critical, a low-value series resistor might suffice.

For example, assuming a 12V supply, and that, when stationary, the motor
resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on
current to 7A.

If the motor takes 1.5A when running, its effective resistance will be 8
ohms. This is relatively small compared with the 0.7 ohms, and when
lightly loaded, and the voltage across the motor will be 11V (the
resistor dropping 1V).

If you load the motor, it will slow down, and draw more current. If you
keep on loading it, the resistor will drop more and more voltage, so the
motor torque will begin to suffer. But, depending on your application,
this might be acceptable.

If you think this simplistic approach might be adequate, you could
measure the actual resistance of the motor, and calculate the correct
value for the required resistor.

Thanks Ian,

it only stirs water at a very low speed (wiper motor out of a car on
slow speed) if i put load on the motor by holding the output shaft the
current increases but not in my application as i don't load the motor
it only draws 1.5amp when running but on start up it is drawing more
and tripping the power supply. *The resistance of the motor is 1ohm
(the faster speed is 1.5ohm...it'd be nice to have the option to
use it too)

What wattage will the resistor need to be to cope with the current
going through it?

Using the 'I squared R' formula, 1 ohm at 1.5A is 2.25W. But you might
be wise to allow for more current. The power rises rapidly (as the
square of the current), so I'd go for 5W or more. Depending one what you
have in you junk box, it might be more convenient to use several lower
wattage resistors in parallel (eg at least five x 4.7 ohms or ten x 10
ohms). This enables you (if you need to) to alter the total value by
adding or removing resistors.

reducing start current, thus reducing starting torque, is ok for
stirring water.


NT
Now that is deep!

Drip-feeding information?
--
Ian

On Sat, 23 Apr 2011 17:36:02 +0100, Ian Jackson
wrote:

In message , H. Neary
writes
On Sat, 23 Apr 2011 08:56:59 -0700 (PDT), Tabby
wrote:

On Apr 23, 11:18*am, Ian Jackson
wrote:
In message , Mr
Sandman writes

i have a 7amp power supply (for a project im doing) powering a wiper
motor off a car. the constant current of the motor is about
1.5amp but
obviously the start up current is rather more than 7 amps and it
sometimes causes the power supply to momentarily shut down. Can
i use *a
capacitor of a reasonable size to hold enough juice for start up
if
so, what size?

You would need a *very* large capacitor. *C = I * dt/dv, where
is the
starting current, dt is start-up time (guess it) and dv is the
permissible voltage 'sag' before the PSU complains, which
be very
small. *I (max) will be determined by the DC resistance of the motor
windings.

A small float-charged SLA battery might be a better bet, or
mods to the
PSU, perhaps.

To hold all the exess startup current, you'd need a big one, but
that's not needed. A cap that holds a yet unknown percentage of the
start surge would be enough to aviod nuisance trips. Of course you
then have the problem that the psu trips at power up from charging the
cap, unless psu has soft start, or unless, as is more likely, the i
and t are within its shutdown limits.

In plain english, put a few thousand mfds on it and see.

I've got an old battery booster than don't work.... maybe thats got
some big
caps in it?

steve

If by that you mean an old car battery charger, it wont have any caps
in.

Does the motor have to maintain its torque when under load? If it's not
that critical, a low-value series resistor might suffice.

For example, assuming a 12V supply, and that, when stationary, the motor
resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on
current to 7A.

If the motor takes 1.5A when running, its effective resistance will be 8
ohms. This is relatively small compared with the 0.7 ohms, and when
lightly loaded, and the voltage across the motor will be 11V (the
resistor dropping 1V).

If you load the motor, it will slow down, and draw more current. If you
keep on loading it, the resistor will drop more and more voltage, so the
motor torque will begin to suffer. But, depending on your application,
this might be acceptable.

If you think this simplistic approach might be adequate, you could
measure the actual resistance of the motor, and calculate the correct
value for the required resistor.

Thanks Ian,

it only stirs water at a very low speed (wiper motor out of a car on
slow speed) if i put load on the motor by holding the output shaft the
current increases but not in my application as i don't load the motor
it only draws 1.5amp when running but on start up it is drawing more
and tripping the power supply. *The resistance of the motor is 1ohm
(the faster speed is 1.5ohm...it'd be nice to have the option to
use it too)

What wattage will the resistor need to be to cope with the current
going through it?

Using the 'I squared R' formula, 1 ohm at 1.5A is 2.25W. But you might
be wise to allow for more current. The power rises rapidly (as the
square of the current), so I'd go for 5W or more. Depending one what you
have in you junk box, it might be more convenient to use several lower
wattage resistors in parallel (eg at least five x 4.7 ohms or ten x 10
ohms). This enables you (if you need to) to alter the total value by
adding or removing resistors.

reducing start current, thus reducing starting torque, is ok for
stirring water.


NT
Now that is deep!

Drip-feeding information?
Using the wetted contact approach!

HN