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Heated jacket temperature regulator

R791945

Jun 19, 2015
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The heat source comes from the 5v 2.1 A (total) output from a battery pack. The heating elements are 5v 8.5 watts. The controller has broken and is long gone. I am wondering how to replace the old controller.

Would this wpm be suitable, if not what do you recommend?

 

Bluejets

Oct 5, 2014
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You could try them as they are cheap enough.
One couldn't start to build one for that amount.
Just might need to add a heatsink on the output element (mosfet)if it starts to get warm.
First one says it's 10A so take that with a grain of salt.......in the same spec it says 30w.
 

R791945

Jun 19, 2015
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Thank you for your response. The reason I asked for advice is because it is claimed to be for an electric motor, which to me means that it expects a high initial draw, usually satisfied by a capacitor. I wondered if that would rule out that WPM for controlling the heat flow to the jacket heating elements.
 

Bluejets

Oct 5, 2014
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No, pwm simply means the power to the load is switched on and off at a high rate with a variable on off duty cycle.
Nothing to do with any high initial draw or capacitors.
The resultant power output if say 50% on and 50% off is 50% speed/ or light output or heat or whatever, BUT with power applied at full rail voltage.
It does assist motors trying to start under load at low revs though as one gets high torque from the pulses of rail high voltage.
In your instance though it is a form of power control without large heat losses in the output driver, therefore more power efficient.
 

R791945

Jun 19, 2015
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I feel this rider is unneccesary, because I am more or less certain of the connections, but I might as well double check:

As the heating elements only take 8.5 watts 5v, I propose:
connect the + - battery power to the wpm power connections.
connect the + - Heater wires to the motor side

Does this make sense? Here is a picture:
pwm.jpg
 

Bluejets

Oct 5, 2014
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Sounds right, although, unless the heater is more than just a resistance device, polarity of the connection to the "motor" would make no difference.

Naturally, as always, the power input MUST be the correct polarity.
 

R791945

Jun 19, 2015
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Sounds right, although, unless the heater is more than just a resistance device, polarity of the connection to the "motor" would make no difference.

Naturally, as always, the power input MUST be the correct polarity.
Thank you for your kind help.
I calculate that there is a 1.7A draw. ie (8.5w/5v)
.
I had a look at the same wpm on AliExpress. This is the comment that now concerns me, because I do not understand why the continuous current appears to be limited to .5A

"This is a speed governor with a maximum current of 2A. If the continuous current is greater than 0.5A, the 775 motor and the stroller motor will not move. Don't buy it wrong. It is not recommended to supply power greater than 12V, such as 12V lead-acid batteries, which may cause damage to the governor."

As I am not intending to use this on a 775 motor / stroller motor, can I ignore the warning or do I need to look for another wpm with higher rating?
 

Bluejets

Oct 5, 2014
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How would anyone know?
You don't provide a link to where you saw this and previous was an Ebay link with no reference to what you say.

FYI...it's a pwm device, not wpm.
 

R791945

Jun 19, 2015
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I apologize for not giving the url. I had assumed this aliexpress item was the same as the one being sold on ebay. I look forward to your reply and hope this helps. Thank you for your patience:

How would anyone know?
You don't provide a link to where you saw this and previous was an Ebay link with no reference to what you say.

FYI...it's a pwm device, not wpm.
Thank you for the correction.
 
Last edited:

Bluejets

Oct 5, 2014
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Tip for posting link you searched for................ anything after and including the "?" can be deleted.
example..... https://www.aliexpress.com/item/4000522614891.html
Put any link you refer to by clicking on the chain link.

I see what you refer to, possibly to do with motors alone as motors tend to draw from 4 to 6 times full load current at startup.
In that instance the maximum current would be in excess of the 2A.
Your load is mostly resistive so would not apply.
 
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