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Need Help with a project

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sal

New Member
i am building a circuit for race cars for my younger brother, but the battery of the circuit has 5V - 1500 mA and i need to put it 1,5V so it doesn't burn the motor. i know that this is a VERY noob question but my head is on another planet this moment... Final Projects Planet... LOL
 

Hero999

Banned
Sorry but this is not a very good idea.

It would be much easier to get a 6V motor.
 

Willbe

New Member
I guess you could chop up the 5v with a 555 timer with duty cycle 1.5/5 driving a BJT or FET.
 

Hero999

Banned
I guess you could chop up the 5v with a 555 timer with duty cycle 1.5/5 driving a BJT or FET.
You mean a duty cycle of 1.5²/5².
 

Hero999

Banned
Because 1.5 on and 3.5 off, doesn't add up to the same DC value.

You need a duty cycle of 1.5²/5² = 9% for an RMS voltage of 1.5V.

Try working out the RMS values and you'll see what I mean.
 

Willbe

New Member
Because 1.5 on and 3.5 off, doesn't add up to the same DC value.

You need a duty cycle of 1.5²/5² = 9% for an RMS voltage of 1.5V.

Try working out the RMS values and you'll see what I mean.
Do we want DC values or RMS values for this application?
 

Hero999

Banned
If you're pulsing it then you can't work on DC values, you must use AC/RMS values.

1.5 on and 3.5 off gives an RMS value of 2.74V which will blow the motor.
 

Willbe

New Member
If you're pulsing it then you can't work on DC values, you must use AC/RMS values.

1.5 on and 3.5 off gives an RMS value of 2.74V which will blow the motor.
I don't see that.
If a DC voltmeter reads 1.5vdc to the motor then it should work as designed.
RMS values have to do with the heat produced; motors work by amps x coil turns.
 

Hero999

Banned
A DC multimeter's reading on a pulsed waveform is meaningless.

It's the RMS voltage across a load that determines the power dissipation, not the average voltage.
 

Willbe

New Member
A DC multimeter's reading on a pulsed waveform is meaningless.

It's the RMS voltage across a load that determines the power dissipation, not the average voltage.
And the performance of the motor depends on average current. The heating depends on RMS values.
 

Sceadwian

Banned
Depends on the multimeter Willbe, you can't be sure you're getting the true average current, it depends effects of frequency on the input stages of the meter itself.
 

Hero999

Banned
I'm not convinced.

As far as I'm aware a motor behaves pretty resistively once it's running. I still think it's the RMS voltage that governs the power dissipation, not the average voltage.
 

Sceadwian

Banned
It's not voltage at all Hero, it's current. Under varying load conditions its static resistance will change DRAMATICALLY.
 

Willbe

New Member
I think now that the heating of the motor will be slightly different on this waveform because of the motor's I-squared-R loss but if most of the input power goes into producing mech. energy then this effect will be minor.

If you know the thermal resistance of the motor (can be guestimated based on shape and volume), surface temp. (use a thermistor) and power in, efficiency could be determined to some level of confidence.

If no thermistor, after a good workout for the motor use your "digital probe" and check the chart below
no burn at 42C
burn in 30 sec at 54C
5 sec at 60C
1 sec at 71C

If
you monitor motor power draw and
drive the car up a ramp and
it takes 1 second to go up 1 foot and
the car weighs 1 pound
then
the car is putting out 1/550 of a hp
{else
it's not!}

As motor effic. drops Hero999 becomes "more right" and the OP would need to use a linear regulator.

I know these little wall 'formers are =>65% efficient and my fist-sized pond pump is 1% efficient at raising water.
 
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