Adapter fails to power DC motor

Thanks, understood. I have a few of these on order to play with:

as a test, connect the motor then get it started by spinning it by hand. does that work?

Thanks a bunch, that fixed it!

With a series resistance of two 0.5 R in parallel, plus some croc-to-crocs and other wires, it starts at once. With no load quickly settled at 2.7A and 3.9V. I couldn't quite stall it but got close, when those became 5.1A and 2.0V.

That was with an identical but little used motor, but I'm fairly confident the installed motor will behave in similar fashion.

Terry, East Grinstead, UK

Thanks George. That fix from you and DemonicTubes gets the seegar! See also my more detailed reply a minute ago.

Terry, East Grinstead, UK

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s power the motor at '3.8 to 4.1 V". This supply runs at 5 volts and he ha s not said anything about reducing that voltage that I saw. They guy is no t very precise in his description of the circuit though. I don't know what he means by "ex-screwdriver 2.4 motor". Is that 2.4 volts? 2.4 amps? Ma ybe 2.4 revision???

ed between the motor input and the PSU output. That would drop the volt at full power. Or maybe use a smaller resistor (~0.1 ohm) and a diode drop. Or maybe two diode drops, one silicon and one Schottky plus optionally a t enth ohm resistor.

t motor from a 5 volt supply.

ggested an NTC ('thermistor' as they are commonly known) is to avoid wastin g power. They are readily available and dirt cheap...of course if you don' t have one on hand and need to get it working now, yeah just throw a low va lue resistor on there and go for it.

hey start out at a few ohms, but quickly drop to almost nothing after a mom ent of current flow.

supply to a 3.8-4.0 volt motor. So the NTC will only set him up for a fut ure motor failure.

e case, a proper supply is a better solution.

You state two batteries provided insufficient current so you upped the numb er of cells which provided more voltage. Do you see the problem with that?

NiMH cells have a higher internal resistance than do NiCd or Li-ion cells o f similar size. So when trying to open the curtain with two NiMH cells you were losing voltage in the series resistance and so the delivered power wa s too low. By adding a third cell the voltage increased to where it should have been and were able to get sufficient power to open the curtain.

Now your cells are showing their age and no longer hold a charge long enoug h to power the curtain with the original charging current. So you pick a p ower supply to directly drive the motor, but at more than twice the rated v oltage. So the peak current is far higher than either required or allowed by the PSU and it doesn't work.

No, a couple of series diodes are not going to work. You need a power supp ly that puts out the correct voltage at the correct current. To charge a b attery power source at 2.4 volts seems problematic since you don't want to use a proper system to do this. I think I would go with a capacitor/super capacitor based system. They can be charged at a higher rate for quick rec harge and are resistant to overcharging as long as your power source doesn' t drive them to an excessive voltage. In fact, there is a capacitive doubl er/halfing circuit you could use to charge caps to 2.5 volts from your exis ting 5 volt supply. It is a simple circuit and would only require some FET s and a 555 timer chip.

Excellent. thanks for the update. George H.

You can use Ohm's law to determine stall current. At 5V it'll be more than twice what it was at 2.4V

From your description it sounded like it was working as designed: the

However, equalizing the voltage across the series arrangement remains a problem for you.

te:

asured at the

th the

o a resistor

niature drill.

tion:

lMotor.jpg?raw=1

er why it

, where it says

function

likely to be

n, recovers,

r-current,

nd try

tronic

ething

llow the

in

the

us

start

trying in

ush current limiter. Cheap, easy, probably works...why not?

s power the motor at '3.8 to 4.1 V". This supply runs at 5 volts and he ha s not said anything about reducing that voltage that I saw. They guy is no t very precise in his description of the circuit though. I don't know what he means by "ex-screwdriver 2.4 motor". Is that 2.4 volts? 2.4 amps? Ma ybe 2.4 revision???

ed between the motor input and the PSU output. That would drop the volt at full power. Or maybe use a smaller resistor (~0.1 ohm) and a diode drop. Or maybe two diode drops, one silicon and one Schottky plus optionally a t enth ohm resistor.

t motor from a 5 volt supply.

ggested an NTC ('thermistor' as they are commonly known) is to avoid wastin g power. They are readily available and dirt cheap...of course if you don' t have one on hand and need to get it working now, yeah just throw a low va lue resistor on there and go for it.

hey start out at a few ohms, but quickly drop to almost nothing after a mom ent of current flow.

supply to a 3.8-4.0 volt motor. So the NTC will only set him up for a fut ure motor failure.

e case, a proper supply is a better solution.

Long ago I learnt that mains power was far more reliable than any battery. The ability to manually open & close will also be far more rleiable than an y battery.

I've not measured anything but suspect the motor is likely drawing way abov e 4A for tiny fractions of time. A 2.4v screwdriver motor that self limited to 4A on 5v wouldn't be a lot of use.

NT

Yes. That motor is an inductive load that takes more than the measured stall current to start.

I see DC motors as a big lossy *capacitive* load. Think about it: Initially, it draws a large starting current, which then progressively drops to more modest values when the motor comes up to speed and develops a back-EMF. The EMF persists when you disconnect, dropping exponentially as the loss consumes the stored energy until the motor is stopped.

Jeroen Belleman

I've never seen a wire wound, Permanent Magnet capacitor.

I suppose you can enlighten us with a reasoned discussion on the apparent impedance of permanent magnet DC motors?

Jeroen Belleman

i shouldn't have to. You have to overcome the inductance to build the required magnetic field. Once the motor starts to turn, the commutator converts the DC input into a crude, chopped AC field that is required for the motor to continue to turn.

Now try to explain your capacitor theory.

I already did, see above. You should read it. Now, I'll admit that a DC motor isn't a pure capacitance. The equivalent circuit might be a parallel RC in series with an inductance. The capacitor models the back EMF and the energy stored in the rotating mass. The resistor models the work that needs to expended to keep the thing rotating. And yes, the inductance models the rotor coil with its iron yoke. My point is that this latter component isn't dominant, except maybe on single- digit milli-second time scales or shorter.

Finer detail would add still more components. My model still ignores rotor winding and commutator resistance, variable reluctance effects, iron losses and magnetic saturation, to name a few.

It's the parallel RC that largely dominates the impedance, except at 'high' frequency. The meaning of 'high' depends on the inertia of the motor and its load.

So again, even though a DC motor has coils, its dominant impedance is a parallel RC. The often heard claim that it's inductive is false.

Jeroen Belleman

PM motors do look like very large capacitors at low frequency.

Cheers

Phil Hobbs

That makes sense. So the 'capacitance' is the energy you have to put into spinning up the rotor?

George H.

Yep! From the motor terminals, it looks like a large leaky capacitor.

Jeroen Belleman

Really big energy storage applications use what are essentially DC motor-generators instead of capacitors for storage. High-field magnets, electromagnetic aircraft launchers, things like that.