Toy Motor on 5V supply

Hi,

I'm faced with a small issue. I basically have a fixed 5V source running a circuit. I want to attach a toy motor to it in parallel.

On attaching the toy motor the voltage across it becomes 4V which is not sufficient to power the existing circuit. So I want to know if adding a 1000uF capacitor would alleviate the issue. Would a resistor also help?


Rough Circuit:




Thanks.

A resistor will not help, all it would do is starve the motor of current and slow it down. It would also make it's speed more dependent on it's loading.
A capacitor might help, it will not increase the voltage though. Where it may be beneficial is that when the motor starts up it draws a much larger current than when it's running, the capacitor will help by providing the extra current until the motor is running. A secondary benefit is that the current drawn by small motors is very irregular as the brushes touch different parts of the spinning commutator, the capacitor will to some extent smooth out the peaks and troughs, leaving the supply cleaner for the other circuit.

Your underlying problem is that the 5V source needs to provide more current than it can at the moment. If you tell us where the 5V comes from, it might be possible to find a solution that can handle the extra load of the motor.

Brian.

Thanks for the reply. The 5V comes from the USB header. I can do away with speed of motor (it just needs to spin to appear "spinning"), but the voltage drop is an absolute kill. Basically, what I have is a flash drive which if deprived of the 5V connects-disconnects infinitely.

A USB port is designed to protect itself by dropping it's voltage as it reaches overload point. They are not intended to drive heavy current loads like motors. The maximum a USB port can provide, if it meets specification (most don't) is 0.5A, a small motor will draw most of that while running and probably twice as much when starting of if stalled. I think you would be better to use a separate power supply, not only because you need the power but also to protect the USB port from damage. If the USB port is on a computer main board it could be VERY expensive to replace!

Brian

Use USB Hub with external power supply, this will ensure usage of USB port with enough power. PC USB port is up to 0,5A max.

Or You can make it :



This can be specially used with external hard disk which uses power supply from USB port. Also can be used when connecting such external disks to CCTV recorders USB ports.....

Thank you both for your wonderful feedback.

@tpetar, unfortunately I want to construct it using minimal effort.
@tpetar, @betwtixt: I want the motor to run only when the header is plugged in, so will using a transistor like this help?




Do you recommend?

PS: Ignore the parallel LED, it should be in serial.

You're heading in the right direction, but I suggest this circuit -

Well your looks better than mine. I have no prior experience with transistors tbh.

But, will this keep the other half of the (parallel) circuit at 5V?

If, by "the other half of the (parallel) circuit", you mean the other load like the USB flash drive, the answer is yes, that part of the circuit will still be operating at 5V. The 9V circuit involves only the motor, LED and transistor, and does not directly interact with anything else connected to the USB port.

BTW, my circuit presents a load of about 11mA to the USB 5V output in addition to the flash drive. This shouldn't be a problem. The motor supply could be any other voltage suitable for running the motor.

Thanks a ton Pjdd for your help. Much appreciated.

BTW wouldn't the 3.3k resistor be perhaps too much for the LED? I'm guessing 330Ohm would suffice fo 2 LEDs in Series, but then again I'm a rookie.

Again much thanks.

I would swap the 330 Ohm and 3K3 resistor over. It shouldn't need 15mA base current to saturate the transistor.

Brian.

So, 3k for base and 330 for LEDs then? I'd have to get things accordingly.

nbaztec said:

Thanks a ton Pjdd for your help. Much appreciated.

BTW wouldn't the 3.3k resistor be perhaps too much for the LED? I'm guessing 330Ohm would suffice fo 2 LEDs in Series, but then again I'm a rookie.

Again much thanks.

Click to expand...

You're welcome. The LED current is (Vs - Vled - Vsat)/R = (9 - 2 - 0.1)V/3.3k = 2.1mA (approx.). This should light up a modern LED well enough to provide a good visual indication. Of course, you can use any reasonable level of your choice. 1k will give you nearly 7mA and so on. This is assuming that the LED is a red one with Vf of 2V.

Also, you drew a single LED in your diagram, so my calculation was based on that. Two red LEDs will drop about 4V, so 3.3k will result in ~1.5mA. 1k will give you 4.9mA and 330 ohms will supply 15mA.

betwixt said:

I would swap the 330 Ohm and 3K3 resistor over. It shouldn't need 15mA base current to saturate the transistor.

Click to expand...

It's 390 ohms, not 330, and the base current will be about 11mA. 330 ohms will supply 12-13mA. In any case, I understand that your point is not about the exact figure, but the order of base current level. This was my reasoning:

We don't know the specs of the motor and the OP probably doesn't either. Simple toy motors can draw quite a bit of current, especially at start-up; many draw some hundreds of mAs of surge current. So, to ensure that the transistor will saturate, I thought it would be best to drive it with a fairly hefty base current. ~10mA seems to be a good compromise - not a heavy load for a USB port but enough to saturate a high-gain transistor like a BC337 at loads up to a few hundred mA.

You can calculate the values fairly easily:

1. Find out how much current the motor draws, hook it up to 5V an
d measure with a test meter. It is probably around 250mA or so.
2. Divide that by the gain of the transistor which my data sheet says is more than 60 at 300mA, if it's higher it doesn't matter. If the motor draws 250mA the result would be 0.25/60 = 4.2mA.
3 That figure is the base current, so the resistor from the USB port has to drop 5 - Vbe volts at 4.2mA, if we say Vbe is 0.7V then (5-0.7)/.0042 = 2150 Ohms.

So a better base resistor value to ensure full transistor conduction under lowest gain conditions would be 1.8K Ohms (1800 Ohms).

For the LED, you know the voltage across it (them) is the same as the motor voltage because they are wired across it. It must be 5V. The series resistor has to restrict the current to a safe level for the LEDs and I would suggest 3mA is suitable. Each LED has a forward voltage drop which you can get from the data sheet, using a standard red LED it would be about 1.6V. If you have more than one LED, add their voltages together so two red LEDs would be 3.2V and so on. The series resistor value is (5V - total of forward voltages)/ LED current so for one LED use (5-1.6)/.003 = 1133 Ohms so 1.2K would be a good standard value to choose.

Brian.

Brian, the minimum gain of 60 is at Vce = 1V, not at Vce(sat). The effective gain drops substantially near saturation. So it would best to provide a base drive current of at least Ic/30. And as mentioned before, we don't know how much current the motor draws, so it would be prudent to provide a fairly high base drive. It does no harm and 11mA is a light enough load for USB.

The OP showed a 9V battery (not 5V) in his proposed scheme, so I based my calculation of LED current on that.

Well, I tried the motor with 9V battery (10.2 V precisely) and the motor ran at 0.70A with a peak of 0.81 A. I'm guessing I'd be needing some resistors there.

P.S. I think I wrecked the motor. Battery now reads 7.9V. Cheap products meh.

So what do you intend to do now? Abandon the project, get another motor or something else? I'm still not clear about what the motor is supposed to do.

BTW, PP3-sized 9V batteries are convenient but are quite puny as a means of supplying appreciable power. Anyway, driving a load near 1A will require a more rugged transistor than a BC337 and will need a base drive of more than 11mA unless you use a Darlington transistor or a logic-level MOSFET.

The motor is just supposed to spin. I'll be getting a new motor now. Will attempt the complete build in a week.

I just need it to spin a toy/cardboard fan which isn't much. Maybe I need something that draws less current. What do you suggest?

EDIT: Would a resistor in parallel help in limiting the current to motor?

@Pjdd - I was going to add that the resistor calculation should be treated as conservative and lower values would be advisable but I got called away and had to close the message quickly. Your observation is quite correct though.

@nbaztec - parallel resistors INCREASE the current, making matters worse, don't go down that route!
What I would do is keep the same circuit but replace the BC337 with a power MOSFET, I use 2SK2989 devices for his kind of application but sadly Toshiba will stop production of them in a few days time. A similar MOSFET should be available. These have lower voltage drop than a conventional transistor so allow more power to the motor and produce less heat at the same time. The way conduction is measured is different between MOSFETs and bipolar transistors but to keep it simple, a BC3378 has a VCEsat (the drop while fully conducting) voltage of about 0.7V while the MOSFET in the same situation has a resistance of about 0.13 Ohms. If your motor draws 0.7A that means you only drop 0.09V. They also draw virtually nothing from the USB supply so you can use a much higher value resistor if you want to.

If your existing motor is damaged, look for one with low current when buying a new one.

Brian.

Brian, what I don't get is why the motor would not run when I connected a 300 Ohm resistor in series to it. Also what are the upper limits on the working currents for the BJT and MOSFET. I'll surely look for a low current motor but once cannot rely on raitings of toy motors here. Isn't there any other way to limit current to motor so it runs at a lower rpm?

Thanks for your help.