DC motor kicker, capacitor

[J4e8a16n]

Hi,

I have  a 1.5 volt dc motor that require 9mA to run.  My problem is that it needs much more mA to start,
I would like to know how have a simple dc motor kicker.


Thanks
JP

[David_AVD]

I think a bit more info is needed.  Can the 1.5V source only supply 9mA ?

[Rerouter]

create a soft start, have an input capacitor, that is connected to the motor rail, via a mosfet
when the capacitor has had a chance to near fully charge then connect the motor

i would imagine a p type mosfet with source to the capacitro and drain to the motor wire, and gate tied to the ground of your circuit
you then find a mosfet that switches full on at 1.5V or slightly less,

this way the capacitor is nice and charged up before you connect the motor,  and the nature of the mosfet will even limit its draw as it passes through its linear region acting as a dropping resistor,

[J4e8a16n]

More info:

From 9 volts battery an a 5 volts regulator I putted a 50 ohms resistor to get 10amps at the motor.

JP

[Rerouter]

umm.. you are aware a 9V battery cannot supply more than about 400mA before seriously dropping in output right?
also fitting a 50 ohm resistor across 5V would pull 100mA not 10A,

also being how your motor is 9mA at 1.5V would imply its resitance is ~160 ohms,  so 360 ohm would be a much better suited resistor (5-1.5)*0.009 = ~380 ohm

hope i am following here? when you said 1.5V motor i was under the impression it was being powered off a conventional AA alkaline or similar,

[J4e8a16n]

I have put the motor at the (now) 7.5 volts batterie and red 19mA. So its resistance is about 400 Ohms...

[TerminalJack505]

It sounds like you just need a lower impedance source to drive the motor.  Rather than using resistors to bring the voltage down to 1.5V do something like shown in the attached schematic. 

You can probably get by with using an appropriately sized resistor instead of a zener diode since your load is going to be constant (and your input voltage is a constant 5V.)  You'd just have to play around with the value a bit.  It will likely be somewhere around 800 to 1K2 ohms.

[TerminalJack505]

Here's a circuit that will boost the voltage (and therefore the current) on start-up of the motor if that's what you truly need.

You can adjust the size of the cap to change the amount of time for the voltage boost.  Note how it takes time for the cap to discharge so you can't use this for PWMing the motor or situations where the motor will be turned on quickly after it has been turned off.

[J4e8a16n]

Thank you very much

JP

[TerminalJack505]

Thank you very much

JP

Yep.  No problem.

If you wind up using that circuit then you might want to change R2 to 22K and R3 to 100K.  This will allow you to make the capacitor (C1) smaller.  With those resistor values a value of 4.7uF for C1 would give you the same timings as the circuit that I posted.

[J4e8a16n]

Hi,

I setted up your first shema and it worked smoother. With 9 volts and a 1k pot , 2n3904 and a .1uF capacitor between the motor poles.   It is easier to start .  I red 2volts for 20mA (100 Ohms).

I equally tried your second shema. It did not start. Probably because I dont have the regulator and replaced it with a 2.2 volts zener. On the other hand  maybe because of the 100 Ohms instead of the 400 Ohms motor resistor.


I will try again when I will have the TLVH431 component.


What is the formula to find the capacitor?


Theese little motors are a basic components in many beginners project and the information I found on how to set them is not astounding over the net.

Thanks again.

JP

[IanB]

Theese little motors are a basic components in many beginners project and the information I found on how to set them is not astounding over the net.

Small electric motors need only to be connected to a battery. They are not very sensitive to voltage, so a variety of voltages will usually be fine. If it's a 1.5 V motor, just connect it to a 1.5 V battery (like a D cell). You might try 3 V and it will run faster and be more powerful, but go too high and the motor will burn out--so you can only vary the voltage within a reasonable range.

It's important not to put any current limiting on a motor as the current required varies with the load applied. If you don't provide sufficient power the motor will stall under load and may be damaged. Stalling a motor is generally not desirable.

[TerminalJack505]

The formula for the capacitor works out to be (approximately)...

C = t_b / 8160

where t_b is the amount of time the voltage is boosted.

So, if you want to boost the voltage for a specified time just plug the time into that formula to find the value for the capacitor.

For example, if you wanted to boost the voltage for 100 ms...

C = 0.1 / 8160 = ~12.25 uF.

This formula is for the input voltage (5V) and resistance values shown in the attached circuit.  (With 22k for R2 and 100k for R3.)  If any of those factors change then that formula will no longer work.  It will probably be easiest to just plug everything into a simulator to calculate the value for the capacitor in that case.

I don't know for how long your particular motor will need to be boosted to get it started.  That's something you'll want to play around with.

Note that I added an optional diode to this version of the circuit.  That will help discharge the capacitor faster.

[J4e8a16n]

I have to redo all the pcb that goes with the motor  (wich will cool a heatsink) constant curent load.
It will take a while
Thanks again.

JP

[TerminalJack505]

I kind of cheated when I first came up with the formula.  I sat down and did it properly.  Here's the revised version:

C = t_b / 8360

If you ever have to figure this out yourself then here's how it is done:

• Determine the boost voltage, V_b.  This is 4.11V as shown in the simulation.  You can calculate this (tedious), determine it by measurement (best) or determine it by simulation.
• Determine the Thevenin equivalent circuit for R2 and R3 with the V_ref node being the output and V_b being the input voltage.  Determine V_th and R_th from this.  Since this is just a voltage divider, V_th = V_b * R3 / (R2 + R3) and R_th = R2 // R3 = R2 * R3 / (R2 + R3).
• Use the following formula to determine C.  This is based on the formula to charge C to 1.25V through a resistance of R_th from a voltage of V_th:

C = t_b / (R_th * -ln((V_th - 1.25) / V_th)

[IanB]

Why all the complication?

[TerminalJack505]

Why all the complication?

So far as the formula goes or the OP's motor driver requirement?

So far as the formula goes, that is the answer to the OP's direct question: "What is the formula to find the capacitor?"  Naturally, the easy answer is "use a simulator to figure it out."

So far as the OP's requirements, I don't know.  I just provided a circuit that did what he asked for.  Does he really need it?  I don't know.

[IanB]

Why all the complication?

So far as the formula goes or the OP's motor driver requirement?

The motor driver requirement of course. It's a 1.5 V motor, so connect a 1.5 V battery to it. Job done.

For the OP, the starting point for all designs begins with the final requirement and works back from there. You have a motor. You want it to turn. What we don't know is why you want it to turn? What do you want your motor to drive? What requirements do you have for that? Does it need to turn at a fixed speed? Does it need to have a limited power output? Does it need to start up slowly? What total power must the motor deliver? From these questions you can determine what level of complication is needed to drive the motor.

If you just want the motor to turn and to run at its normal speed, then connect it to a 1.5 V battery. Everything else is unnecessary.

[J4e8a16n]

Thanks for your answers and .. I learned on motor wich I appreciate.
JP

[J4e8a16n]

do you want your motor to drive? What requirements do you have for that?
Does it need to start up slowly?
What total power must the motor deliver?
From these questions you can determine what level of complication is needed to drive the motor.

In fact the motor is there to cool the heatsink on the current dummy load.
It drives a  small plastic  airscrew.

It should go faster as the temperature rises and slower as it fall.


With a 1.5 v it draws .111 A... it is too slow.
With 5 volts and no resistor it takes .330 A to start.

If I use a 20  Ohms resistor to lower the voltage (kind of dummy voltage divisor)  I get .220 A at minimum for a random starting process. It start or not.

If I put a L meter on the black and red wire I have jumping readings from 7 mH to 0 mH.

[J4e8a16n]

The motor is now on the 9 volts battery.
7 volts where left when I did this working shema.

If I have a 150 Omhs resistor and it takes .200 A to start then it continue at .070 A, can I find the inductance of the motor with  E/di/dt = L ?

I have a look at this simple way to do it.



JP

[J4e8a16n]

Instead of a capacitor, I was wondering if I could ise a self like that way.

[TerminalJack505]

I think I'd use a capacitor and resistor like that guy showed in his video.  That seems like the simplest solution of all.

[J4e8a16n]

Evething is fine now.