Best location for capacitor in following circuit?

Hello,

I have this little setup, which kind of works, but since the Pro Mini and DC motor share the same PSU, upon activating the motor, the Pro Mini loses power for a bit, which causes screwed up timing, and other unwanted side effects.

In the following picture, I've labeled 1, 2, 3; which location would be best to place a capacitor to mitigate the power loss upon DC motor startup?
Capacitor Location.png

Is this a case where a capacitor would be useful? If so, what size capacitor would be appropriate? I fail to understand how capacitors help in certain situations, since a 100 uf capacitor is only .0001 of an Amp, if I understand correctly? In that case, most common capacitors would seem useless...

Capacitor Location.png

I fail to understand how capacitors help in certain situations, since a 100 uf capacitor is only .0001 of an Amp, if I understand correctly?

No, you don’t understand. Capacitance is about charge, not current and you’re missing both time and voltage in the discussion.

As presented, no one can properly answer your question. A capacitor may help. But, we don’t really know what the problem is. How are you switching on the motor? What kind of switch? It could be noise, it could be voltage drop, it could a bad battery. Any and all possibilities exsist as your problem source.

Real information about the components (with links) and how they are actually connected together (a real schematic) will facilitate real answers. Until you’re more forthcoming, all you’ll be getting are guesses.

WattsThat:
No, you don’t understand. Capacitance is about charge, not current and you’re missing both time and voltage in the discussion.

As presented, no one can properly answer your question. A capacitor may help. But, we don’t really know what the problem is. How are you switching on the motor? What kind of switch? It could be noise, it could be voltage drop, it could a bad battery. Any and all possibilities exsist as your problem source.

Real information about the components (with links) and how they are actually connected together (a real schematic) will facilitate real answers. Until you’re more forthcoming, all you’ll be getting are guesses.

It looks like it wasn't getting enough amps. The original supply had an output of 200mA. Connecting it to a 12V car battery did the trick. Would a capacitor, in any location, help with the amp shortage on a weaker power supply?

toxicxarrow:
It looks like it wasn't getting enough amps. The original supply had an output of 200mA. Connecting it to a 12V car battery did the trick. Would a capacitor, in any location, help with the amp shortage on a weaker power supply?

Yes, but for about a millisecond, then you are back to the same problem. Now that you know the problem, FIX IT!!!

Paul

Paul_KD7HB:
Yes, but for about a millisecond, then you are back to the same problem. Now that you know the problem, FIX IT!!!

Paul

How? Supercapacitor? Any other way than "power supply with more amps?"

Use the correct power supply!

The car battery sounds like the "go" from what you describe! With a 13.2 V trickle charger.

A motor at start is practically a short and dropping too much current causing your voltage to go under the regulator voltage causing a brownout.

A "starter" capacitor can be put on the motor as they do with AC motors. Depending on the start stop frequency of the motor and the motor demand is what you size the capacitor for. basically you want the capacitor charging before you hit the motor switch (has to be on power side).

You can try different sizes but as long as on off frequency is low bigger the better. You can calculate it but then you would need an oscilloscope to see what the voltage drop is on the battery when the motor is started. Hope that makes sense. I don't think multimeter is quick enough and it averages voltage out.

Here we go again.

Once again it looks like we'll have to endure countless posts until we are drip-drip-drip fed with the information the OP should have made in his first post. Anyone wanna place bets on how many posts this thread will have before he gets a usable answer? :astonished:

Paul__B:
Use the correct power supply!

Gotcha, thanks.

Paul__B:
With a 13.2 V trickle charger.

Can't tell if that's a joke; i thought trickle charge was the exact opposite of high voltage? :confused:

wolframore:
A motor at start is practically a short and dropping too much current causing your voltage to go under the regulator voltage causing a brownout.

I did not know that; thanks for the info.

wolframore:
basically you want the capacitor charging before you hit the motor switch (has to be on power side).

Can you elaborate on that a bit? Where the "1?" in the picture is? Also, what's the difference between putting capacitor anode to 12V+, cathode to 12V-, versus running 12V+ into anode and out the cathode? Is there a difference?

but this is fun…

have to wire like this. Capacitor that’s wired in series will block DC voltage and will not allow you to start the motor.

You really have to look at the motor spec to determine correct cap size.

  • the starting power requirements are high on motors, but doesn’t mean it’s wrong size battery. Once it starts it becomes much more efficient. The initial loads are usually taken care of with starter caps.

circuit.jpg

wolframore:
but this is fun...

have to wire like this. Capacitor that's wired in series will block DC voltage and will not allow you to start the motor.

You really have to look at the motor spec to determine correct cap size.

  • the starting power requirements are high on motors, but doesn't mean it's wrong size battery. Once it starts it becomes much more efficient. The initial loads are usually taken care of with starter caps.

Appreciate it; thanks!

I haven't been following this thread before now but I do have a suggestion.

I would move the 4700µF to the input of the DC-DC converter. I would change the 1N4148 to a 1n4001 or 4002 or 4004 or 4007.
I would add a 100µF in parallel with a 0.1µf at the motor.

You also need to check your supply is capable to powering the motor with load while it is running.

Reasoning:

Lets assume the startup current averages 1 amp for 100 ms. It would normally be much shorter but we know the supply is dropping to some lower voltage on startup.

So I = C dv/dt where I = 1 Amp and C = 4700 x-6 and t = 0.1 seconds

therefore;
dv = approx 21 volts. Which is not possible in this circuit being a 12V supply.

This shows the dv on startup is too high with just a 4700µF cap. Now in reality it's capacity is added to the powersupply resulting is a dv we can't calculate with the information at hand. So it might work, we don't know.

But:
You can get more benefit from the 4700µF if it is put at the input of the DC-DC power converter. It will hold up the Pro Mini voltage which is much easier than powering the motor.

Diode:

The 1N4148 is a small signal diode. It cannot handle (at least for long) the current going into the DC-DC supply when you first turn on the power supply. It will be even worse if you move the 4700µF to the DC-DC converter side of the diode.

JohnRob:
I haven’t been following this thread before now but I do have a suggestion.

I would move the 4700µF to the input of the DC-DC converter. I would change the 1N4148 to a 1n4001 or 4002 or 4004 or 4007.
I would add a 100µF in parallel with a 0.1µf at the motor.

You also need to check your supply is capable to powering the motor with load while it is running.

Reasoning:

Lets assume the startup current averages 1 amp for 100 ms. It would normally be much shorter but we know the supply is dropping to some lower voltage on startup.

So I = C dv/dt where I = 1 Amp and C = 4700 x-6 and t = 0.1 seconds

therefore;
dv = approx 21 volts. Which is not possible in this circuit being a 12V supply.

This shows the dv on startup is too high with just a 4700µF cap. Now in reality it’s capacity is added to the powersupply resulting is a dv we can’t calculate with the information at hand. So it might work, we don’t know.

But:
You can get more benefit from the 4700µF if it is put at the input of the DC-DC power converter. It will hold up the Pro Mini voltage which is much easier than powering the motor.

Diode:

The 1N4148 is a small signal diode. It cannot handle (at least for long) the current going into the DC-DC supply when you first turn on the power supply. It will be even worse if you move the 4700µF to the DC-DC converter side of the diode.

Thanks for your suggestion! I’ll try it out.

Good call on the diode. The purpose of it is to block reverse current so the filter caps don’t empty and if the cap at the motor is big enough it should handle the start. You could burn out the regulator without it.

toxicxarrow:
Gotcha, thanks. Can't tell if that's a joke

A "trickle charger" is the name for a battery charger that can only charge a battery slowly but not keep up with the full demands that are intermittently placed on the battery.

This is an appropriate approach for your situation where from time to time - but not too often - you want the battery to be able to drive the motor. You do not need the power supply to have enough "grunt" to power the motor, just enough to supply sufficient charge when the motor is not running, to replace what the motor takes from the battery for the short periods it does run.

The battery is your "reservoir"; it is simply not practical to have a capacitor sufficient to power a motor (though this is a concept under development for electric cars - it is a way off at present).

JohnRob:
You also need to check your supply is capable to powering the motor with load while it is running.

It clearly sounds as if it is not. If it is current limiting rather than thermal limiting, then it is not suitable.

Cheers, boys. I feel just a tad more knowledgeable.

What is the stall current for that motor? Post a link to the datasheet or seller's page.

outsider:
What is the stall current for that motor? Post a link to the datasheet or seller's page.

I bought a batch of 25 unmarked motors from a garage sale for 5 bucks. They came with nothing other than the bag they were in.

Just tested what you asked for; at 5V, stall current is 1.10A. At 10V, stall current is .95A. At 12V, stall current is .70A.

toxicxarrow:
Just tested what you asked for; at 5V, stall current is 1.10A. At 10V, stall current is .95A. At 12V, stall current is .70A.

I'd check that statement if I were you. :astonished:

toxicxarrow:
Just tested what you asked for; at 5V, stall current is 1.10A. At 10V, stall current is .95A. At 12V, stall current is .70A.

That’s physically impossible.

The stall current is proportional to the supply voltage, its just the supply voltage divided by the winding
resistance.

So measure the resistance between the terminals.

Or clamp the motor spindle and measure the current (but very quickly, motors tend to cook rapidly on
stall).