Level 2 newby with a few questions

[oobxlr8r]

Hi folks, I have a project I want to get started on that involves 4 dc motors 1.5-3v 270ma and one laser diode which I am assuming is 3v since it runs on 2 AAA batteries I have no idea on amperage as I have no details in that regard. If you haven't guessed it already, I'm trying to build a laser Spirograph.

So what's the big deal right? I want the motors to be variable of course, AND reversible. The device will be manual control only. As in no need for micro controllers or any other fancy doodads.

I know how to wire up a reversible system with DPDT switches, and would like to stick to that. And that's just about the only thing I know. So my question(s) to you guys;

Power supply: since I have 5 loads of 3v max each, how do I determine the voltage on the needed DC power supply? does this mean I need something in the 15v range, or will some kind of 3v power supply work? Do these different loads need to be wired up in series or parallel?

Potentiometers: since the motor speeds will be variable independent of each other, how do I calculate out what pots I'll need once I get the power issue figured out?

I'm actually trying to learn a few things here so I kind of don't just want parts numbers, more at what way(s) I can figure some of this stuff out.

 

[Harald Kapp]

You'll have to wire the components in parallel. Otherwise (in series) the same current would run through all components and you can't control the motors individually. So you'll have to build 4*3V motor control circuit all connected in parallel to a 3V source. Then each motor can be controlled individually.

To control direction you can use the dpdt switch, that's o.k.
To control speed by a potentiometer you will need a wire wound potentiometer that is able to carry 270mA. Finding this may be the hardest part. You could look for surplus controllers for slot racing cars or old-fashioned (meaning electromechanical) controllers for electric RC-cars. A standard potentiometer as used in electronic circuits will not be able to carry that current.

You'll have to inspect the laser, too.
If it is a module with integrated current regulation or current limiting you may be able to attach it directly to 3V. If, however, it is only a laser LED, you will need at least a current limiting resistor in series with the laser. The data sheet of the laser component should give you all the required information.

Harald

 

[oobxlr8r]

Ok so I need a pot to handle 270ma. This isn't a high amperage is it? I don't know what's high. When I went to radio shack I saw pots labeled in ohms, not milliamps (1kohm, 10kohm). How do I figure up what I need?

 

[KJ6EAD]

You won't find the type of potentiometer you need at RS. Pots don't have a current rating, they have a wattage rating and you need at least 2W pots, 5W would be a good choice.

Possibly something like this:

http://www.amazon.com/gp/aw/d/B002IABL6C

If you find that you don't ever need to run the motors at full speed, you can add a resistor in series with the pot to set the upper limit and reduce the stress on the pot.

 

[oobxlr8r]

You won't find the type of potentiometer you need at RS. Pots don't have a current rating, they have a wattage rating and you need at least 2W pots, 5W would be a good choice.

Possibly something like this:

http://www.amazon.com/gp/aw/d/B002IABL6C

If you find that you don't ever need to run the motors at full speed, you can add a resistor in series with the pot to set the upper limit and reduce the stress on the pot.

my question is, how do you come up with the figures of 2 and 5 watts? from the info given, how do you determine these valuse. and why would the 5 watt pot be better than the 2 watt pot?

as previously stated i'm trying to learn just in case i'd like to build something similar in the near or far future. do you try to figure it out using ohms law same as you would a resistor???

 

[(*steve*)]

The power rating for a potentiometer is predicated on the fact that the entire resistive element is used to dissipate the heat.

Effectively it becomes a current limit.

To figure out the maximum current, use the current and resistance, and the relationship P=I^2xR. So for 270mA and a resistance of (let's say 500 ohms), the power rating required would be 0.27*0.27*500 = 36.5 watts.

 

[CocaCola]

To figure out the maximum current, use the current and resistance, and the relationship P=I^2xR. So for 270mA and a resistance of (let's say 500 ohms), the power rating required would be 0.27*0.27*500 = 36.5 watts.

Steve, where do you come up with the 500 Ohm in that equation? I'm trying to wrap my head around the numbers you just pushed and I can't see you needing anywhere near a 36.5 watt pot to do this job, more like the 3-5 watt suggested... Requiring a 36.5 volt pot to restrict 3v to a 270mA motor seems way off, I have used fixed 1/2 watt and 1 watt fixed resistors to tame down vibration motors many times, that are running about the same ratings as the motors listed above... The resistors hardly get warm, and are certainly at no risk of dying anytime soon as would be expected if a 35 watt or so was needed...

 

[oobxlr8r]

The power rating for a potentiometer is predicated on the fact that the entire resistive element is used to dissipate the heat.

Effectively it becomes a current limit.

To figure out the maximum current, use the current and resistance, and the relationship P=I^2xR. So for 270mA and a resistance of (let's say 500 ohms), the power rating required would be 0.27*0.27*500 = 36.5 watts.

So from the looks of it, I need to figure out how much resistance I need to know what pot I'll need to start looking for, right?

He posted a link to a 100ohm pot rated for 5 watts. Using the power calculation you posted, it's underrated by 2.29 watts, is that going to be a problem? And how do I go about figuring up how much resistance I need? Do I have to determine a minimum amount of volts I want to send to the motors? Let's say a minimum of 1 volt.

If I use ohms law to go from 3volts to 1 volt is a 2 volt drop so I would say 2v/.270A= 7.40ohm to go from 3v down to 1volt... So at these values I'd need to calculate a different power rating for this given resistance. P=.27*.27*7.4=.54 watt rating.

I know that 1/2watt pots are pretty common, but to play it safe I should get one rated for more wattage.
But can you find a pot with a resistance of only 7.4ohms?

EDIT: after doing some searching, even for a 10 ohm pot, you just cannot find it, well not one of a physical size that I a, looking for. If I go with the suggested pot of 100 ohms, clearly that is too much resistance right? I mean doing the math in my head that'll mean a voltage drop of 27volts... While It will give me a greater range in speeds, I'm only working with 3v max here.

Let me know if I have anything wron here.

 

[KJ6EAD]

my question is, how do you come up with the figures of 2 and 5 watts? from the info given, how do you determine these valuse. and why would the 5 watt pot be better than the 2 watt pot?

as previously stated i'm trying to learn just in case i'd like to build something similar in the near or far future. do you try to figure it out using ohms law same as you would a resistor???

It's all Ohm's law applied. I took your maximum current and voltage and multiplied them together to get a wattage figure for the motor (0.27A X 3V = 0.81W). Because a resistance in series with the motor will divide voltage and have equal current this wattage is the maximum that the resistive element in the pot will be exposed to. I then rounded up to the nearest full Watt and doubled as a derating to keep the temperature more reasonable and increase the life of the pot. The 5W pot would be better because it will run much cooler.

To come up with a resistance value for the pot I calculated the effective resistance of the motor and multiplied by 10 then rounded. You probably could use a 50Ω pot but 100Ω is a safe starting point for the motor with no load.

I'm not sure what happened with Steve's calculations, possibly the old trap of substituting the control element value for the load value, but in any case you'd have to supply 135V to push 270mA through 500Ω even if the motor were a dead short. If you use the effective resistance of the motor in his formula you get the same results I've had.

 

[CocaCola]

It's all Ohm's law applied. I took your maximum current and voltage and multiplied them together to get a wattage figure for the motor (0.27A X 3V = 0.81W). Because a resistance in series with the motor will divide voltage and have equal current this wattage is the maximum that the resistive element in the pot will be exposed to. I then rounded up to the nearest full Watt and doubled as a derating to keep the temperature more reasonable and increase the life of the pot. The 5W pot would be better because it will run much cooler.

That is how I would have done it (and have done it) as well, that is why I'm perplexed as to Steve's answer above

 

[KJ6EAD]

See my edited post above for more explanation and a speculation.

 

[oobxlr8r]

That is how I would have done it (and have done it) as well, that is why I'm perplexed as to Steve's answer above

Oh boy...

Ok soon my last post is not useable? I mean I thought I had it.

 

[(*steve*)]

Steve, where do you come up with the 500 Ohm in that equation?

I just pulled 500 ohms out of my... ear.

This is a young player error that causes smoke to come out of pots. Note that some high power devices may allow for additional dissipation at low resistances, but this is NOT a general rule.

Imagine you have a resistor that can dissipate 1W. If we were to cut it in half, how much power can it dissipate? (Answer 1/2 Watt). The resistance will also drop to half the value, and Ohms law tells us (quite sensibly) that the current remains the same.

This applies to potentiometers. If you have a pot rated for 1W, then that 1W is assumed to be dissipated over the entire track. As you reduce the resistance, and the length of track dissipating power decreases, the amount of power you can reliably dissipate also drops.

Imagine the 3W pot dissipating 3W over 5% of the length of the resistance track (that is designed to dissipate 5% of 3W). It may not be happy.

If you're using a pot where the current through it squared exceeds the rated wattage divided by the total resistance (I^2 > P/R) then you really need to refer more closely to the specifications to determine if you're going to cook it.

To some extent, thermal conduction into the un-used section of track may compensate a little, but it's not something that you should make assumptions about.

See here.

Power rating: Based on 25°C free air rating. The stated wattage rating applies only when the entire resistance is in the circuit. Setting the lug at an intermediate point reduces the wattage rating by approximately the same proportion. Example: If the lug is set at half resistance, the wattage is reduced by approximately one-half.

 

[(*steve*)]

So, for 50 ohms, you would need a 3.6W pot, and 100 ohms, 7.2W

 

[CocaCola]

Steve thanks for the clarification, it makes sense now and clicks as it should... Before it was just one big huh, and for the life of me none of it made sense especially that 500 value that just appeared

It also explains why my real world use of fixed resistors and their chosen Watt values, didn't coincide with your math, as they obviously don't have the same degrading Wattage rating between values, thus no need to be concerned with such...

 

[oobxlr8r]

I just pulled 500 ohms out of my... ear.

This is a young player error that causes smoke to come out of pots. Note that some high power devices may allow for additional dissipation at low resistances, but this is NOT a general rule.

Imagine you have a resistor that can dissipate 1W. If we were to cut it in half, how much power can it dissipate? (Answer 1/2 Watt). The resistance will also drop to half the value, and Ohms law tells us (quite sensibly) that the current remains the same.

This applies to potentiometers. If you have a pot rated for 1W, then that 1W is assumed to be dissipated over the entire track. As you reduce the resistance, and the length of track dissipating power decreases, the amount of power you can reliably dissipate also drops.

Imagine the 3W pot dissipating 3W over 5% of the length of the resistance track (that is designed to dissipate 5% of 3W). It may not be happy.

If you're using a pot where the current through it squared exceeds the rated wattage divided by the total resistance (I^2 > P/R) then you really need to refer more closely to the specifications to determine if you're going to cook it.

To some extent, thermal conduction into the un-used section of track may compensate a little, but it's not something that you should make assumptions about.

See here.

Power rating: Based on 25°C free air rating. The stated wattage rating applies only when the entire resistance is in the circuit. Setting the lug at an intermediate point reduces the wattage rating by approximately the same proportion. Example: If the lug is set at half resistance, the wattage is reduced by approximately one-half.

but wouldn't the wattage required be relative to the position along the track??? if it was a system rated for 3watts total and your setup was right on the money, as in no room for error, wouldn't the voltage requirement at 50% be 1.5watts???

I mean woulnd't the heat generated be directly related to the amount of resistance being applied to the current? so maybe at full resistance you woud need the 3watts, it won't at anything less right?

But back to the lecture at hand, my calculations back on page one??? any insight, input, concerns, critiques???

 

[(*steve*)]

I mean woulnd't the heat generated be directly related to the amount of resistance being applied to the current? so maybe at full resistance you woud need the 3watts, it won't at anything less right?

That's *exactly* the point.

However you specified a current, and since, if the current remains constant, the power dissipated in a resistor is proportional to the resistance, the issue of changing dissipation per unit length of the track does not come into play.

However in real life, you're likely to find that the current INCREASES as the resistance decreases. In order not to damage the pot, you need to rate it as if the maximum current (which typically occurs at minimum resistance) were flowing through the maximum resistance.

This can lead you to scarily large required power ratings and is a reason that you often find that it is more efficient to find another method (typically you use a small current through the pot to control a larger current through (say) a transistor.

 

[oobxlr8r]

That's *exactly* the point.

However you specified a current, and since, if the current remains constant, the power dissipated in a resistor is proportional to the resistance, the issue of changing dissipation per unit length of the track does not come into play.

However in real life, you're likely to find that the current INCREASES as the resistance decreases. In order not to damage the pot, you need to rate it as if the maximum current (which typically occurs at minimum resistance) were flowing through the maximum resistance.

This can lead you to scarily large required power ratings and is a reason that you often find that it is more efficient to find another method (typically you use a small current through the pot to control a larger current through (say) a transistor.

Ok, I think I understand what you're saying now. I don't thinknim ready to learn about transistor just yet as I'm still not 100% on figuring out what ohm rating on the pot I'll need. I know what was recommended but 100ohm seems like too much voltage drop.

The way I figured it up a 10ohm pot would do the job but I never got any confirmation on that... I thinknim just going to buy it and see if it works for what I want...

There is a 10ohm 5watt pot sold here that should cover all my wattage and restive requirements as far as I know.
http://www.electronicplus.com/content/ProductPage.asp?maincat=RE&subcat=RPO

 

[(*steve*)]

The way I figured it up a 10ohm pot would do the job but I never got any confirmation on that... I thinknim just going to buy it and see if it works for what I want...

There is a 10ohm 5watt pot sold here that should cover all my wattage and restive requirements as far as I know.
http://www.electronicplus.com/content/ProductPage.asp?maincat=RE&subcat=RPO

The best thing to do is to get some fixed resistors and see how they affect the motor speed. You could get (say) 4.7 ohms, 10 ohms, 22 ohms, and maybe 47 ohms -- get say 1W resistors. See what combinations of these resistors in series give you the various speeds you want (also see what happens under load).

Using resistors to limit a motor speed doesn't tend to work well, but going out to buy a pot before you know the resistance you need is even less optimal.

 

[oobxlr8r]

The best thing to do is to get some fixed resistors and see how they affect the motor speed. You could get (say) 4.7 ohms, 10 ohms, 22 ohms, and maybe 47 ohms -- get say 1W resistors. See what combinations of these resistors in series give you the various speeds you want (also see what happens under load).

Using resistors to limit a motor speed doesn't tend to work well, but going out to buy a pot before you know the resistance you need is even less optimal.

So is that the most optimal way of doing it? I mean the whole point of this thread was to find out if there was a way of doing this without wasting too much cash. I used the some of the formulas you guys gave to try to figure out what I might need. It's the reason I kept asking if my calculations were correct based on the formulas and information from the first page. I greatly appreciate all the information and help, I do have more insight into the matter now to make a better informed decision. But I feel like I got lost in the flexing of the minds on here. I guess it's a hazard of the Internet but it is why we come to forums.