Good Morning,

I am stuck on the cusp of understanding something fundamental to electronics but there's something that is just not clicking with me. I am trying to find the "AH HA!" moment from this.

Here is the scenario: I like to go camping and we use those re-fillable air mattresses. The problem is I don't like blowing them up using my lungs so a friend of mine gave me a battery operated blower (Coleman, 4x D cells). I'd like to use my car/solar setup to run the blower as I don't care too much for purchasing D cells.

So. Here's what I have done so far.
Step 1: I hooked it up to 12v and blew the thing apar... WAIT! just kidding

I know that D cells are ~1.5v and running them in series will give ~6v to the motor. I verified this by using my handy dandy volt meter just to make sure it wasn't doing something different.

Now, I suspect if I want to get this to work with a 12v source, I have to somehow split the 12v into 6v to the motor, and 6v to *magic land*.
So I purchased myself a bag of LM317T's to go with my extensive assortment of resistors and bread-boarded out the circuit. I have a 12v source coming in, and with some 1/8th watt resistors (750ohm & 500ohm) I have 6.35v at vOut.

Excellent I say to myself. And I promptly hook it up to the electric motor and it begins to spin. Then I touch the LM317T (because I'm a curious sort) and promptly unplug the setup due to the (what I assume) is a catastrophic melt down about to commence as it was hotter than a hot thing. I then get the idea that what it needs is a heat-sink. So I use thermal paste and a spare aluminum sink and attach it to the LM317T and run the experiment again. As expected, the motor spun, and (also as expected) the heat-sink got hot and I had to end the run after a dozen seconds or so due to a substantial dislike of the smell of burning electronic components. I know I was able to run it for a bit longer but that's 100% due to he heat dissipation characteristics of aluminum.

I then thought it might be prudent to know exactly how much current I am drawing with this little motor as it might be a good bit of info to have. Unfortunately I only have somewhat dead-ish D cells to work with (1.32v,1.46v,1.46v,1.38v = 5.62v @ 1.7A)

Upon inspecting the datasheet for the LM317T I learned that it is good to produce "outputs of over 1.5A".

So I am obviously over driving this component but it makes me think... why is it producing so much heat? Within seconds of turning the blower on, I have to unplug it because its so dang hot. This is where my brain starts to garble. Ultimately I want to make this blower work with 12v, but I MORE want to understand what is going on rather being told how to fix it or "there's a better way of doing that".

I want to try to understand why heat is generated when you drop voltage from a higher source to a lower one. Using my particular example, I'd like to understand the heat generation and then, at the end, figure out a way of either fixing this circuit so it doesn't get so hot, or "do it the better way"

Thanks for the long read, I appreciate you helping me on my journey.

 

Welcome to AAC.

So to make things clear the motor is taking 1.7A at ~6V. OK
Then the LM317 cannot supply that without going into thermal shutdown.

The regulator needs to drop 6V...(12-6 = 6 ) @ say for arguments sake 1.5Amps. i.e 6V *1.5 A = 9W. The regulator dissipates 9W as heat and with a big heat sink and forced cooling it can but not if the current draw is more than 1.5A.

what you need is a current amplifier. A transistor to handle the large current while the LM is doing light work.

I will post a a circuit in a while that you can experiment.

 

See the pdf attached.
On page 17, fig: 23 is the circuit you can try.

Here the TIP transistor will lift the heavy load. You need to mount that on a heatsink.
The bigger the transistor the better in the long run.

 

Thanks very much for the reply. I very much appreciate the circuit (in advance).

So would I be wrong to assume that both the motor and the LM (in this scenario) are consuming 9W each? When it's running, it will be consuming 18W from the battery? Or is the total 9W?

Zach

 

As said if the load is 1.5A at 6V resulting in 9W, and with 12V input, YES. Source will supply 18W (12V * 1.5A ).
9W is wasted.
This is typical for linear regulators.
If you wanna save power you should go for DC to DC, Buck converters.
They are not DIY type circuits.

Just for blowing up something you can spare the waste I believe

 

In a series circuit, for each component you can multiply the voltage drop by the current through the device to get the power dissipated in watts. Semiconductors, like the LM317, always dissipate about the same amount of power for a given configuration. A motor on the other hand dissipates way more power when it starts up than when it is running. Depending on the type of motor you have you may want to reevaluate the need to have a regulated voltage for the motor. Most motors don't care about voltage, what they care about is current.

 

If you wanna save power you should go for DC to DC, Buck converters.
They are not DIY type circuits.

It's worth noting that you can buy a DC-DC converter very inexpensively these days. You can find them on eBay as modules, or you might consider using a 12V USB adpapter. I have one rated to 3A, more than enough, and I know it was only $5 or so. This would be only 5V but the pump would run fine on that. The adapter could be used for other functions. A lot of bang for the buck.

 

@wayneh
One problem.....!
Might not start the motor due to high startup current.

 

I think it would be just as cheap to by a 12 volts pump as making a regulator to drop 12 volts down to 6 volts.
This is one example on Ebay.

Les.

 

@R!f@@ Fantastic! Thank you so much.
I think I understand it now.
The dissipation is spread out over the individual loads to get the voltage down to what the motor requires.
I'm going to continue to experiment to make sure I get this in my head. I'm going to use smaller motors (3v, 1.5v, hobby motors) with the same 12v source and the LM317's to see if I can fully understand it and predict the outcome before measuring.

As far as waste goes, you're correct. I put up solar panels and bring spare SLA batteries for music, lights, electronics charging, etc... so I don't worry about the waste. What I am concerned with is having a circuit that doesn't burn itself out as soon as it turns on due to current draw.

I don't have any TIP transistors (that I know of, I might if I look at some older electronics I have laying around) but I will definitely buy some, just because its good to have, and they're small

@Papabravo - I have a few spare DC-DC Boost converters for my 20w LED's. (the ebay ones with dual heat sinks and the large toroid inductor in the middle) I think I'm going to try using that and turning it way down to see if it is able to do the job. I think they are buck converters... Definitely looking into that this evening.

@wayneh I have a 3A dual USB car charger but have been hesitant to use it as a supply for the blower because I thought it would overheat.

 

I would look for a 5A, 12V to 6V converter, perhaps one of these.
That should give sufficient current so there's no problem starting the motor and it will solve the heat problem.

 

Thanks for the purchase suggestions everyone, I will look into it.

I was more concerned with actually understanding the why's of the thing. Teach me to fish

 

By the way any NPN transistor with enough current capacity and power will do the job.
TIP transistor is not a must.

 

I would look for a 5A, 12V to 6V converter, perhaps one of these.
That should give sufficient current so there's no problem starting the motor and it will solve the heat problem.

I used those and they are not as advertised.
You cannot load a 3A one to 2A without burning the diodes.
They are just junk.
But of course at 20% of stated capacity they are good

 

I was more concerned with actually understanding the why's of the thing.

To summarize, any linear method (linear regulator, resistor, etc.) of reducing the voltage generates heat.
The heat generated (in watts) equals (Vin-Vout) × Iout, where Vin is the input voltage, Vout is the output voltage, and Iout is the output current to the load.
Thus when going from 12V to 6V the efficiency is 50% meaning as much energy is dissipated in heat as energy going to the load.

The only way to get around that is to use a Switching type converter, which can convert voltage levels with much higher efficiency (85%-90% typical).
This method generally uses an inductance to temporarily store the energy, allowing it to be near losslessly converted from one voltage level to another.
The description, Switching regulator, comes from the energy being switched into and out of the inductance at a rapid rate using a semiconductor switch (transistor) to effect the conversion from one voltage to another.

 

@crutschow Thank you as well, that also explains another question I had before I even asked it.

As a thought experiment (I'm not near my equipment right now for testing) ...

Can I run two LM317's circuits in parallel? Would that halve the watts each of them had to dissipate, or would that effectively crate 3 loads of equal value on the entire 12v source? (I may be using terms incorrectly)

 

Can I run two LM317's circuits in parallel? Would that halve the watts each of them had to dissipate

Yes, if their loads were properly balanced by a small resistor on each output (otherwise one will tend to hog the load), each would dissipate half the power, with the total power being unchanged.
But you still have to dissipate the same amount of total power.

 

Ah, good. I was hoping.
Again, this is me just experimenting and learning.

When you say "small resistor on each output" are you talking 100ohm or less? I'm trying to get perspective on "small".

Thanks.

 

I'm trying to get perspective on "small".

In the neighborhood of a tenth of an ohm.

 

Oh...
hmm. I don't think I have any that small lying about.
I do, however, have some Kanthal Wire. I could coil that and make a sub-ohm *resistor* (inductor with resistance)...
The output resistor's won't have to dissipate much heat, will they?