3
\$\begingroup\$

I have a battery that's designed to power a MacBook Pro (it's not an internal battery, it's an external battery meant to act in place of a charger, so it's got all the monitoring electronics internally).

Now, I wish to use this battery to power an iPhone dock. This dock has internal batteries itself, but they run out fast (only 3.6V, 1.8Ah), whilst this MacBook battery can provide 60Wh - which would increase the duration of the playback by many times, and stop the voltage dropping etc.

The only issue is that the iPhone dock DC in wants a 12V, 2A supply, whilst the Macbook battery will supply between 14.5 and 18.5V (4.5A max). I don't wish to play with fire by putting in too high a voltage (there could be no damage, I might have to buy a new dock), so I was hoping to regulate the voltage of the MacBook battery so that it could safely power the iPod dock.

I've been reading 'A designer's guide to the L200C voltage regulator' and it seems to ideal for this purpose, but some aspects seem a bit strange. Can I get someone to answer the questions I have with this circuit? The ST datasheet is here.

enter image description here

Firstly, I'm not sure what the black rectangle at the bottom is. It seems like one plate of a capacitor, but that can't be right... Can that be explained?

Secondly, am I to attach the positive lead of the battery to the top left line and the negative lead to the bottom left line? Is it the same for the output? (top right for positive, bottom right for negative?

Finally, I'm not really sure what Vref means in all of the equations in the datasheet. The values for R1 and R2 given in the documentation for a 12V output are 1 kilo-Ohm for R1, and 3.3kW for R2 (I assume that's a typo, and they mean kilo-Ohm). Do these values seem reasonable, and should I have R2 as a potentiometer (If so, what sort of potentiometer?)

I'm sorry to sound really needy - I wish to get this working, and I also have barely any knowledge of electronics!

\$\endgroup\$
  • 1
    \$\begingroup\$ Stevenh - interesting interaction there - we were both simultaneously editing to add diagram. You "won". \$\endgroup\$ – Russell McMahon Jun 28 '12 at 15:15
  • \$\begingroup\$ @Russell - Ha! And rightly so! ;-) Well, Kortuk and others have been quicker than me a few times too. \$\endgroup\$ – stevenvh Jun 28 '12 at 15:18
  • \$\begingroup\$ @Russell - By the way, it's 20 past 5 in the afternoon here in Belgium. Shouldn't you be in bed? :-) \$\endgroup\$ – stevenvh Jun 28 '12 at 15:22
  • \$\begingroup\$ @stevenvh Answering questions is the best thing to help fall asleep! \$\endgroup\$ – Kortuk Jun 28 '12 at 18:42
  • \$\begingroup\$ @Kortuk - For the answerer or the reader? :-) \$\endgroup\$ – stevenvh Jun 28 '12 at 18:59
4
\$\begingroup\$

The black rectangle is the symbol for ground, the reference level against which all the rest is measured.

Bottom left and bottom right are connected to ground, and are the zero level of input and output. Some people say the minus, but it isn't really negative, so that's kind of a misnomer. Positive input goes to top left, positive output is top right.

Vref is specified in the datasheet, and is 2.77 V typical (page 3). R2 and R1 form a voltage divider, and the L200 will regulate the output so that on pin 4 the 2.77 V appears. That's where the equation top right of the image comes from. So if R1 = 1 kΩ and R2 = 3.3 kΩ then Vout = (3.3 + 1)/1 * 2.77 V = 11.9 V. (The equation says it's Io, for output current, but that seems to be a typo, since the dimension of the RHS is volt.)

You should keep an eye on the input-output difference. Most regulators need a few volts difference, and the L200 is no exception. On page 3 you can read that it can be as high as 2.5 V. Then an input voltage of for instance 14 V may only give 11.5 V out.

edit
Something about power. The L200 and the Micrel are examples of linear regulators, and one of their properties is that the current through them is the same as the current of the load, so that's 2 A. I was keeping an eye on the 14.5 V minimum, and I lost sight of the 18.5 V maximum for a moment. If the input is 18.5 V then there's 6.5 V difference between in and out, times the 2 A flowing through it is 13 W. That's a lot more than any regulator can handle unaided. Aided means a heatsink. At 13 W dissipation the Aavid-Thermalloy 5336XX's temperature will rise by 45 °C. That's already too hot too touch.
There's another alternative, which is not a linear regulator, but a switching regulator. This will only dissipate a few watt, and can do with a modest heatsink. But it's a bit more complex than a linear regulator, and since you already didn't like the L200's resistors this is not something I would let you make yourself. Besides, it requires some experience in switchers to do it properly too.
Small ready-made modules exist for lower power, like this one

enter image description here

which can supply 500 mA. A 12 V/2 A model will be a bit larger, and have a small heatsink for the regulator. I'll look around and see if I can find one.

\$\endgroup\$
  • \$\begingroup\$ The Micrel MIC29300-12BT may be an interesting alternative. Has a fixed 12 V out (no R1, R2), can supply 3 A, but has a much lower dropout voltage: it needs only half a volt between input and output. \$\endgroup\$ – stevenvh Jun 28 '12 at 15:21
  • \$\begingroup\$ So, if I were to use the Micrel part, would I just be able to attach the battery input to one pin, have the output go to the speakers, and connect the grounds from both the Micrel and the speakers back to the battery? Would it be that simple? \$\endgroup\$ – Bakes Jun 28 '12 at 15:38
  • \$\begingroup\$ @Bakes - Yes, but you'll have to add a capacitor to both input and output, just like you would have to on the L200. Look at the schematic bottom left on page 1 of the datasheet. That's all you need for a fixed voltage output. \$\endgroup\$ – stevenvh Jun 28 '12 at 15:43
  • \$\begingroup\$ Many thanks for the help, it's been most useful. I think I'm going to go for the Micrel part, since it seems like less of a hassle (It's also about a quarter of the price). \$\endgroup\$ – Bakes Jun 28 '12 at 15:44
  • 1
    \$\begingroup\$ Finally, what value capacitor do you think I'd need? British A level physics really does not prepare you in any way for actually using what you learn about! \$\endgroup\$ – Bakes Jun 28 '12 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.