Lots of discarded electronics has things like 7805s and white LEDs in it. White LEDs need about 3 volts to light, and the bright illumination ones have a substantially less exponential V–I curve than ordinary indicator LEDs (because of a larger ohmic resistance component and special magic I don’t understand to prevent thermal runaway) so it’s typical to hook two of them in series up to a 6-volt power supply. But they’re not so ohmic that it’s a good idea to hook them up to, say, a 9-volt or 12-volt supply, or that they’ll work on a 5-volt supply. (8 volts — 4 volts per LED — does seem to work with the 6-volt lighting panels I picked up off the sidewalk. At 12 volts — 6 volts per LED — my 12-volt 500-mA switching power supply detects an overload and turns off before burning out any LEDs.)
There is a 7806 6-volt voltage regulator, but it’s much less common than the 7805 and the 7812. But it’s not actually that hard to get a regulated 6 volts out of a 7805.
The 7805 is a three-pin regulator which reduces, according to Fairchild’s datasheet, 7–35 Vdc to 5 Vdc ±4% at up to 1 A (1.5 A from TI) as long as it’s below 125° (150° from TI), with 5°/W of thermal resistance from the junction to the case (3°/W from TI), and so with adequate heatsinking it can dissipate 20 watts or more. It regulates its output pin to 5 V above its ground pin, which sources up to 8 mA. Bypassing with 0.33μF on input and 0.1μF on output is suggested but not generally needed.
The idea is that you float the 7805’s “ground” pin 1 V above the real ground, using a 5:1 voltage divider. This is shown on p. 23 of Fairchild’s datasheet. If we’re satisfied for the 8 mA regulator current through the voltage divider to produce an error of 0.2 V, which is comparable to the 4% error of the regulator itself, the divider circuit needs to be fairly low resistance: 25 Ω down to real ground and 125 Ω up to the output pin. The 125-Ω resistor will be dissipating 200 mW, so you need to use at least a ¼-watt resistor. (Alternatively, you can use an op-amp buffer to set the “ground” voltage for lower power consumption and error, as shown on p. 24.)
In this case, though, you could probably tolerate a larger error and use, say, a 100-Ω resistor and a 330-Ω resistor. 15 mA through the 330-Ω resistor plus 4–8 mA of bias current through the ground pin put 19–23 mA and thus 1.9–2.3 V of offset on the 100-Ω resistor, giving 6.9–7.3 V on the output, or 6.7–7.5 V if we include the 7805’s own error too.
If 500 mA (a guess) is running through the 6-volt LED strings, they’ll be dissipating 3 watts, which is not a big problem since they’re spread out over a large area. If the 7805 is dropping 12 volts (-1) down to 5 volts, it will also be dissipating 3 watts. Though it’s specced to operate at 125°, I’d kind of want to keep it below 70° so it will last longer and I’m not at risk of burning my hand on it, and in environmental temperatures up to 35° here in Buenos Aires, that only leaves ΔT = 35°, so we only have a thermal resistance budget of 11.7°/W, of which the junction-to-case resistance already eats up 5°/W. So we need 6.7°/W or less between the heatsink and its coupling to the TO-220 case.
The heatsink tab is, for better or for worse, connected to the package’s ground pin, since that’s the most negative voltage in the circuit.
A transistor switch (maybe controlled with a second transistor), a small inductor and a small capacitor for an LC filter, a microcontroller, and possibly a resistor or two for analog feedback from the output is a better solution. You can totally run the microcontroller off your 7805, and you get 95+% efficiency (instead of the 50% described here). The ±4% precision of the 7805 may or may not be available from a microcontroller without trimming, but it’s plenty good enough for this. For example, the ATMega328P datasheet specifies its internal bandgap voltage reference as 1 V ±10%, and its ADC contributes about an additional ±1% error; the STM32F103xx is specified to be better with only ±3% error on its reference and ±0.3% ADC error.
Additionally, this gives you programmability and dimming up to a few hundred kHz for free. The dimming is only voltage dimming, and so highly nonlinear for LED illumination.
The downside of this is that you lose the overheating, current-limiting, and foldback protections built into the “virtually indestructible” 7805.
LEDs can general tolerate higher pulsed power than constant power, so the LC filter might seem unnecessary. But if you’re running off a 12-volt or 19-volt DC supply, I’m not confident that even short current pulses through the LEDs won’t destroy the LEDs or the switching transistor, and of course the shortness of the pulses is a function of the microcontroller software.
If you’re dimming at 200 kHz via a IRLML6402 (a 40¢ P-channel power MOSFET rated for 20 V and 3.7 A; see My attempt to learn about jellybean electronic components), you’re dumping 12 nC from its gate to ground every time you turn it on; that might be about 12 volts. That’s 2.4 mA of gate switching losses at 12 V: 30 mW, about 100× less than the linear 7805. Its 65 mΩ on-resistance will dissipate another 16 mW at 500 mA. So you don’t gain much from using a substantially lower frequency like 20 kHz, right?
No, wait — the datasheet also lists a 48 ns rise time and a 381 ns fall time, so each 5-μs-period pulse includes an 0.43-μs transition time during which the on-resistance is significant. If we simplify its behavior to a linear ramp up of current from 0 to 500 mA during that time, while its drain–source voltage linearly drops from 12 V to 0, its linear power consumption follows a parabola from 0 to 0 during that time, with a peak at 1.5 W in the middle. So it might also have some 90 mW or so of switching losses just from being slow. (Its real behavior is somewhat more complex, but that toy model is probably adequate for our purposes here.)
So, although power consumption is acceptable at around 125 mW, you actually might improve it substantially by dropping to 100 kHz or 50 kHz or something, or using a beefier MOSFET.