# LM7805C alternative to produce less heat

I'm currently using LM7805 for circuit which requires 5V for MCU/misc ICs and draws 100-300mA current. Input voltage is 12V from car socket. I don't use heat sink due to space constraints(neither I want to lay down on PCB). Voltage regulator is producing too much heat I understand it's due to power dissipation of 12V to 5V. The question is how can I get this 5V with minimal overheating? Maybe some other common voltage regulator or other technique?

• 7805 comes in many packages. Which one are you using? Can you also check actual current draw from your 7805 regulator output. – Chetan Bhargava Feb 4 '13 at 23:55
• Or you could look at digikey.com/product-search/… if you don't want to change the PCB – PeterJ Feb 5 '13 at 0:14
• You may also use the to-3 package (LM7805K). It may use more area but you may not need a heatsink @300ma. – Chetan Bhargava Feb 5 '13 at 0:15
• How about to put some simple res. voltage divider in power socket to leave some heat there? Then I can get like 6-7V (or whatever value is better) and hopefully less heat on VR. Would you recommend some safe values for that method? – Pablo Feb 5 '13 at 0:24
• @Pablo The conventional understanding of a voltage divider will not apply in this case, as increase in current drawn from the tap-off of the divider will cause the voltage dissipated across the "upper" resistor to increase. So the divider will no longer be a static model. – Anindo Ghosh Feb 5 '13 at 5:37

The practical choice would be to ditch the LM7805C AND the dummy car socket adapter, for a usb charger. By outsourcing the voltage conversion to a car usb charger, you no longer need to worry about pcb space or heat buildup in your device.

You can get a small cigarette lighter sized plug for a dollar on ebay, five dollars in retail stores. A dollar store usb cable, either with a usb adaptor on your board, or just spliced, would plug into the charger without modification. And it would provide a pretty stable 5v from a switching regulator (Unless you get a super cheap knockoff).

The other option would be using diodes to drop the incoming voltage prior to the LM7805C. But the simplest thing is the usb charger.

If you can throw money at the problem, Murata has a switching regulator that is meant as a drop-in replacement for 7805.

Cost will be higher and ripple will be higher, but power loss will be much better.

The Murata part runs close to $10 in distribution. As Anindo points out in his comment, if you can afford a board layout change you can switching regulator modules with similar performance at considerably lower prices. • +1 for integrated buck devices. I use this (less than$1 each) on one design, and this (around $2 in volume) on another - not so too expensive, gets job done. – Anindo Ghosh Feb 5 '13 at 3:03 • If you budgeted for a$0.15 7805, then $1 might look pretty pricy; if your total BOM is over$50, then this solution is probably no problem. – The Photon Feb 5 '13 at 3:12
• @AnindoGhosh, that integrated buck converter looks very interesting, but still uses quite a few additional discrete components. Anyhow, even with those, I doubt that the BOM cost can go beyond \$1.50. – icarus74 Feb 5 '13 at 4:48
• @icarus74 Which of them? I linked two. Also, by discrete components do you mean the reservoir + filter caps? The SD200 datasheet states: No additional capacitor required for input/output. – Anindo Ghosh Feb 5 '13 at 4:50
• @Madmanguruman I like Murata a lot and I use their passives regularly. I do think this particular part is grossly overpriced, but it is a lifesaver when you are trying to fix a design where somebody totally forgot to think about where the power goes in a 7805. – The Photon Feb 5 '13 at 17:20

Minimal heat DC to DC conversion is done with switching regulators. These work by pulsing current into a coil which generates a voltage. The voltage is controlled via feedback, like in the linear regulator. Because inductance generates the output voltage, and because the device which energizes the coil switches on and off (never in between), there is very little heat dissipation.

If you have to do with what you have (the existing PCB), what you can do is help the 7805. Firstly, add a heat sink. Find the space! There is really no excuse.

Next, if that is not enough, try dropping the incoming voltage upstream from the circuit. You can drop some volts using a suitably valued resistor (which has to be selected with regard to the current, and also have an adequate dissipation rating) or perhaps a Zener diode. This voltage-dropping element will dissipate some of the heat, so that the 7805 then has to dissipate less of it. (Do not drop the voltage too far: the 7805 requires some dropout to operate properly.)

The consequence of letting the 7805 get hot is that it will limit current to protect itself from getting any hotter. When it limits current, your device will malfunction.

• I think a series resistor is a nice simple way to move heat away from a 7805 in cases where input voltage is much higher than output voltage. A 15-ohm 2W resistor would drop 4.5 volts at 300mA (leaving 7.5 volts out of 12, which should be plenty for the 7805). Worst-case power dissipation in the 7805 would be reduced from 2.1W at 300mA, to 817mW at 233mA (higher currents would dissipate more in the resistor and less in the 7805). Total dissipation would be unaffected. – supercat Feb 5 '13 at 17:42

If you stick with linear regulators, the excess voltage (7V in this case) will have to be dissipated somewhere; I don't think there's any way to escape this. However, you can share this heat across two regulators connected in cascade:

Use an LM7809 to produce a 9V regulated output; and then feed this signal to the input of LM7805. 7809 will drop about 3V and 7805 about 5V, as opposed to a single 7805 burning 7V.