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I have several projects (clocks) that require 12V and 5V supply. The 5V is for TTL ICs and the 12V powers a SMPS(Switching Mode Power Supply) to give HV(High Voltage) for a tube. The 12V pulls less than 100 mA, the 5V about 400 mA. The power source is a 15V 1A wall wart.

For my first project, I wired both the 7805 and 7812 with heatsinks to the 15V supply. The 7805 heatsink got to around 43.3C (110F). After the first build, I discovered the the heat produced by the 7805 depends on the voltage drop, and the current drawn, and I want to reduce the 10V being dropped over the 7805.

For my 2nd build, I was going to wire the 15V to the 7812, and run the 7805 off of 12V instead of 15V (giving 7V drop instead of 10V). I was wondering if there was a more efficient way to do this? Maybe using a Zener diode, or a voltage divider? I already have the 7805 and 7812 soldered, as well as the output caps, but I haven't soldered the inputs yet.

For the next clock, I am open to suggestions for using other regulator ICs.

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    \$\begingroup\$ Don't use 7805. Use a buck DC-DC converter. \$\endgroup\$ – Majenko Jul 14 '17 at 23:21
  • \$\begingroup\$ That is a good idea I will try to use for the next build. For the current build, is there any better way to reduce the voltage drop other than using the 7812 output instead of the 15V to power the 7805? \$\endgroup\$ – Efram Goldberg Jul 14 '17 at 23:25
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    \$\begingroup\$ Chainig is the only way. Maybe even add a 7809 as well. You can't magically vanish the excess voltage, only convert it to something else, typically heat. All you can hope is to move the heat around a bit. \$\endgroup\$ – Majenko Jul 14 '17 at 23:27
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    \$\begingroup\$ You could use a DC-DC down to 7v to feed the 7805 of course... \$\endgroup\$ – Majenko Jul 14 '17 at 23:29
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    \$\begingroup\$ Whatever you use to drop the 'excess' voltage, no matter if it's a zener, resistor, diode(s), another 78xx, whatever, is going to get hot in proportion to the voltage it's dropping (and the current flowing through it). The only practical way to avoid this is to use a buck DC-DC / SMPS. \$\endgroup\$ – brhans Jul 15 '17 at 2:53
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Simply use an SMPS replacement for the 7805. These are readily available ...here's a commercial variant that plugs into the TO 220 pinout you have with no heatsink required:

enter image description here

There are many variants available on Ebay or from Digikey/Mouser.

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you thought think about a smps for the 5v output.

alternatively, you can use a pre-regulator to off lay the bulk of the heat.

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  • \$\begingroup\$ I thought about a pre-regulator, maybe drop the 12V to 9V, then to 5V. For the current build, I am going to stick with the linear regulators. For the next one, I will try to use a SMPS for the supplies. \$\endgroup\$ – Efram Goldberg Jul 14 '17 at 23:31
  • \$\begingroup\$ without a pre-regulator, use a beefy heatsink on the 7805. \$\endgroup\$ – dannyf Jul 14 '17 at 23:33
  • \$\begingroup\$ if you use a pre-regulator, try to drop as much voltagee on the pre-regulator as you can. One way to do it is to float the 7805 so it drops a fixed voltage, 2-3v. anything else goes onto the pre-regulator. with a mosfet, you are pretty much done. \$\endgroup\$ – dannyf Jul 14 '17 at 23:34
  • \$\begingroup\$ linear regulators cannot reduce losses so learn to use smps regulators \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 15 '17 at 0:33
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The simplest way is to put a resistor between the 7805 power input and +15 volts. Choose the resistor value so that at maximum load the 7805 input is just above the dropout voltage level. Taking it a step further, put a resistor between +15V and +5V to supply the minimum load. You'll still get the same power dissipation but the 7805 will be cooler.

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The same question came up recently, here are a few 5V out buck DC-DC converters which will do the job.

https://electronics.stackexchange.com/a/317550/13616

They are readily available from the usual suspects (farnell, digikey, mouser...)

Also you could use a 12V supply and skip the 7812 altogether.

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