# [sdiy] get 5V from 12V - DC-DC converter or 7805?

Brian Willoughby brianw at audiobanshee.com
Mon Dec 28 00:45:41 CET 2020

```The thing about linear regulators is not the current, so much as the wasted heat.

If I'm reading the various data sheets correctly, the maximum added current draw for a linear regulator is under 6 mA, and typically under 3 mA. Other than that small addition, the current draw on the 12V rail will be the same as the current draw at 5V. The problem is the 7V drop which, multiplied by the current, gives a wattage generation. That heat has to go somewhere (and, of course, the power supply has to provide that wasted wattage).

The advantage of switching is that the current can be cut in half. Even a switching regulator that's only 85% efficient can use only 49% of the current (i.e. the draw on the 12V rail is under half the draw at 5V). Switchers that run at 92% or even 96% efficiency can draw just 45% or 43% of the current. Rough calculations say that the wasted heat (wattage) is 1/7th at 85% efficiency, 1/12th at 92%, and 1/25th at 96% efficiency.

Capacitor pumps can boost, but I don't think they're good for buck. Inductive switchers can generate a lot of noise, both EM and on the rails. So, you'll have to do good layout to place the inductor near the regulator, with fat traces to handle the switching current locally.

You can add heavy filtering on the rails, but then you're sacrificing the efficiency. There's a good chance that by the time you filter out all the noise, you're no longer any more efficient than a linear regulator.

Personally, I wouldn't choose a 7805 pin-compatible module when you're making a custom board anyway, Just place a handful of discrete components and simplify the assembly.

I would agree with Tony: if the current draw is low enough, then don't worry about adding the challenges of a switching regulator. Based on my quick math, if your 5V draw is under 100 mA, then your savings will be under 50 mA draw from the 12V rail if you use a switcher.

Brian

On Dec 27, 2020, at 13:58, Florian Anwander <fanwander at mnet-online.de> wrote:
> I am at designing the pcb for a simple  MIDI-sequencer based on an arduino nano (the basic idea is derived from the JX-3P / SH-101 sequencer). I want to build it as a euro rack module, but want to have it also as standalone unit. So the pcb has the usual eurorack power connector and a socket for a wall wart. I have jumpers to select between 5V from the euro rack connector, or 5V from a regulator which is fed either by the 12V from the eurorack connector or by 9V-12V from a wall wart.
>
> At the moment I implemented the 5V conversion with a 7805. But I think this will cause a lot of "useless" current on the 12V rail of the euro rack. I thought about one of these DC-DC converters, but I am not sure how much (switching-)noise those can insert into the euro racks supply and ground. The module also has voltage inputs (for note length and velocity) and trigger/gate inputs, so electric relations with ground to other modules or external devices are quite likely, and I fear I transfer potential supply noise this way in the external signal paths.
>
> On the other hand the 7805 circuit is quite easy to implement, the DC-DC converter has to be bought as module or requires buying parts I won't have use for anywhere else.
>
> Does anyone have experience with those DC-DC converters?
>
> Regards
> Florian

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