Positive + negative linear regulator

Hello everyone,

Here I was trying to find a way to convert 24v (Industry default) to 5v for an atmega microcontroller.
I know you can use a buck converter, but that adds some costs, space, etc.. etc... You know.
The other option (Linear regulator from 24 to 5) is feasible, as the whole system I'm trying to implement uses only about 50ma, but what bothers me is the heat inside the case where the pcb goes. Also, wasting 19v into heat bothers me.
So it came to my mind about an option I couldn't find anywhere, and I don´t really know why I couldn~t find it.

So, my idea is to first use a -18 regulator:
Gnd + 24 ==> 7918 regulator ==> Gnd + 6v + 24v
Use the Gnd and 6v ==> 7905 Regulator ==> Gnd + 5V.

Like this, you waste only 7v instead of 19v, so, why no one suggests this? Is there any "catch"?

So, my idea is to first use a -18 regulator:
Gnd + 24 ==> 7918 regulator ==> Gnd + 6v + 24v
Use the Gnd and 6v ==> 7905 Regulator ==> Gnd + 5V.

Like this, you waste only 7v instead of 19v, so, why no one suggests this? Is there any “catch”?

The catch is it won’t do what you are expecting. I don’t understand why you think it will only ‘waste’ 7V; it will ‘waste’ 19V, just the same. If I understood why you are making this mistake I’d know what to explain, but I am lost on that point.

Do you know Ohms Law? Do you know the relationship between voltage, current, resistance and power? If not, you need to learn these things, they are basic electrical theory.

Use a buck converter, they are cheap.

When I say "waste 7v" I mean this:

With a 7918, I can make the Gnd and -24v become Gnd, -18 and -24, and by doing this you "wasted" 6v in the regulator resistor.

Now, if I use the -24 and -18 as inputs of a 7805, with the -24 as gnd and -18 as Vcc, I will get an output of -19 and will have wasted 1v in the regulator resistor

If I use the -24 and -19, I will have 5v and wasted 7v total.

It´s not the same as using a single 7805, because in this case the regulator will dissipate energy from the 24v all the way to 5v, wasting 19v.

I'll draw a circuit to make it more clear as soon as I get home.

I attached a diagram of what I mean.

Oh, Thanks!

So, still in the subject, suppose I´m drawing 1A of current
The heat dissipation on the 7918 will be 6w and on the 7805 will be 1w.

If I had a single 7805, the heat dissipation will be 19w.

Having a better look at it, I think I get what´s wrong... Will the 7918 even work if I don´t close the circuit of output to ground? I guess that´s the "catch".
Thinking about it now, it´s really silly if I see the current path, it goes from 0 to 0. Obviously not gonna work.

Using your scheme takes a lot more room than a buck converter because of the capacitors you HAVE to add to the circuit.

Paul

Hi Tadashimori,
I do not see that you understand that
circuit.

  1. The input to the 7918 should be -24V.
  2. You set its output at +6V, so there
    will be 30V across it. You do not
    have 30V.
  3. The actual output of the 7918 will
    be -18V.
  4. The 7805 needs at least +7V input
    for it to regulate. A negative input
    voltage will fry it!
  1. The input to the 7918 should be -24V.
  2. You set its output at +6V, so there
    will be 30V across it. You do not
    have 30V.
  3. The actual output of the 7918 will
    be -18V.

Well, in the circuit I made, the 7918 Gnd is at 24v and the input is 0, so, the input is -24 relative to the ground.
The output is indeed -18, but again, it´s just a matter of reference point. If I connect -24 and -18 I´ll have 6v difference, so I just put the ground reference at the -24.

  1. The 7805 needs at least +7V input
    for it to regulate. A negative input
    voltage will fry it!

The input is not negative in this case, but, yes, 6v wouldn't work. Actually, nothins in this circuit would work anyway.

If you start with 24V and end up with 5V using a linear regulator (or 2 or 3 or 23 linear regulators) then you have dropped 19V across the regulator(s) and the power wasted as heat will be 19 * the load current, divided among however many regulators you have.

You can't use a -ve regulator the way you have it; it expects a negative input and the load between its output and its ground, You have the load, the 7805, between its output and its input. It won't work like that.

I always encourage people to try stuff, if you still doubt what we are telling you get your breadboard out and try it, then come back and tell us the results.

In short, go get a switchmode "buck" regulator and learn the lessons later. It's not as if they are expensive! :cold_sweat:

Well, not expensive, but the component placing would be a tad bit more difficult for me, that's why I was checking other options.
Thanks for the tips anyway, I understood what was wrong.

First of all, don't quietly swap connections on the regulator symbols, only after I noticed that I understood what you had drawn! There's a reason we use standardised components with standardised connection locations... The way you draw it is confusing and can easily lead to errors later in the build.

tadashimori:
Well, not expensive, but the component placing would be a tad bit more difficult for me, that's why I was checking other options.
Thanks for the tips anyway, I understood what was wrong.

I don't understand how you think you can not place a small buck converter inside your case, but you think you can place two TO-220s with heat sinks! Even at your stated 50 mA you'd have to dissipate almost 1W of heat... Buck converters for such small currents are about as small as a single TO220 regulator and its capacitors, they're smaller the moment you add the heat sink.

As you already know your circuit cannot work. Negative regulators pass current from OUT to IN and positive regulators from IN to OUT (and regulate the voltage this way). In your schematic both regulators want to pass current from the "6V point" but there is no way for the current to flow into this point.

Smajdalf:

The 7805 requires at least 7V on its input to work.
The 7918 cannot source any current, it can only sink it.

If it did work (it will, sort of, but still wastes 19V) then the ground of the Arduino is now at +18V relative to the power supply ground. So you can't connect the Arduino to anything else powered by the same supply.

wvmarle:
First of all, don't quietly swap connections on the regulator symbols, only after I noticed that I understood what you had drawn! There's a reason we use standardised components with standardised connection locations... The way you draw it is confusing and can easily lead to errors later in the build.

I don't understand how you think you can not place a small buck converter inside your case, but you think you can place two TO-220s with heat sinks! Even at your stated 50 mA you'd have to dissipate almost 1W of heat... Buck converters for such small currents are about as small as a single TO220 regulator and its capacitors, they're smaller the moment you add the heat sink.

Haha, well, I did draw it quite quikly so didn't bother finding the standar, as long as I indicated where each pin is I thought it would be fine to understand. I'll take note of this.
I wasn´t thinking about using heat sinks, that´s why I wanted to use the two TO-220s. But, my main issue with switch regulators are the availability here. I can only find some like this: