# Better to choose a switching voltage regulator or a fixed voltage regulator

I am designing a circuit which has a 24V input. The circuit however uses 8V and 12V supplies. This whole circuit requires approximately 1A.

I have decided on using fixed voltage regulators that give the respective outputs with nothing much to add apart from a few capacitors at the input and output of these regulators. I have the 12V voltage regulator connected to the 24V input, and the 8V is connected to this 12V voltage regulator. My question is, if this is an appropriate circuit design or would it better to use a switching regulator? This circuit is used to drive a ZVS system and the subsequent circuit is switching at 125kHz.

i am using LM7812SX for 24V to 12V and a UA7808CKTTR for 12V to 8V

Edit: This is the whole circuit. The circuit drives a ZVS circuit.

Just to add more information. The 24V to 12V voltage regulator should be able to flow upto 1A, the 12V to 8V regulator has less than 0.1A. (I had connected these points to a power supply and measured how much current flows at these two points).

• You can get fixed switching regulators. I think you're debating between LDO style vs switching style regulators? Commented Sep 8, 2023 at 10:05
• A linear regulator will generate 12W of heat dropping 24V to 12V at 1A. That's a massive waste and an engineering problem to get rid of it. Commented Sep 8, 2023 at 10:16
• @Puffafish, umm, both the regulators are fixed voltage regulator. I am debating whether to use this or go for a switching regulator such as L5973D. im worried as this might actually add noise to the circuit, but i feel the fixed voltage regulators will generate a lot of heat. Commented Sep 8, 2023 at 10:32
• @Finbarr. yeah that was my concern. Hence the question as to what would be a better fit for this applicaiton. Commented Sep 8, 2023 at 10:33
• There are two types of regulator, switching and linear. Both types come in fixed and variable (also called "programmable") variants. Your question should read "Better to choose a switching voltage regulator or a linear voltage regulator" Commented Sep 8, 2023 at 14:55

The added complexity of a switching regulator is worth it if you have a large voltage difference to span between input and output. This is because switching regulators "bridge" that voltage gap in a very different, more efficient, way than linear regulators (which is what I think you mean when you say "fixed").

The two types perform the same function, to derive a certain voltage $$\V_{OUT}\$$ at their output, given a certain voltage $$\V_{IN}\$$ at their input. Output current $$\I_{OUT}\$$ is decided by the load, not the regulator, and is therefore the same in both cases.

The difference between linear and switching regulators concerns input current $$\I_{IN}\$$. A Linear regulator's input and output currents are equal (almost):

simulate this circuit – Schematic created using CircuitLab

Power in and power out for a linear regulator are:

$$P_{IN} = I_{IN} \times V_{IN}$$ $$P_{OUT} = I_{OUT} \times V_{OUT}$$

By conservation of energy, whatever input power is not appearing at the output must be dissipated by the regulator itself. In other words, power dissipated by the regulator is the difference between output and input power. Remembering that for a linear regulator input current and output current are equal, $$\I_{IN} = I_{OUT}\$$, and determined by the load:

$$P_{REG} = P_{OUT} - P_{IN} = I_{IN}V_{IN} - I_{OUT}V_{OUT} = I_{OUT}(V_{IN}-V_{OUT})$$

For your requirements, where $$\V_{IN}=24V\$$, $$\V_{OUT}=12V\$$ and $$\I_{OUT}=1A\$$:

$$P_{REG} = 1A \times (24V - 12V) = 12W$$

That's a lot of power, and without serious heat-sinking (and probably forced-air cooling) would destroy a small TO220 linear regulator very quickly if it didn't shut itself down first.

A switching regulator works by transferring tiny packets of energy from the source of input voltage, very rapidly, very frequently, and very efficiently to a capacitor at the output, doing so until the energy accumulated in the capacitor is exactly enough to produce exactly the required output voltage. It then adjusts the rate at which energy is transferred from input to output to exactly match the rate at which energy is removed from the capacitor.

This means that input power is equal to output power, which is 100% efficient in an ideal switching regulator, and therefore the regulator does not dissipate any power itself. Of course, they are never really 100% efficient; typically they are between 80% and 98% efficient in reality.

With some efficiency factor $$\E\$$ (0 is 0%, to 1 is 100%), output power, input power, and regulator heating power are related like this:

\begin{aligned} P_{OUT} &= E \times P_{IN} \\ \\ P_{IN} &= \frac{P_{OUT}}{E} \\ \\ P_{REG} &= P_{IN}-P_{OUT} \\ \\ &= \frac{P_{OUT}}{E}-P_{OUT} \\ \\ &= P_{OUT}\left(\frac{1}{E}-1\right) \\ \\ \end{aligned}

Input current will also depend on efficiency:

\begin{aligned} I_{IN} &= \frac{P_{IN}}{V_{IN}} \\ \\ &= \frac{P_{OUT}}{E \times V_{IN}} \\ \\ \end{aligned}

For your requirements, where $$\V_{IN}=24V\$$, $$\V_{OUT}=12V\$$ and $$\I_{OUT}=1A\$$, and regulator efficiency $$\E=90\%\$$:

\begin{aligned} P_{OUT} &= I_{OUT} \times V_{OUT} \\ \\ &= 1A \times 12V \\ \\ &= 12W \\ \\ P_{REG} &= P_{OUT}\left(\frac{1}{E}-1\right) \\ \\ &= 12W \times \left(\frac{1}{0.9}-1\right) \\ \\ &= 1.3W \end{aligned}

Maximum input current will be:

\begin{aligned} I_{IN} &= \frac{P_{OUT}}{E \times V_{IN}} \\ \\ &= \frac{12W}{0.9 \times 24V} \\ \\ &= 0.56A \end{aligned}

simulate this circuit

Clearly then, for the conversion from 24V to 12V, a switching regulator dissipating 1.3W is preferable to a linear regulator dissipating 12W. If you also derived 8V from that same 24V source, you should also use a switching regulator for that.

However, since the biggest obstacle is the difference between input and output voltages, you should consider deriving 8V from your existing 12V source. The setup would look like this:

simulate this circuit

Here we ask a little more of the switching regulator; not only do we need it to supply up to 1A to the 12V load, we also require an additional 0.1A for the 8V load, for a total of 1.1A.

This increases input current and switching regulator power dissipation a little, but it's a small price to pay for the simplicity of using a linear regulator to obtain 8V.

The power dissipated by the 7808 linear regulator is:

$$P_{REG(8)} = 0.1A \times (12V - 8V) = 0.4W$$

The 7808 will have no problem handling this, without any need for cooling.

• Wow, this is a detailed explanation. It has cleared up the doubts i had regarding the design and also answered my next question relating to if i could use a linear and switching in series like you have depicted above in the image. Commented Sep 11, 2023 at 9:03

If the circuit uses 1A, it means the 7812 has to pass 1A with 12V drop.

That's 12 watts it needs to dissipate. It cannot handle that amount of power that needs to be dissipated.

Also if the device takes in 1A at 24V, it consumes 24W, and half of it is already wasted as heat in the 7812.

So to answer the question, this is not an appropriate circuit, and even in the case it would be an appropriate cirvuit, in any case a switching regulator will be better - assuming that better means less wasted power as heat.

If you can tolerate the energy loss a linear regulator has some advantages in some applications. These may include cost, complexity, noise, PCB area, more ... . And may not in a given case.

You can substantially reduce regulator dissipation by dropping the input voltage with a resistor, which is dimensioned to provide Vout + Vdropout_max + a little more at maximum current.

A 7812 / LM340 (datasheet here ) has a dropout voltage of 2V typical at 25 degrees C at 1 A.

If we allow say 3V Vdropout, and assume that 24V is the minimum Vin_24 then
Rseries = V/I
= (24-3-12)V/1A = 9 ohms.
Max resistor dissipation is I^2 x R = 9 watt.
Using 2 x 10 watt aircooled "ceramic" resistors allows use with no resistor heatsinking.

See the many white rectangular ones here.
And specific examples here

Max regulator dissipation = (Vdropout_design) x Imax
= 3V x 1 A = 3 watt.
For the TO220 package Rth-JA is 24 C/W - you could use it with a PCB only heatsink at 25C* - but a modest added heatsink would be sensible. (* and notionally to about 70 degrees C ambient if brave and somewhat unwise :-) )

• yeah i see you mean, but i have a some size constraints, especially to the size of the heat sink that i would probably be needing if i was to use a few resistors in parallel to account fot the voltage drop. but non the less this is also an interesting way of going about the issue. Commented Sep 11, 2023 at 9:01
• @EmeraldMonk The resistors do not ned to have heatsinks. See the many white rectangular ons here. And specific examples here Commented Sep 11, 2023 at 12:49
• @EmeraldMonk see added image and links. Choose resistor value to suit. Commented Sep 11, 2023 at 12:57
• I had no idea about these ceramic resistors. i was under the impression that you meant resistors like de.rs-online.com/web/p/durchsteckwiderstande/8220198 . These are 30W resistors but better to have a heat sink. Commented Sep 12, 2023 at 14:12
• @EmeraldMonk Yes. The resistors you linked to say they have a 2.5W rating in free air! || The ones I mentioned are widely available (many different brnds) and nominally able to dissipate rated wattage without heatsink (and have no formal means of adding a heatsink). As I noted above, operating them at no more than 50% of their rating is "wise". They are generally "reliable enough". Commented Sep 13, 2023 at 1:11