# Switching Regulators vs Linear Regulators: A worked example

It seems that every time linear regulators are mentioned online, someone is quick to pop up and describe them as inefficient - advocating their replace with a switching alternative. This doesn’t answer the obvious questions though: just how inefficient are they? and how much better are switching regulators?

Armed with littlle more than two datasheets, some primary school maths, and some spare time - I’m going to walk to through examples, explaining how to calculate efficiency and limits, before exploring just how big a deal the differences are. (Spoiler: a lot bigger a deal than I thought!)

## Comparing Thermal Performance#

Voltage regulators - like the humble resistor - dissipate wasted energy as heat. This gives us a key metric that we need to measure to understand their limits: given xW in power, how much is wasted as heat?

All the data we need can be found in the datasheet:

• Thermal Resistance: measured in Kelvin (K/W) or Celsius per Watt (°C/W) - $$R_{\varnothing}$$.
• Effiency: Percentage of power output vs input. (Only provided for switching regulators) - $$\bigcap$$

Whilst the only maths we really need is our trust Power Formula - $$P = IV$$ - and $$T = R_{\varnothing} * P$$.

In our examples we’ll look at the wastage at 500mA running at 5V from a 12V supply, and the maximum amount of output current we can safely achieve at 5V. We’ll work on the assumption that there’s no heatsink in use, and that the ambient temperature is room temperate - or 20°C.

### LM7805: Linear Regulator#

The LM7805 Linear Regulator specifies that it has a Rθ1 value of 50 °C/W, and a maximum of junction temperature of 125°C.

The power expended in dropping 7 volts for a 500mA load is 3.5W.

P = V * I
P = (12V - 5V) * .5
P = 3.5W


To dissipate 3.5W through a thermal resistance of 50 °C/W we’ll be looking at a surprising 175°C of heat!

T = Rθ * P
T = 50 * 3.5
T = 175°C


This would leave us 50°C above our maximum junction temperature before we even account for the ambient temperature.

If we re-arrange these variables we can calculate just what current the regulator can provide at room temperature before it’s liable to fault. First of all we need the maximum temperature increase that the regulator can handle - inclusive of ambient temperature.

T(max) = 125°C - 20°C
T(max) = 105°C


From this temperature we can use the Rθ to determine what amount of power this equates to.

P(max) = T(max) / Rθ
P(max) = 105 / 50
P(max) = 2.1W


Finally, we can use the voltage to find the associated current:

I(max) = P(max) / V
I(max) = 2.1 / 5
I(max) = 0.42A (420mA)


There we have it: at a room temperature of 20°C, the LM7805 can deliver a total of 420mA whilst staying within it’s operating temperature range.

### LM2576: Switching Regulator#

The LM2576 switching regulator specifies two different values for thermal resistance (Rθ): one with a heatsink and one without. The difference is quite large too: 45°C/W with a heatsink vs 65°C/W without! In the interests of fairness we’ll take the larger number though.

The first thing to note is the calculations are a bit different for a switching regulator: as it essentially functions as a high frequency switch, it doesn’t dissipate in the same way. This means we calculate P as the output power - i.e the power that’s consumed by our load, in this example 500mA at 5V.

P(out) = V * I
P(out) = 5 * .5
P(out) = 2.5W


Next up we need the efficiency (η) value from the datasheet - this is the percentage of input power which is used as output power. For the LM2576 this is approximately 80 - meaning 80% of input power is directly used to drive the load, with 20% wastage.

P(in) = P(out) / η
P(in) = 2.5 / 0.8
P(in) = 3.12W


This means that the wasted power is the difference between the input and output - Pin - Pout - and this is the value that needs to be dissipated as heat.

P(heat) = P(in) - P(out)
P(heat) = 3.12W - 2.5W
P(heat) = 0.62W


Lastly, we can use the thermal resistance value Rθ to find the actual heat generated:

T = R(θ) * P(heat)
T = 65 * 0.62
T = 40.3°C


Even with an ambient temperature of 20°C, we’re still well below our maximum operating temperature of 150°C - at a cool 60.3°C. If we re-arrange the maths slightly, we can see that we can dissipate up to 2W at room temperature before we hit the operating limit.2

T(max) = 150°C - 20°C
T(max) = 130°C

P(max_heat) = T(max) / R(0)
P(max_heat) = 130 / 65
P(max_heat) = 2W


Knowing that an efficiency (η) value of 80 means that 20% of the energy is wasted as heat, and we can afford to lose a maximum of 2W, we know that we can have a maximum output of 8W power - or 1.6A. (8W = 1.6A * 5V)

P(out) = V(out) * I(out)
P(out) = 1.6 * 5
P(out) = 8W

P(in) = P(out) / η
P(in) = 8 / 0.8
P(in) = 10W

P(heat) = P(in) - P(out)
P(heat) = 10W - 8W
P(heat) = 2W


So at a room temperature of 20°C, the LM2576 can deliver a total of 1600mA whilst staying within it’s operating temperature range.

## Conclusion#

The switching regulator is able to deliver around 3x the current as the linear one, and it still manages to do so with less energy wasted - as it dissipates 2W vs 2.1W! That’s quite something.

It’s common knowledge that linear regulators are inefficient, but it wasn’t until doing these numbers whilst I wrote this that I realised how inefficient they were. It looks like I have some schematics to go update…

1. Specifically, we’re using the thermal resistance between the junction and air - or RθJA in the datasheet. ↩︎

2. These calculations are somewhat condensed - in that they don’t have the explaining text. I decided not to make you read the same stuff… again! ↩︎