# Load regulation

## What do you need to know to understand this topic?

## Sections

## What is load regulation?

In linear regulators, the output voltage is affected by the load/output current. Load regulation measures how much it is affected:
$$\mbox{Load regulation} = \frac{\Delta V_{out}}{\Delta I_{out}}$$
We would like a small value for this specification. There are several ways of finding out this relation, but the most systematic one, that can be applied the same way to all circuits, is the small-signal analysis. Basically, a circuit is working at some operating point and we analyze it for small variations of some parameter, this case the load current $i_{out}$. Since we are analyzing around a specific point, we approximate the circuit to a linear version (remember Taylor series?), just with resistances, capacitors, inductors and sources. From here we can get a closed-form relation between any two parameters of the circuit. Constant voltage sources, such as $V_{in}$ become short-circuits and constant current sources, such as the average value of the load current, become open-circuits.

## Examples

### Shunt regulator

Let's try with the shunt regulator. For a given operating point, the zener diode can be replaced by a resistance $r_z$ (that is the small-signal model of the diode). Our load is replaced by a current source $i_{out}$, because this is the parameter we are changing. The input voltage, since it is constant for our analysis, becomes a short-circuit. Therefore, we have:

$$v_{out} = -(R_D||r_z)i_{out}$$
$$\frac{v_{out}}{i_{out}} = -(R_D||r_z)$$

### Series regulator with feedback and MOSFET pass transistor

Now consider a series regulator with feedback. This one is more complicated: we have the small-signal of a MOSFET transistor. Say it is a PMOS: the most simple small-signal model for it is a voltage-controlled current source such that $i_{sd} = g_m v_{sg}$. The opamp can be represented as a linear gain stage with gain $A_o$ and the input voltage source becomes a short circuit. Now we can represent the small-signal model of the regulator and find the transfer function from the output voltage to the output current, which is:

$$i_{out} \approx g_m v_{sg} = -g_m A_o \frac{R_2}{R_1+R_2} v_{out}$$
where we assumed that $R_L \ll R_2 + R_1$ and all current flows to the load (we wouldn't like our sensing circuit to consume as much as the load, right?). We end up with:
$$\frac{v_{out}}{i_{out}} = -\frac{1}{g_m A_o}\frac{R_1+R_2}{R_2}$$
We can see that the load regulation of the regulator with feedback is much lower than the shunt regulator, for a large $A_o$. In fact, the opamp has a fundamental role in pushing this spec down.

### Datasheet

This is the load regulation spec in the 78xx series datasheet, named *Output voltage regulation* and measured for two current ranges

and this is in the LM317 datasheet, measured in several conditions and indicated in %, since it is an adjustable regulator.