How are voltage dividers useful? I can't seem to get enough current for them to be useful
A voltage divider circuit looks like the following:







simulate this circuit &ndash; Schematic created using CircuitLab

The left side is the physical circuit (except that I am adding a switch that is initially OFF or open.) The right side is the equivalent circuit once the switch is ON or closed.

The following equations are generated by treating \$R_1\$ and \$R_2\$ as a voltage divider (and where \$SW_1\$ is OFF):

$$\begin{align*}
V_{TH}&amp;=V_S\frac{R_2}{R_1+R_2}\\\\
R_{TH}&amp;=\frac{R_1 \cdot R_2}{R_1+R_2}
\end{align*}$$

So without the load added, \$R_1\$ and \$R_2\$ just make up a simple voltage divider. Like you already know and understand.

But once the switch is engaged (ON), then it becomes the right side circuit. And guess what? Now you have yet another voltage divider! I think you can see that the voltage, \$V_L\$ as seen by the load, will be smaller than the original voltage divider's unloaded (switch = OFF) voltage, \$V_{TH}\$. And the lower the value of \$R_{LOAD}\$ (the more current it requires) the smaller the voltage for \$V_L\$.

Just as you probably observed.

I won't harass you with the differential equation that expresses the rate of change of \$V_L\$ with respect to \$R_{LOAD}\$. But it's not a very good way to go, most of the time. The bottom line is that your load must be lighter than the original divider by at least a factor of 10 but usually even more than that. So if your load needs \$100\:\textrm{mA}\$ for example, then your resistive divider should be using perhaps several amps! Which is usually considered "crazy."

A simple solution is to use an emitter follower:







simulate this circuit

In this case, you are letting the BJT's collector handle most of the work with only a much smaller "load" now on the resistive divider. Since a BJT can often provide \$\beta\ge 100\$ times (or more) collector current than the required base current, the base current will be \$\beta\$ times smaller than your load current. (Larger current BJTs may have a smaller \$\beta\$, but I'm just trying to get the general idea across for now.) So, now the resistive divider only needs to be requiring perhaps 5 to 10 times less current than the load. Which is almost good, now.

But you will need to compute a resistive divider that provides about \$V_{BE}\approx 700\:\textrm{mV}\$ more than your load uses. And there will still be some variation on the load voltage as the load draws more or less current. But it will be much, much better than just using two resistors to set the voltage.

An additional step that can be taken is to use a zener for \$R_2\$. And/or to replace the BJT above with a Darlington, instead, and further adjust the divider voltage to accommodate it's still larger \$V_{BE}\$. But I don't want to dwell too long on all this. So I'm stopping here.