# How are voltage dividers useful? I can't seem to get enough current for them to be useful

I must be doing something wrong here, because when I make my simple voltage dividers and do the math, the most current I could draw through my voltage divider would be 33.3mA. Here are my factors:

Vin = 5V

R1 = 100Ω

R2 = 200Ω

Vout = 3.33V

Now, using ohms law, i=v/r, so 3.33/100 = 0.0333, 33.3mA.

Am I doing something wrong here? Thanks :)

• Voltage dividers are not intended to serve as power supplies. – Eugene Sh. Jul 26 '17 at 19:05
• So, what exactly are they used for? – Bacon Jul 26 '17 at 19:06
• To convert a signal level for a high impedance input, for example. – Eugene Sh. Jul 26 '17 at 19:06
• Ah okay, makes sense. I've had to make one before for my arduino to ESP8266 serial bus, since one was 3.3v and the other was 5v. – Bacon Jul 26 '17 at 19:07
• R1 current limits causes load regulation errors in high current, a linear feedback reg is better – Tony Stewart Sunnyskyguy EE75 Jul 26 '17 at 19:30

A voltage divider circuit looks like the following:

simulate this circuit – 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}&=V_S\frac{R_2}{R_1+R_2}\\\\ R_{TH}&=\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.

It seems you have this:

With no load, that will produce the voltage (Vin R2)/(R1 + R2) = 3.33 V. However, this comes at the impedance of R1 parallel with R2. R1 // R2 = 67 Ω. Put another way, from the point of view of something connected to OUT, your circuit is equivalent to:

From this, it should be more obvious that the voltage at OUT will drop when you try to draw current from it. In fact, if you short OUT to ground, the voltage is completely across the resistor and the current is (3.33 V)/(67 Ω) = 50 mA.

You can make the output stiffer (be able to supply more current) by lowering both resistors, but that increases the quiescent current. And, the output voltage will still sag proportional to the current, just less so.

This is why voltage dividers are not very useful for providing a lower voltage you intend to draw significant power from. A better answer is a voltage regulator. There are plenty of 3-terminal 3.3 V fixed voltage regulators available that can take 5 V in and produce a reasonably steady 3.3 V out from 0 to 100s of mA.

Your current calculation should depend on how you have the resistors connected.

If you have them in series (as most voltage dividers are) then the current is the same through both resistors

I = V/(R1 + R2)
= 5V/300
= 16.6mA


Then, if you wanted to attach a load at the 3.3V voltage, you need to calculate the current through each of the branches (the load and the resistor in parallel to it). If you want to get more current you could use a couple transistors as described here: current amplifiers & Buffers to build a current amplifier instead!

You ask "How are voltage dividers useful?"

Well, every time you need to scale down a voltage!

The voltage comparator above compares both voltages on its "+" and "-" outputs and sets its output to the positive or negative supply voltage as a result of the comparison. (output positive if IN+ > IN- for example).

So if you want to compare the input with a threshold voltage, you need to generate this voltage first. For simplicity, a divider from the supplies works fine. Adjust R1 and R2 (or use a pot) to have the threshold you need.

This is just an example, you find voltage dividers everywhere, in the feedback of an opamp or voltage regulator, etc.

Now if you want to create 3.3V from your 5V supply, then you need a voltage regulator. There are tons of those, depending on input to output voltage, current, etc.

Am I doing something wrong here?

You used too big of resistors. Use smaller ones for bigger current flow.