# Voltage Divider w/ Varying Load Current

I have an easy question that I feel I might be overlooking.

The question I was given was:

A 15V power supply is used to power a 6V, 500mA device. The device draws 500mA at the rated voltage, but will operate properly with +/- 20% voltage variation. Design the appropriate voltage divider. Note that the load current requirement is variable and not always 500mA (edited again for clarity on what I was told about the question).

simulate this circuit – Schematic created using CircuitLab

My question is: What do I take into consideration when designing the circuit?

1. Do I design the circuit such that the "Ohm's Law $V_t/R_t = I$ current" is at least 500mA and also make sure that the device receives 4.8 - 7.2V at the same time by choosing the right resistor values?

2. Or, can I design the circuit (ignoring the current, but making sure the overall current in the circuit is greater that 500mA) taking just voltage in consideration? By that I mean I assume that at 500mA in the circuit, there will be a voltage drop of so and so much based on the resistor values?

If this is the case, do I ignore the "Ohm's Law $V_t/R_t = I$ current" ?

As a follow up question: Do I have to consider the "unloaded voltage divider equation" i.e. $V_{out} = V_{in} \left(\frac{R2}{R1+R2}\right)$ or can I just work with the voltage drops at 100mA and 500mA, making sure that the voltage drop on $R2 / RL$ is at between 4.8 and 7.2V?

• Now when you removed the current requirement, R1 is 18 $\Omega$ and R2 is $\infty$. – venny Sep 17 '14 at 2:48
• electronics.stackexchange.com/a/83659/8627 – jippie Sep 17 '14 at 5:40
• The question you were given is stupid, and misleading, because there are an infinite number of R1R2 combinations which can cause the voltage across R3 to be 6 volts with 500 milliamps through it. – EM Fields Sep 18 '14 at 0:02

What do I take into consideration when designing the circuit?

Think it through.

• The load voltage will be highest when the current is lowest
• The load voltage will be lowest when the current is highest.
• You're given the high and low load current specification as well as the high and low load voltage specification.

So, you have two points on the load line. This will allow you calculate the equivalent resistance (Thevenin resistance) of the voltage divider as well as the voltage division factor $\frac{R_2}{R_1 + R_2}$.

Can you take it from here?

I edited the question to remove the current fluctuation part. I was just warned to keep it in mind, it wasn't part of the actual question.

Since there are two resistor values to pick, you need two independent specifications. Certainly, one of those specifications is that the voltage must be within 20% of 6V when the load current is 500mA.

If you form the Thevenin equivalent circuit of the 15V source and voltage divider resistors, the voltage across the load resistor is seen to be given by

$$15V \frac{R_2}{R_1 + R_2} - I_L\cdot R_1||R2$$

So, you need another specification. One approach would be to limit the open-circuit voltage to be 20% higher than the nominal voltage.

$$15V \frac{R_2}{R_1 + R_2} = 7.2V$$

Now you have two equations and two unknowns:

$$\frac{R_2}{R_1 + R_2} = \frac{7.2}{15}$$

$$R_1||R_2 = \frac{7.2 - 6}{0.5A}$$

This is just an example of choosing an additional reasonable constraint in order to uniquely specify the resistor values. You might reasonably choose a different additional constraint.

• Hi - I edited the question to remove the current fluctuation part. I was just warned to keep it in mind, it wasn't part of the actual question. So - can I just calculate the equivalent resistance for R1 and R2//RL and make sure that at least 4.8V is across R2? I am confused because it seems simple if I consider the Thevenin representation, but I always try to "make sure" and use Ohm's Law. If I do use Ohm's Law, everything breaks down. – metropolis Sep 17 '14 at 2:51
• @SDQ, I've updated my answer – Alfred Centauri Sep 17 '14 at 17:16