Voltage Dividers

Given 1 to 3 loads of parallel 1kΩ resistors, we calculated the Vsource and Rsource of an unregulated power supply to make the voltage variations on the loads + or - 5% of 5V.

We recognized that the R_EQ,max and the R_EQ,min were 333Ω and 1kΩ, respectively, and that the larger resistance corresponded with the larger V_BUS. I worked out then the following:

• V_S = 5.54V
• R_S = 55.5Ω
• I_BUS,max = 14.26mA
• I_BUS,min = 5.25 mA

Below is collected data. (I'll fix the format later.)

Color Code	Nominal Value	Measured Value	Wattage (W)
Brwn,Blck,Red,Ag	1 kΩ	0.995	kΩ	0.25
Brwn,Blck,Red,Ag	1 kΩ	1.001	kΩ	0.25
Brwn,Blck,Red,Ag	1 kΩ	0.989	kΩ	0.25
Resistor Box	55 Ω	56.4	Ω	0.3
Power Supply	6 V	6.06	V	NA

We made sure nothing would be broken by the circuit we were planning to make, so we put it together. Here's a picture with all 3 loads connected.

These are the results:

Configuration	REQ (Ω)	VBUS (V)	I BUS(mA)	Pload(mW)
1 Load	1001	5.75	5.74	33.0
2 Loads	499	6.74	10.9	66.3
3 Loads	332	5.20	15.6	81.2

The power of 2 loads was found by P = V^2/R = (5.46 V)^2 / 499Ω = 66.3 mW

The actual % variations were quite different from the theoretical because we used 6.06 V instead of 5.54V.

There would be theoretically a -9.4% variation from 5V if we were using a 5.54V power supply and added a fourth 1kΩ load.

If we redesign it for a ±1% variation of V_BUS from 5V, then we would have a R_S = 10.2Ω and V_S= 5.10V.