Voltage Dividers

In this experiment we had to study how does the concept of voltage dividers work in a circuit. We will have three resistors connected in parallel which will serve as loads. Such circuit can be redrawn using Req (three parallel resistors):

Problem Statement: If R1=R2=R3=1000ohm and 1-3 resistors may be on at any time, determine Vs and Rs that will guarantee bus voltage to satisfy: 5.75<Vbus<6.25.

Step 1: Design calculations

Req, max (1 resistor) = 1000ohms
Req, min (3 resistors in parallel) ~ 333.333ohms

Vbus, max = 6.25V - Req, max
Vbus, min = 5.75V - Req, min

Vbus equations:

Vs = 6.53V               Rs = 45.45ohms

Ibus, max:
I = 6.25/1000 = 6.25mA

Ibus, min:
I = 5.75/333.333 = 17.25

Step 2: Laboratory equipment

•  Each load will be a fixed resistor
• Rs will be a variable resistance box
• Vs will be a unregulated power supply
• The switches will be short cables

Step 3: Measure the components

Color Code				Nominal Value (ohm)	Measured Value (ohm)	Wattage (W)
Br	Bl	Bl	Br	1000	978+/-001	1/8
Br	Bl	Bl	Br	1000	976+/-001	1/8
Br	Bl	Bl	Br	1000	979+/-001	1/8
Resistor Box				45.45 ohms	45+/-0.2 ohms	10
Power Supply				6.53V	6.50+/-0.01V	x

Is the maximum bus current within the capability of the power supply? Yes
Is the maximum current within the power capability of the resistor box? Yes

Step 4: Building circuit on the bread board

Mounted resistor in parallel (Req)

Rs

Step 5: Data collection

Config	Req (ohm)	Vbus (V)	Ibus (mA)	Pload (W)
1 Load	978+/-1	6.08+/-.05	6.04+/-.01	0.0378+/-.00044
2 Load	488.7+/-0.80	5.65+/-.01	11.27+/-.02	0.0653+/-.00020
3 Load	325.9+/-0.69	5.33+/-.02	15.72+/-.01	0.0871+/-.00050

Step 6: Calculations

a) Calculate the power delivered to the load for the three configurations and record the values in the table. Show the calculation for the 2 loads.

Pload2 = V^2/R = (5.65+/-.01)^2/488.7+/-.80 = 0.0653+/-.00020W


b) Calculate actual percentage in load voltage variation achieved (how much % up and how much % down, assuming 6V is nominal value). Why is this different from the +/-5% design goal?

 % up =100*|6 - 6.08+/-.05|/6 = 1.3+/-.83%

% down = 100*|6 - 5.33+/-.02|/6 = 11.2+/-.33%

The difference between these % in load voltage and the +/-5% design goal is probably due to some possible flaws in the circuit bread board and/or the fixed resistors not being exactly 1000 ohms (~976 ohms).


c) Given our source parameters (Vs and Rs the same), what would be the new load voltage variation if we added a fourth 1k ohm load?

With 4 - 1000 ohm resistors in parallel, Req will be 250, so;
Vbus = (Req)(Vs)/(Req + Rs) = (250)(6.53)/(250+45.45 = 5.53V


d) What would be the new source parameters required for the 3load case if we wanted to reduce the voltage load variation to +/-1%?

with 1% variation, we have 5.95 < Vbus < 6.05
so we use the Vbus formulas;

6.05 = 1000Vs/(1000+Rs)           and         5.95 = 333.333(Vs)/(333.333+Rs)

Solving the system of equations yields: Vs = 6.10V and Rs = 8.48 ohms.