Calculating voltage droop

[ScuzZ]

How does one calculate the voltage droop when a load is placed on a voltage divider circuit?

I have wired up a 12v supply with two 1k resistors in series, to give 6v output inbetween the resistors. But when I place a 6v motor in parallel with the second resistor the voltage droops dramatically.

The motor itself has an 8ohm resistance. Do I have to calculate for that resistance when creating the voltage divider circuit?

 

[ljcox]

The simplest way to do it is to use Thevenin's Theorm.

The Thevenin voltage is 6 Volt and the Thevenin resistance is 500 Ohm (the two 1k resistors in parallel).

So a 12V source with a 1k, 1k voltage divider is equivalent to a 6 Volt source in series with a 500 Ohm resistor.

But you need to know what current the motor draws when it is running. The static resistance is only of use if you want to calculate the starting current.

If the motor is 8 Ohm, then the starting current will be 6/(500 + 8) = 6/508 = 11.8 mA.

The starting voltage will be 8 * 0.0118 = 0.0994 Volt.

So it is VERY unlikely that the motor will start!!

What you need is a 6 Volt voltage regulator that will maintain the voltage at about 6 Volt regardless of the current drawn.

The starting current will be 6/8 = 0.75 Amp. So you need a regulator that can dissipate at least 6 * 0.75 = 4.5 Watt.

 

[Hero999]

Treat the motor as a resistor.

Measure the current flowing through it when it's driving the load you want (from a 6V supply) and calculate its resistance, then put a resistor in series with it of the same value. Always checkt that the power dissipation in the resistor is within its rating, P = I²R.