Nuts and Volts - September 2007

BY STUART BALL

■ FIGURE 1. Schematic of the fan controller circuit. T1 is a thermistor mounted to the component needing to be cooled.

inaudible, so the circuit always supplies 5V to the fan, through U1B. As the temperature rises so that the output needs to rise above 5V, U1A begins to sink current, raising the fan supply voltage. Operating the fan in this way provides two advantages: First, there is always some airflow through the box. Second, it simplifies the circuit since there is no need to add circuitry to turn off the output when it would be below the minimum fan operating voltage.

U1C provides fault detection. When the temperature rises above approximately 65°C, U1-8 goes low. U1D is just an inverter to drive Q2, providing an open-collector output. This allows multiple fan circuits to share a common fault signal. U1C includes a small amount of hysterisis through R9 to insure that the fault output is a well-defined transition.

Note: R13 is not installed for a fan that can operate at 5V. For a fan with a minimum operating voltage higher than 5V, R13 is needed to provide that voltage to the fan. See the sidebar “Working the Equations” for details.

Calculating Circuit Values

In an op-amp that is linear (the output is not saturated at either supply rail), the inputs (U1-3 and U1-2, in this case) will be at the same voltage. The voltage at U1-2 is controlled by the voltage divider comprised of T1 and R1 and varies with temperature. The voltage at U1-2 is controlled by the

voltage divider comprised of R3 and R2, and varies with fan voltage.

The equation for U1-2 can be written as:

(Vo - 5V) x R2

+5V

R2 + R3

The equation for U1-3 can be written as:

12V x R1 R1 + RTh

I defined two endpoints for the circuit, with the fan operating at 5V up to about 40°C, where the thermistor resistance is about 7K, and applying the full 12V at about 60°C (thermistor resistance about 4K). By setting R3 to 10K and then setting pin 3 equal to pin 2, we have the following equations:

5V output:

12V x R1 (Vo - 5V) x R2

= +5V R1 + 7K R2 + 10K

Since Vo is 5V, the right side of the equation reduces to 5V and R1 is 5K.

At the other end of the range, the equations are:

12V x 5K ( 12 - 5V) x R2

= +5V 5K + 4K R2 + 10K

which makes R2 equal to 3.12K. Since I didn’t need precise temperature control, I used the nearest 5% resistors; a more precise circuit would use 1% or better parts. Note that R1 reduces to a simple equation only because the reference for pin 3 is equal to the minimum fan voltage. A different reference value (such as a precision 2.5V reference) would result in solving both equations simultaneously (remember high school algebra!) for the values of R1 and R2. Also, many 12V fans will not operate as low as 5V, and a different minimum fan voltage would also change the result.