Electrical Tips and Tricks## SHUNT REGULATORThe lab manuals for many DC circuits courses, including the ones that come with popular text books, have experiments with circuits like the one shown in figure 1. The problem with them is that sometimes the measured values of voltage and current don't agree with the calculated values. It seems like a mystery: does circuit analysis not always work? Of course it does! The problem is likely to be in the power supply you're using. Circuits like the one in Look at E V1 - V2 12 - 6 I = ---- = ---------- = -------- = 6 mA. R R 1000 But if you are using a typical power supply instead of batteries, you will measure 0 mA. What's more, you will measure 0 Volts across the resistor. What's going on? The answer is that the typical power supply uses a series regulator. A simplified schematic of a series regulator is shown in
If you apply a voltage to the emitter that is greater than what the supply is set to put out, then you reverse bias the transistor. That means that current can flow out the emitter of the transistor, but current can not flow into the emitter. In fact, if too much reverse bias is applied to the transistor it will be damaged. So often a diode is put in series with the output as protection. Is there some way to get a power supply to sink current? Yes there is! You can use a circuit called a shunt regulator.
If there is no load on the supply, all the current goes through the transistor. If there is a resistive load, some current goes through the load and the rest goes through the transistor. But here's the important part: if something tries to drive current back into the supply, the transistor will shunt that current to ground as well. Look at
You calculate the resistor values as follows: SUPPLY VOLTAGE - MAX OUTPUT VOLTAGE R1 = -------------------------------------- MAXIMUM OUTPUT CURRENT 2 WATTAGE of R1 = (SUPPLY VOLTAGE) / R1 MAXIMUM SINK CURRENT BASE CURRENT = -------------------------- MINIMUM BETA of TRANSISTOR SUPPLY VOLTAGE - DIODE DROP R2 = ----------------------------- BASE CURRENT 2 WATTAGE of R2 = (MAX CURRENT) x R2
20 V - 10 V 10 V R1 = -------------- = ----- = 100 Ohms 100 mA 0.1 A WATTS = (20) x (20) / 100 = 4 W (use a 5 Watt resistor) 200 mA BASE CURRENT = ------ = 4 mA 50 18 V R2 = ----- = 4.5 K Ohms (Use 4.7 K Ohms) 4 mA WATTS = (.004) x (.004) x (4700) = 75 mW (use 1/4 Watt) You can use a value as low as 1 K for R2 to provide some over-drive capability since a 741 can supply up to 20 mA. If you use a CMOS op-amp, check it's maximum current output. To develop a voltage for the adjustable set-point, we used a 15 V, 1 W zener diode and a 4.7 K trim-pot. To calculate the series resistor for the zener, we just used: VOLTAGE DROP (20 - 15) V R = ------------ = ------------- = 250 Ohms. We used 200 Ohms. ZENER CURRENT 20 mA WATTS = (5V) x (5V) / 200 = 125 mW (Use 1/4 Watt) |