# Not understanding shunt regulator

Hello i was just wondering if someone could explain to me why the shunt regulator is used in this circuit. I get the zener diode keeps the 12 v constant to the motor but does it have to be there? Couldnt just the resistor be used in place without the zener diode to drop the 24 v to 12 v? Also only 1 motor is on at time

• – The Photon Jan 10 '17 at 5:08
• Thanks for some reason i kept thinking that the resistor could have varying currents to maintain the output. As in when the motors turned off it could handle 24 V since the current would just change. So the zener diode controls the current in a sense so that the resistor will get 0.19A even though the motor might draw more when under load, hows that possible if the motor is drawing more than 0.19 A?? Or is that why when working with motors use maximum efficiency – I have no clue Jan 10 '17 at 5:22
• V=IR. If the current through the resistor changes, the voltage across it changes. If the motor draws more than .193 A, then the voltage it gets will be less than 12 V. – The Photon Jan 10 '17 at 5:26

## 2 Answers

Could this work without the Zener? .....only if the motor current is constant, which seems quite unlikely.

Motors draw current related to their load yet in the circuit shown they have 190 mA flowing through the motor and just 5 mA through the Zener.
If the load on the motor caused more current to flow then voltage will drop below 12 V and make it easier to stall the motor.
And if the motor for any reason is disconnected or lightly loaded then more (or all) current will flow through the Zener.

If the motor current varies with a design maximum of 190 mA @ 12 V, then the Zener is needed and could maximally be required to dissipate 2.34 W.

This seems to be what the circuit is designed for, with a maximum expected motor load current of 190 mA @ 13 V.
If the motor is disconnected, all current (195 mA) flows through the Zener while maintaining 12 V.
If the motor is connected, the load current can range up to 195 mA and still be within design spec at 12 V.

If the load current increases beyond 195 mA due to extra mechanical load on the motor, the voltage will start to fall. Stall (or locked rotor) current for the motor could potentially approach 350 mA (you'd need to know the rotor resistance to calculate accurately) or so with close to 9 W dissipated in the 62 Ohm resistor.

Using the resistor to create a simple voltage divider would only work if you are connecting to a circuit with infinite input impedance, because any network connected in parallel to the resistor would draw current away from it, causing the the voltage across the resistor to consequently decrease. The zener diode, on the other hand, is operating in reverse breakdown. Like in the forward active region of a diode, the voltage across a diode in reverse breakdown is largely stable to changes in current.