Electronic Design

Low-Quiescent-Current Regulator Withstands High Input Voltages

In automotive applications, more and more on-board modules need to meet stringent requirements for ignition-off current, with many modules requiring less than 100 mA of current. A controller-area network (CAN) interface IC and a microprocessor in sleep mode can easily consume half of this current, leaving little operating leeway for the power supply. On top of this, the module's power supply must withstand automotive battery voltages of 24 V during jumpstart and around 40 V during load dump.

What's needed is a way to get extremely low quiescent currents (IQ) and still withstand the high breakdown voltages needed for automotive applications. The circuit (see the figure) harnesses the NCV553 fixed-output linear voltage regulator to do the job. This device has an IQ of 2.8 mA typical, 6 mA maximum. The only problem is that the regulator can't accept an input voltage greater than 12 V. The solution is to limit the input voltage seen by the regulator. Adding a 6.8-V zener diode (D1) from the regulator output to the gate of a 2N7002 n-channel MOSFET (Q1) does just that.

With the output of the regulator at 5 V, the voltage at the cathode of zener D1 will be approximately 11 V. The MOSFET has a high-value pull-up resistor (R1) to VBATTERY, which turns on Q1 but keeps it operating in its linear region. Because the 2N7002 has a 1-V threshold voltage, the voltage at its source will be around 10 V. That's 2 V less than the maximum input voltage of the regulator.

The value of gate resistor R1 allows the circuit to supply the minimum load current at the maximum input voltage. Otherwise, as the input voltage increases, the current flowing through R1 and D1 would be higher than what the load is demanding, and the output would start to rise, taking it out of regulation. The value of resistor R1 is determined by:

RMIN = \[(VIN max) - (VZ + VOUT )\]/IMIN

Because MOSFETs are voltage-driven devices, almost no current is used to keep the gate turned on. What's more, because all of the current to set up the zener voltage on the gate is fed directly to the load, there's no wasted current going to ground.

With this circuit, it's possible to produce 50 mA of output current with less than 5 mA of quiescent current. The only precaution needed is to monitor the heat generated by the MOSFET. Because Q1 is dropping all of the voltage above 10 V, it will dissipate some power. The higher the output current, the more heat will be generated.

Hide comments

Comments

  • Allowed HTML tags: <em> <strong> <blockquote> <br> <p>

Plain text

  • No HTML tags allowed.
  • Web page addresses and e-mail addresses turn into links automatically.
  • Lines and paragraphs break automatically.
Publish