Electronics lab experiments often use low-power control cards that can fail even in expert hands. These cards contain ICs that fail if the supply voltage exceeds a certain value or has an unsuitable polarity. Also, current must be limited to avoid damage to the card. Therefore, protection against inverted polarity, overvoltage, and overcurrent is needed. In our case, the control card that we wanted to protect consumed less than 50 mA, a negative input voltage wasn't allowed, and a sensitive IC would fail if the input voltage was greater than 20 V.
The usual way to protect sensitive ICs is by clamping the IC supply voltage to a permissible positive value. Figure 1 shows the easiest solution. Resistance R1 must be low to avoid a voltage drop that's too high. But then the power handled by zener diode D2 could be excessive in case of overvoltage. Other more complex solutions also clamp the IC supply voltage.
Figure 2 shows a simple, low-cost solution that accomplishes all of these requirements. In our application, uninterrupted operation isn't necessary, and in case of a failure, VSUPPLY will drop to zero. Other applications may require users to know at any moment if the system is working well or if there's any problem. The LEDs supply that capability.
A series diode (D1) supplies polarity protection. For a lower voltage drop during normal operation, an 11DQ10 Schottky diode was used instead of the cheaper 1N4007. Applications with higher input current would require polarity protections that are more complicated to reduce the voltage drop. Diodes D4 and D5 ensure that the reverse voltage across LEDs D3 and D6 is lower than their breakdown voltage, normally 5 V, when applying a negative input voltage.
In this circuit, F1 is an RXE005, a positivetemperature-coefficient device. At 20 C, the RXE005 will hold 50 mA with a maximum resistance of 11.1 Ω. If current increases, its resistance rises rapidly.
With a positive input voltage below 18 V, there's a small difference between VSUPPLY and VIN (VIN - VSUPPLY = VD1 + VRF1 + VD7 is around 1.4 V for 50 mA) and the green LED shines, indicating proper operation. If the input voltage rises above approximately 18 V, zener diode D2 turns on and the red LED starts to shine, indicating an overvoltage condition. Zener current flows through the 680-Ω resistor, producing a base voltage that turns on Q1.
That action initiates a rapid sequence: Q2 turns on and input current rises. Current through F1 reaches (VIN - VD1 - VCE(Q2)/RF1. With a resistance of 11.1 Ω, this means around 1.5 A for an input voltage of 18 V. If this resistance doesn't increase, input current will grow with the input voltage, reaching 2.5 A at 30 V. But because the value of RF1 increases with higher input voltages, it limits the input current to lower values.
If diode D7 weren't used and the input power supply were suddenly removed, capacitor C1 would discharge itself across transistor Q2 (which is not overcurrentprotected) and ultimately break down. Diode D7 avoids this current path, and the output capacitor discharges across the load presented by the circuit.
F1 must have a low voltage drop (small enough resistance) with the normal operating input current flowing through it. The value of zener diode D2 depends on the clamping input voltage. This clamping voltage (VCL) is:
VCL = VZ + VRed_LED + VBE(Q1)
where VRed_LED is close to 2 V. R1 determines the bias current of D2 and of the red LED:
IBIAS1 = \[(VIN - VD1) - VCL\]/R1
R4 determines bias current of the green LED:
IBIAS2 = \[(VIN - VD1 - VRF1) - VGreen_LED\]/R4
where VGreen_LED is also close to 2 V.