Circuit adds foldback-current protection
Enhance three-terminal regulator with foldback oversight.
Rafael García-Gil and JM Espí, Electronic Engineering Department, University of Valencia, Spain; Edited by Brad Thompson and Fran Granville -- EDN, July 21, 2005
For many applications that require power-supply currents of a few amperes or less, three-terminal adjustable-output linear voltage regulators, such as National Semiconductor's LM317, offer ease of use, low cost, and full on-chip overload protection. The addition of a few components can provide a three-terminal regulator with high-speed short-circuit current limiting for improved reliability. The current limiter protects the regulator from damage by holding the maximum output current at a constant level, IMAX, that doesn't damage the regulator (Reference 1). When a fault condition occurs, the power dissipated in the pass transistor equals approximately VIN×IMAX. Designing a regulator to survive an overload requires conservatively rated—and often overdesigned—components unless you can reduce, or fold back, the output current when a fault occurs (Reference 2).
The circuit in Figure 1 incorporates foldback-current limiting to protect the pass transistor by adding feedback resistor R4. Under normal conditions, transistor Q2 doesn't conduct, and resistors R1 and R2 bias MOSFET Q1 into conduction. When an output overload occurs, Q2 conducts, reducing the on-state bias applied to Q1 and thus increasing its drain-source resistance and limiting the current flowing into regulator IC1, an LM317. Adding R4 makes Q2's bias current dependent on the output voltage, VOUT, which decreases under overload conditions.
For the circuit in Figure 1, you can calculate the maximum foldover and short-circuit currents, IKNEE and ISC, respectively, as follows:
EQUATION 1
EQUATION 2
In a practical design, you select values for IKNEE and ISC and equal values for R3A and R3B and then use equations 1 and 2 to calculate resistors RSC and R4. For the circuit in Figure 1, the output's maximum and short-circuit currents are fixed at 0.7 and 0.05A, respectively. With R3A and R3B set to 100Ω, solving the equations yields values of 0.73Ω for RSC and 4.3 kΩ for R4. You can demonstrate the circuit's performance by applying a variable-load resistor that's adjustable from 0 to 200Ω. As Figure 2 shows, the output's simulated and measured voltage-versus-current characteristics, VOUT and IOUT, respectively, are in close agreement.
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Hi,
I've some comments about the circuit of Figure 1. The schematic shows a MOSFET pass transistor, but the part number BC557 refers to a bipolar PNP transistor.
I think Q1 and Q2 Part numbers are swapped.
Additionally, the mosfet of figure 1 seems to be a N-CH, so I think that in order to turn the mosfet on the gate voltage have to be higher than the source voltage. It seems not to be possible in the circuit shown.
is IRF9130 a P-CH?
Augusto Musolino - 2009-14-10 07:23:00 PDT -
Looks like you guys have a couple of mistakes here:
1- The most obvious: Q1 is a MOSFET, so it is the IRF913, and Q2 is the PNP, so it is the BC557
2- The MOSFET is a P-channel, and the drawing is for an N-channel
And one omition:
- What is Vsense ?
Looks like it is the transistor's Vbe, and doing the reverse calculations yields a close enough value.
Sergio R. Caprile - 2005-18-10 04:37:00 PDT -
Sorry, my previous comment applies to the power supply interrupter idea, and not to the foldback current limiter
Mario Pazzini - 2005-21-7 23:41:00 PDT -
I think you should verify more carefully the published circuit diagrams before publishing, otherwise reading a design idea becomes just a waste of time. In this case the value of R6 (47K)is quite obviously wrong. This is defined as a current sensing resistor...
Mario Pazzini - 2005-21-7 23:35:00 PDT
