Optically isolated overcurrent detector works from ac mains
The circuit uses two opto-transistors you arrange for bipolar sensing.
José M Espí, Jaime Castelló, Rafael García-Gil, and Marcos Arasa, University of Valencia, Valencia, Spain; Edited by Paul Rako and Fran Granville -- EDN, October 6, 2011
The circuit in this Design Idea detects short-circuit faults or excess-current conditions in line-operated equipment, such as a UPS (uninterruptible power supply), across an isolation boundary. If the load current rises higher than a predetermined level, the circuit generates a 5V pulse to immediately shut down the power inverter. This compact circuit provides a reliable means for protecting power semiconductors and batteries against overcurrent conditions. It features an isolated output and a fast response. The ac mains self-power the sensing network. The circuit requires a 5V power supply to the output-enable circuit on the other side of the isolation boundary. The circuit detects bipolar over-currents through the load, using shunt resistor RSHUNT’s resistance.
During the positive cycle of the ac
source, components R1, IC1, R5, C1, Q1,
D1, and C3 work to detect an overcurrent
(Figure 1). If the load produces
an overcurrent, the voltage drop at
the shunt resistor generates a positive
voltage on the base of Q1 that turns it
on. Consequently, capacitor C1, which
R5 initially discharges, receives charge
from current across R1 and IC1’s photodiode.
IC1’s photoresistor saturates,
charging capacitor C5 and providing
a 5V output at VSENSE, indicating the
overcurrent detection. D1 increases
Q1’s base-voltage turn-on threshold
to approximately 1.4V. Capacitor C3
filters any noise or peak voltage that
may accidentally turn on the transistor,
thus enhancing the system’s detection
reliability.
A circuit, comprising R2, IC2, R6,
C2, Q2, D2, and C4, for fault detection
during the negative cycle of the ac
source works similarly. In this case, the
direction of current flow through the
shunt resistor generates a positive voltage
on the base of Q2. D5 and C6 rectify
the ac source, providing the isolated
supply voltage to polarize the photodiodes.
The inverter logic’s power supply
supplies the 5V source. When the load
currents are within limits, the output
at VSENSE extinguishes due to the discharge
of C5 through R8.Talkback
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For a truly simple circuit I would use a dual LED coupler across the current sense resistor (with 1K or so in series). The output transistor will turn on whenever the voltage exceeds 2 or so.
Einar Abell - 2011-20-10 00:09:41 PDT -
It is possible to replace one 4N25 by using a wired OR configuration made of two common anodes 1N4148 diodes or its equivalent. Connect the common anode of this new diode pair to the single 4N25 LED cathode and each of their cathode respectively to the "positive" (R5//C1) and "negative" (R6//C2) leg of the current peak detection circuit.
PF Thibault
PF Thibault - 2011-19-10 02:25:31 PDT -
As well as the points previously made, altogether rather big and clunky, power hungry in both shunt and power droppers and prone to the Vbe drift of the sense transistors and emitter diodes. Personally I'd use an ultra-cheap LMV324 or Cmos quad opamp configured as a ground sensed rectifier and following with comparators from the remaining opamps for level and pulse extension driving a low-power digital isolator. Fraction of the size, cost and power, and much more stable with temp.
A. Macfarlane - 2011-11-10 14:36:31 PDT -
1.It is a very bad idea to use BUT56A 1kV 8A 100W transistors Q1 and Q2 to sense and drive opto-couplers such as 4N25.
2. Needs to find more appropriate transistors in couple watt range with high gain and low leakage currents.
3. Like Brian wrote only one opto-coupler should be used and resistors in series with bases must be mandatory.
4. It is hard to believe that this circuitry could drive opto-couplers 4N25 when collector currents of Q1 and Q2 are limited by so huge 1Mohm resistors R5 and R6.
Vladimir Doubovis - 2011-11-10 14:17:30 PDT -
The circuit seems to use a lot of unnecessary parts. Capacitors C3 and C4 are in parallel, and could therefore be reduced to one component. The collectors of Q1 and Q2 could be connected together, obviating IC2, C2, R2, and R6. Also, if resistors were added in series with the bases of Q1 and Q2, so that nearly the full overload current didn't flow through them, perhaps Q1 and Q2 could be a bit smaller than the 100W TO220 devices specified.
Brian Rhodefer - 2011-10-10 12:12:49 PDT
