Solid-state overcurrent-protection ICs such as USB switches and card-slot power switches offer a simple and robust protection for pins at risk of overloads and short circuits. Such switches, however, are at risk for overvoltage and even destruction if the energy stored in the stray inductance is not accounted for. Thus, the degree of protection against fault conditions is not without limits.
For a current limit of 1.2 A, one would think that a circuit-protection IC would maintain complete control in the event of a short circuit or other fault (Fig. 1). However, the current limit usually allows a delay time before the circuit is actually interrupted. During a hard short, the current can rise rapidly: first reaching the dc limit, then initiating a turn-off of the internal switch. (Typical dc limits are associated with an accurate but slow-response threshold, which limits the steady-state current that can be drawn from the switch.)
A short time later the switch opens, but not before the peak input current reaches a limit that can be much higher than the dc limit. Low-inductance leads can cause the current to rise even faster. If a hard short is applied to the output, the input current rises rapidly until it reaches a limit.
Current Limiting by Resistance
In Fig. 1, to analyze the response of the circuit to a fault condition, consider a USB switch with low-inductance leads (such as the MAX1558 or equivalent) subjected to a hard short at the output. (The same analysis applies to other semiconductor power switches with fast turn-off capability.) In this example, input currents are resistively limited by the internal protection switch. When the protection circuit finally opens, the peak input current (IIN_PK) can be measured as shown in Fig. 2. With peak current flowing through the stray input inductance (LSTRAY), the stored energy (E) is:
E = ½ 3 LSTRAY 3 I2.
Where does this energy go when the circuit breaker or protection switch finally opens? As shown in Fig. 2, the input current (IIN) rises rapidly to 48.8 A, then limits due to resistance. When the switch turns off 12 µs later, the rate at which the current slews back down can be measured. With IIN slewing down at 20 A/µs and the input voltage (VIN) soaring to 8.6 V (VMAX), the circuit inductance can be empirically calculated using this equation:
(VMAX - VIN) = di/dt × LSTRAY.
With VMAX - VIN = 3.6 V and di/dt = 20 A/µs, LSTRAY = 180 nH.
So with E = ½LSTRAY × IINPK2, LSTRAY stores 214 µJ at the end of a fault. Bypass capacitance must be added to absorb this energy and limit the voltage rise. An input capacitor of 10 µF with an initial charge of 5 V stores some initial energy, ½ × C × V2 = E. Assuming that all the stored energy in LSTRAY will transfer to the input capacitor (CBYPASS), then initial energy + stray energy = final energy:
125 µJ + 214 µJ = 339 µJ.
This final energy, 339 µJ, in the input capacitor is:
½ × C × V2 = E, or ½ × 10 µF × V2 = 339 µJ.
Solving for V yields 8.23 V, which closely agrees with the 8.6-V value measured in Fig. 2.
If the input bypass capacitor is now reduced to only 0.1 µF, the input voltage can rise to destructive levels, initial energy + stray energy = final energy:
1.25 µJ + 214 µJ = 215 µJ.
Again, ½ × 0.1 µF × V2 = 215 µJ. Solving for V yields 65.6 V.
This hard short (Fig. 3) would damage a part rated for 5.5 V. Note that the output also soars to 9.8 V. The output rises because the short is removed before the switch turns off, and that action also accounts for the high di/dt in this test. Normally, di/dt is governed by the turn-off characteristics of the power device, but a USB port is subject to the behavior of end users — who are anything but controlled. An extremely fast turn-off like this can be caused by an intermittent cable, a bad connector or, as in this case, contact bounces associated with a mechanical connection.
Clearly, the voltage does not rise to 66 V. A protection diode in the part has zenered, thereby clamping the voltage rise and likely suffering damage from the absorbed energy. During such overvoltage events, excess energy is absorbed by the silicon chip.
A 10x-faster sweep speed for the same circuit (Fig. 4) shows that a larger-input bypass capacitor is necessary for protection from the stray energy following a hard short. Note that a pc board with a continuous ground plane has much less stray inductance than do the test leads used in this test, or that found in a typical laboratory prototype. It can be especially difficult to reduce the strays from test leads and test equipment in a lab test.
The Limiting Factors
Even with as much as 1.3 µH of input lead inductance, the part can survive if the bypass capacitor is 10 µF. As shown in Fig. 5, the larger inductance causes the input current to ramp up and down slowly. It's important to note that the current cannot change as fast when the inductance is higher. Because energy stored in the inductor is proportional to the inductance value and also to the square of the current, high peak currents carry far more energy. Energy stored in the 1.3 µH inductor is only 419 µJ:
125 µJ + 419 µJ = 544 µJ, and ½ × 10 µF × V2 = 544 µJ.
Again, solving for V yields 10.43 V.
Although the part survived this hard short, a larger bypass capacitor should be used — one that limits the maximum voltage to less than the absolute maximum specified in the data sheet.
Without accounting for the stored energy in stray inductance, USB devices can be subject to overvoltage and even destroyed. Fig. 5 shows that input inductance can be the limiting factor for peak current, but Fig. 2 shows that resistance can also limit the current. One could conclude that a lower-input inductance might perform better, but only if the current is limited to nondestructive levels. If current is not limited, the energy for a low-inductance situation can reach damaging levels very fast. The response time of the switch dictates the degree of protection. Consequently, one must exercise care to ensure that this does not happen.
In most pc-board applications, for which a ground plane extends under the protection switch and under the input and output traces, the inductance should be much lower than 180 nH. A 1/16-in.-wide pc-board trace over a ground plane has approximately 10 nH of inductance per inch. However, the required value of the input bypass capacitor can vary with each application. Measurement and analysis of the expected inductance and the switch response time might indicate that a larger bypass is needed to ensure reliability, or that a lower value is permissible.