We’ve used both a fuse and PTC
because the fuse reacts faster to very
high currents, protecting the rest of
the circuitry on the board if there’s a
serious fault, but the PTC does not need
to be replaced if it ‘trips’ and helps the
circuit to handle moderate overloads
without damage.
PTC1 normally has a low resist-
ance – around 100mΩ below 1A – but
if the current through it increases, its
resistance rises, limiting it at around
2A (given enough time for it to heat
up). This would normally only occur
if the 5V line rises above 5.5V and the
USB Port Protector is shunting current
in order to prevent it rising further.
In fact, the USB Port Protector does
very little as long as the USB supply
voltage is in the normal range of 0V
to 5.25V and the D– and D+ lines are
in the normal range of 0-3.3V. Green
LED1 lights up to indicate power is
present, but that’s about it. The USB
Port Protector draws around 3mA in
this condition.
If the 5V rail is pulled negative,
ie, below 0V (eg, you’ve accidentally
shorted it to the output of a trans-
former or some other supply rail)
then schottky diode D3 will conduct.
This prevents VCC from going below
about –0.5V.
D3 is a high-current diode, capable of
handling 15A continuously and 275A
for around 5ms, so it makes a very ef-
fective clamp. It limits the voltage on
VCC to –0.55V at 15A, so your PC is safe
from damage from negative voltages on
the supply line.
Should the overload condition per-
sist, either PTC1 will limit the overload
current to a safe level or F1 will blow,
disconnecting the compromised cir-
cuitry from your computer.Clamping positive voltages
It’s even more likely that you might ac-
cidentally short the 5V rail to a higher
voltage; eg, 12V from a car battery. Just
think of the heavy currents which will
fry anything connected to it! The USB
Port Protector has active and passive
systems to handle this situation.
The active system is the first line
of defence. It comprises high-current
PNP transistor Q1 and shunt voltage
reference REF1. The 1.2kΩ/1kΩ resis-
tive divider across the 5V supply feeds
45.45% of the supply voltage to the
adjust terminal of REF1. It sinks current
into its cathode terminal as soon as this
adjust terminal exceeds +2.5V.
So, given the voltage divider, that
means that it will sink current when
the supply exceeds 5.5V (2.5V ÷
45.45%). This will cause a voltage to
develop across the 470Ω resistor and
once that voltage exceeds around 0.7V
(Q1’s base-emitter voltage), Q1 will
switch on and shunt the 5V supply
rail, pulling it down.
In this manner, REF1 and Q1 act to
limit the 5V supply rail to just over
5.5V. Q1 is capable of handling more
than 10A, but since there will be 5.5V
between its collector and emitter, it can
only do that for a very short time before
it overheats. But at the same time PTC1
will rapidly heat up and increase itsFig.1: the circuit diagram of the USB Port Protector. Diode D3, zener TVS1 and transistor Q1 are all connected between
VCC and GND and shunt current when an excessive voltage is applied, while polyswitch PTC1 and fuse F1 prevent large
currents from flowing if the fault is serious. Diodes D1, D2 and zener TVS2 protect the D+ and D– data lines.resistance, to limit that current. And
in any case, if the current exceeds 3A,
for example, the fuse will very quickly
blow before Q1 is damaged.
So REF1/Q1 act together as a very
precise and very fast clamp. When
REF1 is sinking current from Q1’s
base, Q2 will also normally switch on
as its base is also pulled around 0.7V
below its emitter, via the 10kΩ resistor.
This will light up red LED2, indicating
that the clamp is operating and that
you have a problem. LED2’s current
is limited by its low base current and
relatively fixed gain (hFE).
REF1 can sink up to at least 100mA
and Q1 has a current gain (hFE) in the
hundreds, so Q1 is more than capable
of passing its full peak current rating
of 24A in this circuit.
Note that LED2 may go out if there
is a persistent overload, since when
Q1 heats up, its base-emitter voltage
will drop and it may drop low enough
below Q2’s base-emitter switch-on
voltage that it will no longer switch
on. But chances are that PTC1 and/
or F1 will have acted to limit the fault
current by that stage anyway.
The only problem with the clamp
provided by Q1 and REF1 is the reac-
tion time. It takes a short time for REF1
to react to an increase in the feedback
voltage and it also takes time for Q1
to switch on – around a microsecond.Passive clamping
This is why we also have a transient
voltage suppressor (TVS1) connected
across the 5V supply rail. It’s a passiveUSB Port Protector