As a=20
newly hired employee recently out of college, previously known as =
=E2=80=9CJohn the intern=E2=80=9D, I thought my learning =
days were over.=20
However, I quickly found out that is not the case and had a lesson on=20
electro-static discharge (ESD) diodes of a device and the unintended =
effects=20
they may cause in an application.
I needed=20
to do some testing on the OPA192, so I grabbed my circuit board, headed to =
the lab and=20
connected everything up. I noticed some very odd behavior when I =
observed=20
the output on the oscilloscope. The device was in a unity gain =
configuration, so=20
I expected the output to be equal to the input. Figure 1 shows the input =
(channel 1) and output (channel 2) of what I saw.
&n=
bsp;
After=20
some troubleshooting, I realized I made a rookie mistake. I forgot to =
turn on=20
the power supplies! This got me thinking, how can there be an output =
without=20
powering the device?
I=20
researched this phenomenon and found Figure 2 in the OPA192 =
datasheet=20
that shows the internal ESD diodes of the device.
I found=20
that if an input signal is greater than the supply voltage by =
approximately one=20
diode drop then the positive supply diode would conduct and route =
current=20
around the device and to the power supplies. With the power supply =
turned off,=20
all that is present is the internal impedance of the supply. Figure 3 =
shows a=20
simplified version of Figure 2 with an equivalent model of the power=20
supplies.
Routing=20
the input signal to the supply will create a voltage drop across the =
impedance=20
of the power supply and this will cause a voltage to be present on =
the=20
power supply pins. Table 1 shows the voltage measured at the output, =
positive=20
supply and negative supply for different input =
voltages.
As you=20
can see, with a positive voltage at the input the V+ supply has a =
measurable=20
voltage. This is because ESD diode D1 is conducting and creating a =
voltage on=20
the positive power supply. Similarly, with a negative input, a voltage =
is=20
present at V- supply because ESD diode D2 is conducting. The voltage at =
the=20
supplies turns on the internal transistors of the device and it begins =
to=20
operate as it would with the supplies turned on. This is why I saw a =
distorted=20
sine wave at the output with no power to the device.
In a=20
typical system, there are multiple devices that share the same power =
supply.=20
These devices may turn on unexpectedly because of the voltage at the =
power=20
supply pins caused from turning on the ESD diodes of the device. Figure =
4 shows=20
a block diagram of how sharing the same supply can potentially power =
other=20
devices, such as analog to digital converters (ADC) or logic gates.
Furthermore, having an unexpected output =
similar=20
to the one shown in Figure 1 is just as problematic. The output might be =
connected to a comparator that interfaces with a microcontroller (MCU). If the comparator and MCU =
are powered=20
with a separate supply, the output of the amplifier could cause the =
comparator=20
to output false data to the MCU. Figure 5 shows a block diagram of how =
having an=20
unwanted output to a comparator can cause a MCU to receive false data.=20
Both of=20
these situations will give false results in the system, which can cause =
the=20
entire system to fail. The effects of this vary but can be as drastic as =
a car=20
turning off while driving or disabling a piece of medical equipment.=20
If your=20
design can have an input with the device not powered, you may want to =
consider=20
placing a switch between the input signal and input pin of the device.=20
Eliminating any possible input to the device will prevent the internal =
ESD=20
diodes from conducting and powering the device (and possibly other =
devices) with=20
the input signal.