E-5-2
Constant Current Circuit Operation
Assume that terminals I and I of Figure E-3
hi
lo
are shorted, and 1.0 volt is applied to
E
so that
in
I
is positive. To equalize the 1.0 volt applied to
hi
E
in
, the inputs of IC202, IC201 must be driven
to zero. This condition occurs only when the
voltage drops across R212 and R222 are equal
to the drops across R213 and R221. For these
voltage drops to be equal, the output of IC202
must be at +1.0 volt. Since the output of IC201-
8 must be zero, the drop across R213 is 0.5
volts, making the inverting input 0.5 volts. The
drops across R212, R221 and R222 will also be
0.5 volts. Since the inputs to IC201 are
essentially equal, its output is zero (offset by the
few microvolts required to drive IC202 to +1.0
volt). Under these conditions the sum of the
voltages across R212, R213 , R221 and R222
equals the sum of
E
plus the output of IC202.
in
Consider now that the short is removed from the
I
and
I
terminals and a 100-ohm resistor (R )
hi
lo
L
is connected in its place. The current through R
L
increases the voltage at the input to IC201. A
balanced condition will be reached when the
output of IC201 is equal to the non-inverting
input of IC202. Again, this condition occurs
when the voltage drops across R212 and R222
are equal to the voltage drops across R213 and
R221. At this time the output of IC202 is 1.0
volts. The voltage drop across the range resistor
is 1.0 volt, just as it was when the output
terminals were shorted. The current through R
L
is 5 milliamperes, just as it was through the
jumper when the output terminals were shorted.
E-6.
Failsafe Design
Reference to the AMPTEC 620ESTester Igniter
Tester schematic will show that the output of
IC202-6 is actually applied to the base of
transistor Q202,
which acts as a current limiter
.
The worst-case component failure that could
occur in this circuit would be a Q202 short
,
which would effectively connect the -5 volt
supply directly across R218, D202, the range
resistor and R .
L
D203, however, acts as a 1.6 volt zener diode,
limiting the voltage that can appear across these
components. Even if every component in the
amplifier circuit shorted, the current through the
igniter could not exceed safe limits, because the
-5 volt and +5V supplies includes inherent
current limiting. Because of the design of both
supply isolation transformers T101 and T102,
the ±5 volt supplies can only deliver 20 to 25
milliamperes before the DC/DC converter
disengages, dropping the -5 volt output to zero.
See Section D.
620LM Failsafe Current Calculation - Worst
case Component Failure
V
oltage limiting diode, D203, provides a 1.6
volts maximum across R227 (10
W
) and R223
(100
W
), which are 110 Ohms across the Igniter
Testers output terminals.
(I
max
)
Current Maximum Calculation
1.6V/110 ohms = 0.014 Amperes
(15mA) max
The AMPTEC 620LM Tester is powered by a
rechargeable internal battery pack and cannot be
operated directly from the battery charging
adapter. This is to
eliminate the possibility
of
an electrical short to/from the AC line.
Only
when
the 620 POWER switch is in the “
OFF/
CHARGING
” position are the batteries
connected to rear panel charging jack.
When
the POWER switch is in the ON position
, the
batteries are disconnected from the battery
charger and connected to the internal circuits of
the AMPTEC 620 Igniter Tester.
R212
100K
R213
100K
R222
100K
R221
100K
R215
100K
RANGE
ATTENUATION
RESISTOR
RL
I hi
I Lo
E
in
Figure E-3 Constant Current Circuit
IC201
IC202
This applies to all Resistance ranges/mode except for Diode Teste Mode
