E-5-2
Constant Current Circuit Operation
Assume that terminals I and I of Figure E-3 are
hi
lo
shorted, and 1 volt is applied to
E
so that
I
is
in
hi
positive. To equalize the 1 volt applied to
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 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 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 1.1
volts.
The voltage drop across the range resistor
is 1 volt, just as it was when the output terminals
were shorted. The current through R is 10
L
milliamperes, just as it was through the jumper
when the output terminals were shorted.
E-6.
Failsafe Design
Reference to the AMPTEC 630 Tester 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-7.
The AMPTEC 630 Testers are 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 630 POWER switch is in the “
OFF/
CHARGING
” position are the batteries are
connected to only 2 of the possible 8 pin/socket
contacts of the “Connections” panel mounted
connector on the front panel to allow for
recharge (see J1 Connections diagram in section
E-6 of this manual).
For safety reasons, none of
the test lead connections (outside pins A thru
D) can ever make the pin-socket contact with
center Pin H
. If for some reason, an abusive
630 Tester user tried physically jambing the test
lead connection onto the J1 connector (i.e with a
hammer) it is vitually
physically impossible due
to the inner and outer align ring and drop into
sleeve construction of J1
. So even if this
impossible connection did some how occur, and
make contact with pin H, remember all battery
wiring is disconnected from the Main
Connection Jack “
J1”
if the AMPTEC 630
Tester is turned ON.
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 630 Igniter Tester.
CONNECTIONS
TEST LEADS
OR
BATTERY CHARGER
A= I High
B= V High
C= V Low
D= I Low
A
B
C
D
E
F
H
G
“J1” 630 Series Main Connection Jack
E=
F= Ground (- neg Battery Charger)
G=Chassis Ground
H= +5 V (Battery Charger)diode protected
