Calibration - Charge Converter Amplifiers
8500C/8500C+ System Maintenance Manual
D-13
Procedure
1. With nothing connected to the input of the Charge Fixture, connect its output the 8520CS input.
2. Measure the 8520CS input capacitance (between TP1 and TP2 of the Charge Fixture). Record this
value as C
in
(pF) (should be about 0.1 mfd = 100,000 pF).
3. Determine the capacitance and charge sensitivity of the intended acclerometer:C
aac
(pF) and
S
Q
(pC/g).
4. Connect the input of the Charge Fixture to a signal generator. Configure the generator for a 100
+0.1 Hz sinewave. Using a floating DVM, monitor the voltage between TP1 and TP2. Adjust the
generator for a DVM reading of 100 +0.1 mV
rms
.
5. Connect the 8520CS according to Figure D-8.
6. Remove the cover from the 8520CS and set the DIP switch for the charge sensitivity closest to that
of the intended accelerometer. This switch is set according to the Decal inside the instrumenct. See
Table D-4 for channel switch and potentiometer adjustments. See Figure D-10 for DIP switch
settings. See Figure D-9 for potentiometer and DIP switch locations on the Signal Conditioner PC
board.
7. Calculate the required 8500C/C+ ips reading:
8. Set up the 8500C/C+ for
BALANCE
. In
SETUP
set the balance frequency to 100 hz, and the
balance type to
STROBEX
. Also select
MANUAL
STOP
.
9. Press
ENTER
softkey, wait 10 seconds, then press
START
and adjust the 8520CS trimpot (see
Table D-4) for an 8500C/C+ reading of the value calculated in step 7. Ignore the
STROBEX
instructions; press
STOP
when finished.
EXAMPLE: You wish to calibrate and 8520CS which is intended to be used with an
Endevco 6233C-50 accelerometer connected to 8520CS.
You’ve measured the input capacitance of the 8520CS: C
in
= 100400 pF.
You’ve set the generator for: V
in
= 100 mV
rms
between TP1 and TP2 (f = 100 hz sine wave).
Assume the 6233C-50 accelerometer has:
SQ = 50 pC/g, C
acc
= 1350 pF.
Hence,
N O T E :
S e n s o r
C
a c c
( p F )
S
Q
( p C / g )
6 2 3 3 C - 5 0
1 3 5 0
5 0
7 7 0 4 A - 5 0
2 8 0 0
5 0 ( N o m i n a l )
i p s
8 5 0 0 C
= V
i n
2
6 1 . 4 5 ( C
i n
+ C
a c c
)
f S
Q
(
)
W h e r e : V
i n
:
C
i n
:
C
a u
:
f :
S Q :
i p s
8 5 0 0 C
= 0 . 1 0 0 2
6 1 . 4 5 ( 1 0 0 4 0 0 + 1 3 5 0 )
(
)
1 0 0 5 0
= 1 7 6 . 8 i p s ( p K ) + 1 %
