Loading Effects - Keithley 6430 Instruction Manual

F-10 Measurement Considerations
Overload protection
The Model 6430 may be damaged if more than 200V is applied to the input. In some appli-
cations, this maximum voltage may be unavoidably exceeded. In these cases, additional over-
load protection is required to avoid damaging the input circuitry of the instrument.
Figure F-4 shows a protection circuit consisting of a resistor and two diodes (IN3595). The
leakage of the 1N3595 diode is generally less than 1pA even with 1mV of forward bias, so the
circuit will not interfere with measurements of 10pA or more. This diode is rated to carry
225mA (450mA repeated surge). Since the voltage burden of the ammeter is less than 1mV, the
diodes will not conduct. With two diodes in parallel back to back, the circuit will provide pro-
tection regardless of the input polarity.
The resistor (R,) must be large enough to limit the current through the diodes to prevent
damage to the diodes. It also must be large enough to withstand the output voltage. A good rule
of thumb is to use a large enough resistor to cause a one volt drop at the maximum current to be
measured.
The protection circuit should be enclosed in a light-tight shield that is connected to input/
output low.
Figure F-4
Overload protection
for ammeter input
High impedance voltage measurements

Loading effects

Circuit loading can be detrimental to high-impedance voltage measurements. Fortunately,
the input resistance of the Model 6430 voltmeter is very high (>10
a very large load resistance to cause voltmeter loading.
To see how meter loading can affect accuracy, refer to Figure F-5 where the SourceMeter is
configured to measure voltage only. R
while R
be calculated using the formula in the illustration. To keep the error under 0.1%, the input resis-
tance (R
tance of Model 6430 is >10
of the measured voltage must be <10
R
HI
LO
represents the input resistance of the voltmeter. The percent error due to loading can
IN
) must be about 1000 times the value of the source resistance (R
IN
16
Ω. Therefore, to keep the error under 0.1%, the source resistance
To Ammeter
represents the resistance component of the source,
S
13
Ω.
16
Ω), therefore, it would take
). The input resis-
S