Keysight Technologies B1505A Handbook page 65

B`. Measurement Procedure: Ic-Vceo_R Application Test
- Starting from pre-defined test setup of My favorite Setup
This test approach starts measurements by using a pre-defined test setup saved in
My favorite Setup instead of starting from a scratch by using an Application Test
Library.
Setup and execute the test by following the next steps. The same instruction steps
are illustrated by the numbers on the "1`. Starting from Application Test Library"
side of figure 3-15.
Step 1. Click the Preset group of My Favorite Setup.
Step 2. Select Example_AT preset group.
Step 3. Select Ic-Vceo_R100k (Click the Ic-Vceo_R100k).
Step 4. Click "Recall" button.
Then go to step 5 of "Starting from Application Test Library" above, and continue by
following the test step numbers.
Note: 100 k Ω Ω Ω Ω resistor error and the calibration:
There is roughly 5% maximum error in the 100 kΩ resistor used in the N1259A Test
Fixture. The calibrated 100 kΩ value of this parameter is used in the compensation
routine of this application test, and more accurate Vce value can be obtained by
using the calibrated resistor value.
The Vce calculation when using the 100 kΩ resistor is made by the following for-
mula:
Vce = Vsmu - Ic * 100 kΩ value
where, Vsmu is SMU output voltage and Ic is the output current from the SMU, and
within
they are about 0.03% error in the range of application voltage and current.
The resistor can be compensated within 0.1 % by using the calibration routine: "100
kohm CT" Classic Test example routine introduced in Section 3-2-3.
The following shows two examples of BVceo error when using a resistor with 5%
and 0.1% accuracy for measuring the MJL4281AG example.
As can be seen from the calculation, using the 0.1% calibrated resistor results in
within 0.2% BVceo error, but non-calibrated 5% error resistor shows about 10% er-
ror.
It is important using the calibrated resistor when using this application test example
with 100 kΩ resistor..
0.1% resistor error example:
Vce = 1,189.97V - 7.98569 mA * 100 k (1 +- 0.001)
= 391.4 +- 0.8 V
(-0.2% ~ +0.2%)
= 390.6 V ~ 392.2 V
5% resistor error example:
Vce = 1,189.97V - 7.98569 mA * 100 k (1 +- 0.05)
= 391.4 +- 39.9 V
(-10% ~ +10%)
= 351.5 V ~ 431.4 V
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