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test-and-measurement:analog-discovery-pro-5250:transistor-tester [2022/01/13 09:34] – [Testing] Álmos Veres-Vitályostest-and-measurement:analog-discovery-pro-5250:transistor-tester [2022/10/04 11:38] (current) – external edit 127.0.0.1
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   * [[test-and-measurement:analog-discovery-pro-5250:start|]]   * [[test-and-measurement:analog-discovery-pro-5250:start|]]
   * [[https://www.diodes.com/assets/Datasheets/ZVN3310A.pdf|ZVN3310A]], or other transistor   * [[https://www.diodes.com/assets/Datasheets/ZVN3310A.pdf|ZVN3310A]], or other transistor
-  * 2x 10W 100Ω ceramic resistors+  * 2x 10W 51Ω ceramic resistors
  
 === Software === === Software ===
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 In this test, the drain current versus the drain-source voltage of the selected MOSFET will be plotted for varying gate-source voltages. Connect the gate of the transistor to the waveform generator and the source to the ground. In this test, the drain current versus the drain-source voltage of the selected MOSFET will be plotted for varying gate-source voltages. Connect the gate of the transistor to the waveform generator and the source to the ground.
  
-To limit the maximum drain current, a 50Ω resistor is needed in series with the transistor. The maximum output current of the ADP5250 on the +25V supply is 500mA, so the maximum power dissipated by the resistor, in the case of an ideal switch instead of the MOSFET (with 0Ω resistance), would be $P_{R}={I_{R}}^{2}*R={(0.5A)}^{2}*50Ω=(0.25*50)W=12.5W$ which is higher then the rated power of the 10W ceramic resistors so two pieces of 100Ω resistors will be used in parallel (the current is divided equally between them, so the dissipated power will be only half, which is 6.25W, below 10W).+To limit the maximum drain current to the maximum output current of the supplies, a $R=\frac{V}{I}=\frac{V_{+}-V_{-}}{I}=\frac{25V+25V}{0.5A}=100Ω$ resistor is needed in series with the transistor. 
 + 
 +The maximum output current of the ADP5250 on the +25V and -25V supplies is 500mA, so the maximum power dissipated by the resistor, in the case of an ideal switch instead of the MOSFET (with 0Ω resistance), would be $P_{R}=I_{Rmax}*V_{Rmax}=0.5A*50V=25W$ which is higher then the rated power of the 10W ceramic resistors. To protect the circuit, two pieces of 51Ω resistors will be used in series, so the voltage is divided equally between them, the dissipated power being only half, which is 12.5W. This is still above 10W so the maximum voltage should also be reduced. If the maximum voltage drop is only 40V, the dissipated power by the resistors will be $P_{51R}=\frac{I_{Rmax}*V_{Rmax}}{2}=\frac{0.5*40V}{2}=10W$
 </WRAP><WRAP half column> </WRAP><WRAP half column>
-{{ :test-and-measurement:analog-discovery-pro-5250:transistor_tester_sc.png?600 |}}+{{ :test-and-measurement:analog-discovery-pro-5250:transistor_tester_sch.png?600 |}}
 </WRAP></WRAP> </WRAP></WRAP>
  
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 </WRAP> </WRAP>
 </WRAP><WRAP half column> </WRAP><WRAP half column>
-{{ :test-and-measurement:analog-discovery-pro-5250:transistor_tester_bb.png?600 |}}+{{ :test-and-measurement:analog-discovery-pro-5250:transistor_tester_bbd.png?600 |}}
 </WRAP></WRAP> </WRAP></WRAP>
 ---- ----
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 ===== Writing the Script ===== ===== Writing the Script =====
 <WRAP group> <WRAP group>
-The full script can be downloaded {{|here}}.+The full script can be downloaded {{ :test-and-measurement:analog-discovery-pro-5250:transistor.zip |here}}.
 </WRAP> </WRAP>
  
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 # parameters # parameters
 Vd_resolution = 500 # drain voltage resolution in mV Vd_resolution = 500 # drain voltage resolution in mV
-Vd_limit = 25       # maximum drain voltage in V+Vd_limit = 40       # maximum drain voltage in V
 Vg_resolution = 2   # gate voltage resolution in V Vg_resolution = 2   # gate voltage resolution in V
 Vg_limit = 12       # maximum gate voltage in V Vg_limit = 12       # maximum gate voltage in V
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             print("\tDrain voltage: " + str(drain_voltage) + "V")             print("\tDrain voltage: " + str(drain_voltage) + "V")
             # set the drain voltage             # set the drain voltage
-            supplies.switch_25V(device_handle, True, False, drain_voltage, 0)+            if drain_voltage > 25: 
 +                supplies.switch_25V(device_handle, True, True, drain_voltage - 25, -25) 
 +            else: 
 +                supplies.switch_25V(device_handle, True, True, 0, drain_voltage * (-1))
  
             # delay             # delay
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 </WRAP><WRAP half column> </WRAP><WRAP half column>
 +{{ :test-and-measurement:analog-discovery-pro-5250:output_zvn3310a.png?600 |}}
 </WRAP></WRAP> </WRAP></WRAP>
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