- #1
mzh
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Dear Physics Forum users
Is it true that conventionally, [itex]V_{SD}[/itex] is defined as the difference in potential between the drain and the source. Meaning, [itex]V_{SD} = V_S - V_D[/itex] with the voltage on the source as the reference.
As an example: if [itex]V_{SD}[/itex] is given as -5V (note minus), then this means the drain is at a potential of 5 Volts lower than the source. Right?
Thanks for hints.
Is it true that conventionally, [itex]V_{SD}[/itex] is defined as the difference in potential between the drain and the source. Meaning, [itex]V_{SD} = V_S - V_D[/itex] with the voltage on the source as the reference.
As an example: if [itex]V_{SD}[/itex] is given as -5V (note minus), then this means the drain is at a potential of 5 Volts lower than the source. Right?
Thanks for hints.