I am using a P-channel MOSFET as a switch in my design & I have connected the drain to the load. The load needs 0.8mA of current. How can I decide whether the MOSFET is conducting that much current if I connect the source of the MOSFET to 3V ? The MOSFET used is BSS84 from Fairchild & the link for BSS84 datasheet is given below.
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\$\begingroup\$ More info needed. LOad is what> resistor ? what value ? If you connect load from 3V to ground what curr ent does it draw? have you got atest meter ? \$\endgroup\$– Russell McMahon ♦Commented Mar 16, 2012 at 7:52
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\$\begingroup\$ Madhu, are you sure that you have to drive 0.8 mA? It's a pretty small current, can I ask you what are you driving? \$\endgroup\$– clabacchioCommented Mar 16, 2012 at 10:08
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1 Answer
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I think this is the graph of your datasheet that gives all the information you need:
So assuming this circuit
V_GS is -3V which allows I_D to be up to ca. 0.5 A.
EDIT:
clabacchio is right pointing out that the graph is actually for V_DS = 5V.
Of course a graph for V_DS = 3V would be better, but there isn't one in the datasheet.
So the question is: Will the resistance R_DSON be different for very low drain currents?
Looking at Figure 2 ("On-Resistance Variation with Drain Current and Gate Voltage") gives the answer:
it does change vs. I_DS, but only for currents above ca. 0.1 A and: resistance even gets lower (=better) for lower I_DS.
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\$\begingroup\$ The current spec is 0.8 mA (may be wrong?) and the VDS for that graph is 5V, too high for this application \$\endgroup\$ Commented Mar 16, 2012 at 10:07
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\$\begingroup\$ I think Curd is write with that 2 figures. From the graph, Id for 3V Vgs is approximately 0.5A to 0.6A. But I need only 0.8mA for programming an EEPROM (i.e, Programming Current). Now, I want to know how much current that EEPROM will take? \$\endgroup\$– MadhuCommented Mar 16, 2012 at 10:46
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\$\begingroup\$ @clabacchio: concerning I_D: The OP wants to use the MOSFET as switch (not as current source). Since 0.5 A >> 0.8mA this is ok. Concerning V_DS: see my EDIT \$\endgroup\$– CurdCommented Mar 16, 2012 at 11:16
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\$\begingroup\$ @Madhu ok, my assumption about current sources was wrong, I read a different thing :) \$\endgroup\$ Commented Mar 16, 2012 at 12:30
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\$\begingroup\$ Anyway I think it would be better posting Fig 1 and saying that at the specified current level the VDS is very small (close to 0). \$\endgroup\$ Commented Mar 16, 2012 at 15:57