ishwaryasampath
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Only by changing the circuit. The DC output of a source follower is reduced by a voltage somewhat greater than the gate-source threshold voltage.i was connected n channel mosfet (ZXMN6A08K) gate voltage as 5v and drain connected to 5v, the load connected at source. The problem was there was a voltage drop at the output ie., source. i want 5v output at the source. is there any solution to rectify this condition.
MOD:Threads merged, please don't cross post
transistor switching is not a problem. the problem is voltage drop in the 2nd p channel MOSFET.
for p channel mosfet:
test 1:
gate voltage = 0v
source voltage = 5v
output voltage(drain)=5v
test 2:
gate voltage = 0v
source voltage= 1.2v
output voltage(drain) = negative voltage
the actual output is 1.2v but we get negative voltage at output. we want to get drop less 1.2v output. is any solution for that?
for n channel mosfet:
gate voltage = 5v
drain voltage =5v
output voltage = 3.5v
the actual output is 5v but we get 3.5v dropped output. we want to get drop less 5v output. is any solution for that?
What is the voltage of the battery?
What is the voltage of the battery?
What is the voltage of the battery?
OK. In your circuit, change the bottom transistor to an N-channel, logic-level type MOSFET with the drain connected to V1.2 and the source as the output. Connect the gate to the 9V battery voltage to turn the MOSFET on and to ground to turn it off (which can be done with an NPN transistor and a resistor to the 9V). The 9V will be enough to fully turn on the logic-level device for the 1.2V drain voltage so the voltage drop across it will only be due to the the ON resistance of the MOSFET.the battery voltage is 9v
that logic not suitable for our application.the condition for our application is both logic outputs get at the same time when the kbt battery goes to ov.OK. In your circuit, change the bottom transistor to an N-channel, logic-level type MOSFET with the drain connected to V1.2 and the source as the output. Connect the gate to the 9V battery voltage to turn the MOSFET on and to ground to turn it off (which can be done with an NPN transistor and a resistor to the 9V). The 9V will be enough to fully turn on the logic-level device for the 1.2V drain voltage so the voltage drop across it will only be due to the the ON resistance of the MOSFET.
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