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MOSFET

cheniour

Member
I need to used it in 2 switch half bridge rectifier that's mean Replacing the diodes with Nchannel MOSFETs as a diodes M3 and M4
1606690316282.png
 

crutschow

Well-Known Member
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That doesn't tell us what you are trying to do.
Exactly what are the inputs and outputs to this "bridge"?
 

rjenkinsgb

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The FETs need a significant gate voltage (relative to the source) to turn on. Without a separate gate control for the top FETs, they will be far less efficient that schottky diodes and never conduct at 0.5V.
I cannot read the FET numbers in the diagram so cannot look up the specifications.

Note that with 0.5V RMS in and a hypothetical perfect rectifier, you would only get 0.707V out of that circuit and not 1.5V
 

gophert

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I think we need to know what your load is, how you plan on stepping up a 0.5vAC source to 1.55vdc. Rectifying, even with an ideal rectifier, does not get a 0.5 v sine wave (rms, peak, peak-to-peak, average) to 1.55v. I don't know how you are getting there.

how much current are you hoping to deliver to your load - and what is your load?
 

rjenkinsgb

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Most Helpful Member
@rjenkinsgb How you find 0.707 v please
The ratio of the RMS value of a sine wave to its peak value is 1.414 (or the RMS is 0.707 x the peak).

0.5V x 1.414 = 0.707, as it happens.

It's nothing to do with the square of the voltage, if the other answer seems to imply that.
 

gophert

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Most Helpful Member
The ratio of the RMS value of a sine wave to its peak value is 1.414 (or the RMS is 0.707 x the peak).

0.5V x 1.414 = 0.707, as it happens.

It's nothing to do with the square of the voltage, if the other answer seems to imply that.
play with the numbers a bit more and you'll realize that 1/1.424 is 0.707 or 70%. And, therefore, rms is 70% of peak.

it is a great number to memorize. It is the square root of 1/2 and therefore, the reciprocal of 1.414.
With a few more decimal places = 0.70710678
 

crutschow

Well-Known Member
Most Helpful Member
My virtual HP-42s gives 0.707106781186547524400844362104849. :happy:
 

gophert

Well-Known Member
Most Helpful Member
gophert peak to peak 1v sine and the load current equal 640 µA
ok, but are you trying to...
- build an energy harvesting system (hand-cranked generator or similar) to rectify and then use the power? In this case it wouldn't make any sense to add battery operated components to actively rectify the signal.
Or,
- monitor an AC signal from another system so you can add batteries or other power supply to power microcontroller, Mosfets and more?
 

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