MOSFET

[cheniour]

hello

someone know a reference from a mosfet ideal diode better than DIODE BAT54

 

[ChrisP58]

The BAT54 is a Schottky diode which, while better than a standard silicone diode, still has forward voltage losses.

A mosfet can be used to make a near ideal diode, but you need to turn it on and off depending on the direction of the current.

Analog devices has parts that will do that. The LT8672 is one such part.
https://www.analog.com/en/products/lt8672.html#product-overview

 

[gophert]

hello

someone know a reference from a mosfet ideal diode better than DIODE BAT54

what are you trying to do? Do you need to rectify an AC signal? Peak detection, or something else?

 

[cheniour]

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

 

[crutschow]

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

 

[cheniour]

The input voltage 0.5 sinusoidal and the output 1.55V direct

 

[rjenkinsgb]

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]

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?

 

[cheniour]

gophert peak to peak 1v sine and the load current equal 640 µA

 

[cheniour]

rjenkinsgb How you find 0.707 v please

 

[gophert]

rjenkinsgb How you find 0.707 v please

,707 is reciprocal of 1.414. it is the square root of 0.5

 

[rjenkinsgb]

@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]

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]

My virtual HP-42s gives 0.707106781186547524400844362104849.

 

[gophert]

My virtual HP-42s gives 0.707106781186547524400844362104849.

And Sheldon Copper from The Big Bang Theory was completely wrong about the best number being 73The best number is, 0.5^0.5. Or, as crutschow approximates,

... 0.707106781186547524400844362104849.

 

[gophert]

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?