thanks man, pretty clear and with the pictures and all i finnally got it. Really helped that you covered the topic so extensively (covering all states and what not)
I dont understand. R already limits the current. Led has additional resistance. FET is shorting line to ground, so current flow through R. When FET is closed resistance of line is R + Led. Is current enough to power Led? If so, why R doesnt burn when shorted to Gnd?
this configuration is used for example on an I2C-Bus to prevent a current flow between two devices if they are set to output mode (one device could give Vcc on the line while one device could give GND on the line -> 3.3V to 5V voltage difference between those two transistors results in a big current which can destroy the devices). If both devices are used in an open drain configuration the output pins can only drive the whole data line to GND and never drive it to a higher voltage by themselves (data line is only on high if both devices would close their transistors). So even if two devices are set to output mode there is no state where there is a voltage difference between the devices so no current will flow.
This is not how you drive an LED. The LED should be connected from VCC, through an appropriate current limiting resistor and to the GPIO pin. It then DRAINS into the MCU. In the setup you show, the device would draw more current with the LED off, than with the LED on! That is not a good way to set this up.
thanks man, pretty clear and with the pictures and all i finnally got it. Really helped that you covered the topic so extensively (covering all states and what not)
Thanks for that. It's the first really good explanation of open drain output that I've seen.
short and clear explanation. thanks a lot greattings from Argentina..
I dont understand.
R already limits the current. Led has additional resistance.
FET is shorting line to ground, so current flow through R.
When FET is closed resistance of line is R + Led.
Is current enough to power Led?
If so, why R doesnt burn when shorted to Gnd?
If set R enough large, current will reduce, the R does not burn. I hope you can understanding.
Very useful video
Thanks for this
Hi sir, Is there any advantage of using this configuration over normal configuration?
this configuration is used for example on an I2C-Bus to prevent a current flow between two devices if they are set to output mode (one device could give Vcc on the line while one device could give GND on the line -> 3.3V to 5V voltage difference between those two transistors results in a big current which can destroy the devices). If both devices are used in an open drain configuration the output pins can only drive the whole data line to GND and never drive it to a higher voltage by themselves (data line is only on high if both devices would close their transistors). So even if two devices are set to output mode there is no state where there is a voltage difference between the devices so no current will flow.
when we output "1" state it means that it output "0" to the nmos transistor ? as always
in Diagram of GPIO Mode with Open drain Pullup you have written P instead of N correct that
This is not how you drive an LED. The LED should be connected from VCC, through an appropriate current limiting resistor and to the GPIO pin. It then DRAINS into the MCU.
In the setup you show, the device would draw more current with the LED off, than with the LED on! That is not a good way to set this up.
can you please share your udemy course link i am interested and how much it cost
good one...
thank your sir
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