Hello!
There is such a button with a backlight, when connecting the backlight to a 24V power supply, it consumes 6-7 mA.
After trying to control this backlight using stm32f407, the pin in open-drain mode burns out. Now in open-drain mode, in a high state, there is a potential of 1.2 V on the leg, and it should hang in the air. After that, in push-pull mode, the pin works fine, although I did not measure the consumption without a load.
The question is, why is the pin lit? According to the datasheet, the maximum load on the pin is 20 mA, in output mode, the pin resistance is close to zero, so the current should be the same as when connecting the backlight to the power supply, i.e. 6-7 mA. Why is the pin lit? If I made a mistake in the calculations, then how can I correctly calculate the permissible load for operation in this mode?
I want to understand how exactly the currents flowed and where the malfunction occurred.
Here is the driver diagram.
As I understand it, the port burns out because the open drive mode is not turned on immediately after power is supplied, but after the initialization of the gpio. Before initialization, the output resistance is very high, so when 24V is supplied through the LED and resistor, the current will flow through the protective diode to VDD_FT. The pin will have a voltage of VDD_FT + voltage drop on the protective diode, which does not exceed the permissible value of VDD+4.
The current is 6 mA. The datasheet states that positive leakage current is impossible, the main thing is not to exceed the voltage VDD+4 on the pin.
What happens next, how do currents flow in the driver, what elements are destroyed and why? Why is there now a voltage of 1.2 V on the pin in the high state in open drain mode? What resistor value would not burn out the controller pin at 24 volts?
The circuit is very simple, 24V is supplied from the power supply to the LED and resistor, and connected to the open drain pin of the stm32. GND of the power supply and stm32 are connected. I understand that I have exceeded the input voltage and that such an LED must be connected through a transistor. I do not understand why there was 24V on the pin, and not vdd_ft + voltage drop on the protective diode. If this happened after the protective diode burned out, then how to calculate the value of the protective resistor so that it does not burn out? If the protective diode burned out from 6 mA, then from what current will it not burn out? @gbm said that perhaps a 200k resistor is enough, what voltage will be on the pin then?
So, combining your diagram with Fig 25 from your opening post, we have:
Or:
If we take VDD_FT as 5V, and 2V across your LED, that leaves 17V across the 4k resistor - giving ~ 4.3mA
@Snetka wrote:If the protective diode burned out from 6 mA, then from what current will it not burn out?
Table 12 says 5mA
Why 17v, there is also a voltage drop on the protective diode, say 0.6v.
In fact, the resistor is most likely smaller than the nominal value, it is soldered in the button and I can not measure it separately. According to Ohm's law, about 3.15-3.6k.
If vin>vdd_ft, then this is not a positive injection current? In the table it is generally equal to zero, and in the note it is written that a positive injection current is impossible and you must not exceed vdd+4 on the pin.
If this equivalent circuit is correct, then there will be 5.6V on the pin with a current of 6mA. Because of this, the protective diode burns out, and instead of 5.6V on the pin there is almost 24V due to the closed transistors.
If you double the resistor value, the current will be 3mA. Will the port burn out with such a current?
