How should we understand the OD gate, which can be used for both input and output?

小太阳yy

How should we understand the OD gate, which can be used for both input and output? [Copy link]

 

The following is the chip's fault feedback FAUTL pin, which is inside the chip. It is an OD gate with a weak pull-up inside. The FAULT pin can be used as both input and output.

FAULT pin as output:

When the chip detects a fault, the internal MOS is turned on and a current of 750μA flows from the pull-up source through the MOS tube to GND. Is this what you mean?

When the chip does not detect a fault, the internal MOS is not conducting. From the table below, we can see that the weak pull-up current is 8μA. Is this 8μA the leakage current flowing through the MOS tube? Is this true?

Then the manual says that the actual voltage of FAULT depends on the external circuit. How should I understand this sentence?

Then the FAUTL pin can be used as an input: when it is detected that the voltage of the FAULT pin is lower than 0.7V, all parallel chips will be turned off. When it exceeds 2V, all chip channels will be turned on. So how does the current flow at this time?

For example, in the figure below, the upper chip detects a fault. After the MOS tube is turned on, will the lower chip have a current injection? I don't understand this.

LeoMe

The characteristics of the OD gate + pull-up resistor combination itself are probably determined by the characteristics of the OD gate + pull-up resistor. It is a multi-input AND gate. When all outputs are 1, the OD gate outputs high impedance and the bus is pulled up to 1. When any OD gate outputs 0, the entire bus is 0. This is the output

As for input, if a chip itself outputs a fault, then it knows that its output is low, and this situation does not matter. If a chip itself does not fault, and its OD output is high impedance, its own bus should be pulled up to 1, but the chip detects that the bus is low, then there must be a chip on the bus that has a fault and its OD output is 0, then this chip can be shut down along with everyone else.

As for the two currents, it may be because the pull-up inside the chip is a weak current source, providing a high level for itself when the fault pin floats.

Each OD gate internal pull-up provides 8uA current. If the ODs output low level, the MOS will provide 750uA pull-down current after turning on. These two parameters should be used as a reference for you to design external pull-up and pull-down.

maychang

It seems that you have studied 51 single-chip microcomputer and asked questions about the 51 single-chip microcomputer assembly program.

The P1, P2 and P3 ports of the 51 microcontroller are weak pull-up OD gates, which can be used as both output and input.

chunyang

The OD structure itself is an output circuit. To be used as a bidirectional input and output port, OD itself is not suitable, but it can be connected in parallel with the input circuit. When the OD drive is invalid, it can be used as an input port normally.

小太阳yy

chunyang published on 2023-5-11 13:47 The OD structure itself is an output circuit. To be used as a bidirectional input and output port, the OD itself is not feasible, but it can be connected in parallel with the input circuit. When the OD drive is invalid, it is...

I am a little confused about what this weak pull-up current means. Where does the current flow to?

小太阳yy

LeoMe published on 2023-5-11 09:20 The characteristics of the entire combination of OD gate + pull-up resistor are determined by itself. It is a multi-input AND gate. When all outputs are 1, the OD gate outputs high impedance, and the bus is pulled up to 1, ...

Isn't there only voltage when pulling up? Why is there current?

maychang

Xiaoshangyangyy posted on 2023-5-11 22:05 I am a little confused about what this weak pull-up current means. Where does the current flow to?

[I am a little confused about what this weak pull-up current means and where does the current flow to?]

When the MOS tube of the OD gate is turned off, the positive end of the chip power supply is connected through the pull-up resistor, then through the pin (whether the pin is an output end or an input end depends on what you want the pin to be used for), and then through the load (the pin as the output end) or the signal source (the pin as the input end) to the ground (the negative end of the power supply).

When the MOS tube of the OD gate is turned on, of course the positive end of the chip power supply is connected to the ground (negative end of the power supply) through the pull-up resistor and then through the MOS tube.

小太阳yy

maychang posted on 2023-5-12 11:54 小太阳yy posted on 2023-5-11 22:05 I am a little confused about what this weak pull-up current means, where does the current flow to [a little...

There is another sentence in the chip manual, saying that you can use an external strong pull-up to cover the weak pull-down inside the chip. What does this mean, teacher?

小太阳yy

maychang posted on 2023-5-12 11:54 [I am a little confused about what this weak pull-up current means, where does the current flow to?] When the MOS tube of the OD gate is turned off, the chip power supply is positive...

What is the signal source you are talking about? Can you draw a simple picture?

小太阳yy

maychang posted on 2023-5-12 11:54 [I am a little confused about what this weak pull-up current means, where does the current flow to?] When the MOS tube of the OD gate is turned off, the chip power supply is positive...

I really don't understand how the weak pull-up current in this chip flows. I understand that the pull-down is the current that flows to the ground when the MOSFET is turned on, but how does the pull-up current flow? It is directly connected to the power supply, right? If it is directly connected to the power supply, there is no loop, only one voltage, so how can there be current?

maychang

小太阳yy posted on 2023-5-15 19:16 There is another sentence in the chip manual, saying that an external strong pull-up can be used to cover the weak pull-down inside the chip. What does this mean, teacher?

[There is another sentence in the chip manual, saying that you can use an external strong pull-up to cover the weak pull-down inside the chip. What does this mean?]

I don't know what this means. Maybe the author of the chip manual made a mistake. I guess you are reading the Chinese manual.

There is a way to add a stronger pull-up externally. You only need to add a resistor outside the chip, with one end connected to the positive power supply and the other end connected to the pin. This is actually to connect another resistor in parallel at both ends of the pull-up resistor inside the chip, reducing the value of the internal pull-up resistor, but I don't know what [covers the weak pull-down inside the chip] is.

maychang

Xiaoyangyy posted on 2023-5-15 20:49 What is the signal source you are talking about? Can you draw a simple picture for me? = =

What signal source do you mean?

The chip pin is used as an input terminal, and of course a signal source is needed to provide a signal (high level or low level) to the pin. This is what I call the "signal source". It may be the output terminal of other chips.

maychang

小太阳yy posted on 2023-5-15 22:14 I really don’t understand how the weak pull-up current in this chip flows. As for the pull-down, I understand that it flows to the ground when the MOS tube is turned on...

The pull-up provider inside the chip can be a resistor, but in fact it is often a more complex circuit, usually a constant current circuit composed of several tubes. This is because in the chip manufacturing process, the resistor takes up a larger area (possibly several times larger) than the tube, so it is cost-effective to use several tubes to form a constant current circuit to replace the resistor in terms of occupied area. The so-called [FAULT internal pullup current] is the output current of this constant current circuit when the pin is short-circuited to the ground. If the pin is used as the input end of the chip and is suspended, the constant current circuit will of course have no output current. But when the pin is used as the input end and an external signal is input to the pin, when the external input is low, the constant current circuit will output current (the size is the value marked in the manual) through the pin and the signal source to the ground.

小太阳yy

maychang published on 2023-5-16 08:03 The pull-up provider inside the chip can be a resistor, but in fact it is often a more complex circuit, usually a constant current circuit composed of several tubes...

Teacher, let me talk about the fault pin of this chip. This pin can be used as both input and output. When this pin is used as output, when the chip detects a fault, it will pull down the pin by 750 microamperes. When the device fault is restored, it will pull up the pin by 8 microamperes. Then he said that the actual output voltage is determined by the external circuit. I don’t quite understand the last sentence. How is the actual output voltage determined by the external circuit? Indeed, in the manual below, when used as an output, external pull-ups and pull-downs are added to obtain low-level and high-level output voltages of 0.7 and 2v. When the output is low, an external pull-up of 500 microamperes is used, and when the output is high, an external pull-down of 1 microampere is used. Why do we need to add this test condition? How is the external circuit configured for this condition? I don’t understand

maychang

Xiao Taiyangyy posted on 2023-5-16 08:49 Teacher, let me talk about the fault pin of this chip. This pin can be used as both input and output. When this pin is used as output, to the chip...

As an output, the OD gate usually does not need an external pull-up or an external pull-down. As long as the load input resistance is large enough, it can output a high level or a low level. Unless the load input resistance is not large enough, then it is necessary to add an external pull-up, as shown in the second picture on the 14th floor, add a 10 kilo-ohm resistor between the pin and the positive power supply. The pull-up current provided by this resistor when the pin is short-circuited to the ground is indeed 5/10=0.5 (mA). As for [when outputting a high level, an external 1 microampere external pull-down is used], it is not shown in the picture, and I don't know.

小太阳yy

maychang published on 2023-5-16 09:01 OD gate as output, usually do not need to add external pull-up or external pull-down, as long as the load input resistance is large enough, it can output high level or low level...

The internal circuit is probably like this. I don't understand why the pull-down current is larger than the pull-up current. The pull-up current is 5 microamperes. If the MOS tube is turned on, the current should be smaller than 5 microamperes. Why is it 500 microamperes?

maychang

小太阳yy posted on 2023-5-16 13:13 The internal circuit is probably like this. I don’t understand why the pull-down current is larger than the pull-up current. The pull-up current is 5 microamperes. If the MOS tube is turned on...

[The pull-up current is 5 microamperes. If the MOSFET is turned on, the current should be smaller than 5 microamperes. Why is it 500 microamperes?]

When the MOS tube is turned on, it can be regarded as a wire connecting the drain and source of the MOS tube together.

When the MOS tube is not turned on, it can be regarded as an open circuit. At this time, the 0.5mA current source has no effect (no current), and the pin can output 5uA current (flow out), so the pull-up current is 5uA.

After the MOS tube is turned on, the pin can absorb 0.5mA current (flow in), and the pull-down current is 0.5mA.