Basic driver capability Problems

Basic driver capability Problems

I have never understood what the driver is. in general, if you want to drive a load and your voltage reaches a threshold, you can drive the load. Why do you sometimes say that the driving capacity is insufficient? Why can't an op without an output buffer drive small resistors and large capacitor loads? What is the impact of loading a large capacitor or a small resistor directly behind this op? When I use an op-amp to drive a large MOS tube, why do I need to connect several revertors? Can I directly connect the op-amp output to this large MOS tube?

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I have raised this question. I try to answer it with my own understanding.
First, the so-called driving capability refers to the output current capability. For example, the maximum output current of a single-chip microcomputer General I/O port is 20mA in High-Power periods. This 20mA indicator represents the driving capability of the I/O port.
Second, if the load is too large (as the landlord said in the small resistance), the load current may exceed its maximum output current, then we say that the driving capacity is insufficient.
Once again, when the drive capacity is insufficient, the direct consequence is that the output voltage is reduced. For a logical circuit, it is unable to maintain its high level, resulting in logical disorder and the expected effect cannot be achieved. This phenomenon is generally not allowed.
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Because MOS have internal resistance, the maximum current that each MOS can provide is limited.
Small MOS provide small current, large MOS can provide large current.
The landlord said, "I think you can drive this load as long as your voltage reaches a threshold ",
The key is that when the load is too large (that is, the resistance is too small and the capacitance is too large) and exceeds the driving capacity of the output tube, the output voltage will be pulled down,
The change range of Gnd-VDD is not reached.
Limit: R is infinitely small, and C is infinitely large (Gnd). Then the output will not change!

Driving Capability of P0 port of 51 Single Chip Microcomputer

P0 is an open-circuit interface with drain, so it is no problem to drive the led when the current is filled. That is: VCC> throttling resistance> led> P0.

Fill current: When the IO port is low, the current from the outside of the IO port "fill" into the microcontroller opposite is "pull current" when the IO port is high, the current flows out of the microcontroller to power the load.

The P0 port can work in two ways: Bus and Io. Simply put, P0 is used as an address data multiplexing line (commands such as movx movc), and P0 port is a push-pull output, that is, PMOS and NMOS are used to drive the output, therefore, at this time, no matter whether the input is 1 (high level) or 0 (low level), the driver is strong (the internal resistance is small, and the output current is large ), no one of you has ever seen The P0 port in the ram hardware accessing the OSS 51 connected to the pulling resistance!
The second method is the I/O method, which is used as a normal I/O. In this case, it is different from the other three, it has no internal pull-up resistance (that is, the other three ports have internal pull-up), which is an open-drain output (OD), so its output is low (0) and high-impedance state (z), you only need to connect a light emitting tube from the pin and the throttling resistor to the power supply. At this time, the pin is equivalent to a switch, closed to the ground or disconnected, this should be a simple connection method with low power consumption.
Of course, you can also use a high level to drive, that is, pull the resistor on the outside, and the upper part of the connection method has already said that this method is not good, and the power consumption is high, you can find out the reason for this. Especially when the load is heavy and several luminous tubes are driven at the same time, the pull-up resistance must be reduced to provide sufficient driving current. However, A small pull-up resistor causes a high current when the load does not work, or even a constant load (the luminous tube cannot be extinguished ).
Other ports are connected to the pull-up resistor, which is required by a pseudo bidirectional port.

About 51 Single Chip Microcomputer I/O port drive capability

Why when the P0 port is high, the output current is 400ua, while the output current is 3.2maat (0.45 V), instead of the high level as we think, the output current is large. p1, P2, and P3 ports are also. why is the output current low ?????
In addition: 51 Single-Chip Microcomputer output high power and low power is usually a few volts?

Answer:

51 Single-Chip Microcomputer I/O port is the output mode of the collector, the high output current is equal to the current of the upper resistance, this current is relatively small, the low output is the current absorbed by the internal transistor, the maximum value is 10mA, but the total current of the whole port cannot exceed 24mA.
A level lower than 0.7v is a low level, and a level higher than 1.8v is a high level. There is no doubt between them.

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Conservation of energy
The conservation of uit should not be done if both of them are big or both are small. Otherwise, where is the energy? Where did it come from? Ha
5 V and 0 V

Driving Capability of AT89C51

I have been reading a book for more than a month, but I have never been an experiment. I have a problem with this kind of mobile phone. Please tell me the following question. It is best to detail it.

What is the driving capability of P0, P1, P2, and P3 single-chip microcomputer! What is the driving current? (AT89C51)

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The output drive current is small.

He mainly looks at the size of the pull-up resistance.

For example, if the upstream resistance of the P0 port is 10 K, the output current is only UA.

However, it absorbs a large amount of current, so if you want to connect to a digital tube or led and do not want to buffer it, you can use it to absorb the current, such as when you connect to the led, the LED is connected to the positive power supply, negative end string traffic limiting resistor connected to P0 Port

In my opinion, I generally use a buffer. In these few ports, P1, P2, and P3 are half of P0's driving capabilities. It seems that you can read the books on your own.

About LED Driver/answer: AT89C51 I/O Load Capacity

Many types of drivers can be used to drive the public pole and data end of the LED. For example, 7407 or something. That is not to say that the common TTL circuit or CMOS circuit can have this driving function, as long as it can develop a certain voltage or current?

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Line! However, use the low-level lighting mode whenever possible.

However, it is best not to directly use the I0 port to drive the LED. In practice, the driver chip is generally used. This is a mature practice.

Use a digital tube directly driven by a transistor. Adding IC in the middle not only increases the cost but also increases the volume. I think it is better to make it as simple and cheap as possible.

 

The I/O port drive current of the C51 series chip is not exactly the same. Please check the technical information.

1. under steady state (non-transient) conditions, ECHO must <br>
Be externally limited as follows: <br>
Maximum intraocular lens per Port Pin: 10 Ma <br>
Maximum intraocular lens per 8-bit port: <br>
Port 0: 26 Ma <br>
Ports 1, 2, 3: 15 MA <br>
Maximum total ior for all output pins: 71 MA <br>
If echo exceeds the test condition, vol may exceed the <br>
Related specification. pins are not guaranteed to sink <br>
Current greater than the listed test conditions. <br>
2. Minimum VCC for power down is 2 v.

The following translations may be inaccurate. They are for reference only.

1. In a stable (non-transient-long term) state, the output load (driving) Current (IOLs) of the I/O port must be smaller than the following values:
The maximum size of IOLs for each port terminal is less than 10mA;
The maximum size of 8-bit ports: P0 is smaller than: 26mA; P1/P2/P3 must be smaller than: 15mA;
For the maximum intraocular lens of the entire output terminal (PINs): the total current is less than 71mA.
If the output load current i0l exceeds the test conditions, the output load voltage (VOL) may exceed the above specifications. The suction current of the Output Terminal cannot be greater than the above value.
(My understanding: If the output current exceeds the threshold, the output terminal voltage may exceed the threshold, which cannot be ensured to be able to inhale the specified current value .)

2. The minimum drop value of the power supply voltage VCC is 2 v.