74HC573 Is there a current limiting resistor for driving digital tubes? How big is the sink current

It depends on what the power supply voltage of your digital tube is, the commonly used small 7-segment digital tube is 2 LEDs in series in series, 20mA, so each section of the voltage is about, the driving current of 74HC573 can reach 35mA, the saturation voltage drop is, the saturation voltage drop and the LED working voltage add up, when the power supply voltage is 5V, the voltage will be higher, and when the current limiting resistance is not added, the current will increase to more than 35mA, so that the saturation voltage drop and LED voltage are increased, until the addition is equal to 5V. The temperature of the device increases. At this time, the sink current can not be determined, it should be beyond the normal range, and the current is related to various factors such as the size of the LED, the ambient temperature and heat dissipation, the manufacturer and batch number of the device.

Over-range use is also not good for the reliability and longevity of LEDs and 74HC573. It is not recommended to use this except in special emergency situations.

If a voltage of 5V is used, assuming that the LED voltage is 2V and the current is 20mA, then R=U I=(5-2).02 = 150 ohms.

It is recommended to use close to 300 ohms first, and in actual operation, if it feels too bright, the resistance can be increased, and too dark can be reduced.

The above is a normal red LED, but it is different if it is a white LED (with a voltage drop of about 3V) or a high-current LED.

It depends on how many pieces of 74HC573 you use, if you use two pieces, one to latch the segment code, one to latch the bit code, you can drive the 8-bit digital tube, as for the common yang, the common yin digital tube is OK. If there are more digital tubes, you will need to use a few more pieces.

That's not the case, in the absence of a current-limiting resistor, your LED itself has a resistance value current I=U R U is 5V, R = LED internal resistance, in addition, it is best to install the current-limiting resistor, even if there is no in the specification of the chip, it is common sense to add resistance to the LED circuit.

And a lot of chips take out the protection or something, because it's described in an ideal state.

Hehe, IC=200mA is calculated by IB, and the actual IC is also related to the collector load.

According to the digital tube, 80mA can be calculated, which is actually 80mA.

When this number is less than 200mA, the triode 9012 is in a saturated state, and the CE junction voltage will be less than, which is an ideal "on/off" state.

When this number is less than 200mA, the triode 9012 is in a saturated state, and the calculation formula of IB is invalid, and the formula is only suitable for the "amplification" state.

This is the knowledge of "modular electricity".

The transistor here should be used as a switch, not for amplification.

At low currents, the 9012 magnification becomes linear.

The source current and sink current of each output pin of 74HC573 are the same, both can reach 35mA, and it is still possible to drive the digital tube. However, to use it to control an 8-bit digital tube, you need to raise the level at the 11-pin CLK end each time, then input data, and then pull the CLK down to latch. It is best to use a common cathode digital tube to control the 8-bit digital tube, with two pieces of 573 latch, one piece of 573 latch output segment code, the output pin is the current source current, each section should be connected in series with a 510 beat current limiting resistor, and the other piece latches the output bit code, and the output pin is the current inflow.

You have a problem with your design and don't have any current-limiting resistors.

Since 138 is a decoder, there is only one output at a time, so the actual current is only one LED lit, and if the current of the 8th segment is lit at this time, it is very large, but due to the limitation of the internal driving ability of 138 and other devices, the actual output voltage rises, resulting in a significant decrease in the actual current. However, if you do this, you are actually accelerating the aging of the device, and it is difficult to ensure the reliability of the circuit.

Usually the IO port chip of the single-chip microcomputer is not to increase his output current, but to have a fixed purpose, such as 138 decoding, 573 latching. Some chips integrate the function of increasing the output current (probably because the chip is mostly used to drive the circuit), but its main role is not to increase the driving ability.

The current flowing into the IO port is too large, of course, it may cause damage to the MCU, and the bearing capacity of different IO ports may be different for different MCUs. But why do you want to apply a high current to the IO of the microcontroller, or why not add a current limiting resistor.