Can you guys answer my question about 74HC04?

hyj966

Can you guys answer my question about 74HC04? [Copy link]

1. The power supply is 5V constant voltage, the maximum output current is 0.35A, and the resistance of the DC motor is 62.6 ohms. In the circuit below, if pin 1A inputs a low level, 1Y will output a high level, but the current should not be enough to drive the DC motor. However, the simulation shows that the 74HC04 current is overloaded and burned. Why is this?

2. If 1Y outputs a low level like this, the current overload problem can be avoided. However, the current of the DC motor directly connected to the power supply is higher than that of the motor connected in this way. Is it because of the internal resistance of the IC or other reasons?

I hope you guys can answer my questions.

maychang

"But the simulation showed that the 74HC04 overloaded and burned out."

That's right. 74HC04 is not allowed to output such a large current and it will burn out easily.

Another reason for damage is that the strong inductive load such as the motor is not voltage clamped, causing the induced electromotive force of the inductor to break down the 74HC04.

maychang

2. If 1Y outputs a low level like this, the current overload problem can be avoided. However, the current of the DC motor directly connected to the power supply is higher than that of the motor connected in this way. Is it because of the internal resistance of the IC or other reasons?

I don't understand what your second question is about.

hyj966

maychang posted on 2022-2-13 12:45 2. If 1Y outputs low level like this, the problem of current overload can be avoided, but it is also powered by a power supply, and the power supply is directly connected to the current of the DC motor...

That is to say, if I connect the power supply directly to the motor, the current can reach about 79mA, which means it can be fully loaded.
However, if it is connected like this now, the current of the motor is only 49mA. Is it because of the internal resistance of the IC or other reasons that it cannot be fully loaded?

maychang

hyj966 posted on 2022-2-13 12:57 That is to say, if my power supply is directly connected to the motor, the current can reach about 79mA, which means it can be fully loaded. But if it is connected like this, the current of the motor is only...

Chips like the 74HC04 have different output current sourcing and sinking capabilities, although the difference is not as great as with chips like the 74LS04.

maychang

hyj966 posted on 2022-2-13 12:57 That is to say, if my power supply is directly connected to the motor, the current can reach about 79mA, which means it can be fully loaded. But if it is connected like this, the current of the motor is only...

When the load is connected to the positive terminal of the power supply, it is still possible to damage the chip because the load is inductive but no voltage clamping circuit is added.

hyj966

maychang posted on 2022-2-13 13:21 If the load is connected to the positive terminal of the power supply, it is still possible to damage the chip because the load is inductive but no voltage clamping circuit is added.

Got it, thank you very much for your answer

maychang

hyj966 posted on 2022-2-13 14:22 I understand. Thank you very much for your answer

I don't know why such a "fragile" chip as 74HC04 was chosen to drive the small motor.

It is recommended to use ULN2001_2004. If the signal of the driver chip can output 1-2 mA or more, you can even consider using discrete transistors.

wangerxian

It will be less problematic if you choose a dedicated drive for the motor.

Fred_1977

There is something wrong with the design of this circuit. Such a large current must be driven by an external transistor or MOS tube.