Regarding this question, VBsemi's editor did some research online and found that most answers focused on ensuring the BJT's "reliable cutoff," "current equalization," and "current limiting."

Today, VBsemi delves deeper into this question based on the aforementioned answers. Here's a popular Q&A found online:

Question: In the circuit shown, with a high input voltage (set to 5V), the voltage at the voltage divider point Vb is 3.4V. However, when the transistor conducts, the voltage Vbe is around 0.6V. Given these two different voltages, what is the voltage at the base?

Answer: With a 5V input, the statement "the voltage at the voltage divider point Vb is 3.4V" assumes a voltage division between two resistors without the presence of a transistor. Once the transistor is connected, the Vbe junction will conduct, and the voltage at Vb will be clamped at around 0.65V, no longer behaving as a simple voltage divider. It should be understood as "a 4.7K resistor in series with a forward-biased diode connected to ground," with the 10K resistor simply paralleled with the diode.

As for the role of the 10K resistor, it ensures reliable cutoff of the transistor when there's no input voltage (or when the input terminal is floating).

VBsemi's Editor:

In switch circuits composed of transistors, the resistor paralleled with the emitter typically has a value ranging from 1K to 10K. This value's significance is closely related to the signal level and frequency of the preceding stage circuit.

What purpose does this resistor serve?

In the circuit diagram, adding a 10K resistor ensures that when the input terminal is floating, the base of the transistor is effectively grounded. Additionally, when the transistor accelerates out of saturation, the rising edge of the output pulse will closely match the rising edge of the input waveform.

Are there any other explanations?

Let's take an NPN transistor as an example. When the signal circuit is at 0, the collector potential equals VCC. If there's a leakage current flowing from the collector to the base, without a paralleled resistor, this leakage current might cause the transistor to exit the cutoff region and enter an erroneous conducting state.

The resistor acts like a drainpipe connected to the emitter, diverting the leakage current to ground, thereby enhancing cutoff reliability.

In conclusion, with advancements in technology, the leakage current from the collector of the transistor has become quite small. Some may opt to eliminate or decrease the value of this resistor, but this approach has both advantages and disadvantages, and it's not prudent to make excessive judgments.

In certain DC and low-frequency applications, the PN junction of the transistor may exhibit resistive characteristics (on the emitter junction or the input side of an optocoupler). However, in high-frequency applications, due to the relationship with the junction capacitance, if the discharge isn't thorough enough, the charge stored in the junction capacitance won't discharge properly, leading to sudden breakdown of the optocoupler or transistor, interrupting signal transmission.

In such cases, adding a resistor helps accelerate the discharge of the charge stored in the junction capacitance.