Basic idea of how to calculate the base current and input voltage of a bias resistor built-in transistor (BRT)　

BRTs are generally used as switches. Ideally, their on-state voltage (collector-emitter voltage, V_CE) should be as close to zero as possible.
The base current should be considered to achieve this condition.
To minimize V_CE, BRTs are used in the saturation region shown in Figure 1. Typically, bipolar transistors are considered to provide an h_FE of 100. However, since BRTs operate in the saturation region, it is necessary to assume that they provide an h_FE of 10 to 20. This means that the base current should be 1/10th to 1/20th of the collector current.

Suppose that we need a V_CE of 0.2 V and an I_C of 10 mA. Here, the RN1402 (with R₁=R₂=10 kΩ) is used as a BRT. Look at the V_CE(sat) – I_C curves shown in the RN1402 datasheet (Figure 3). When h_FE＝I_C / I_B＝20, V_CE(sat) can be read as 0.05 V to 0.06 V at I_C=10 mA. So, let’s assume these conditions.

The internal base current (I_b) of the basic BRT circuit shown in Figure 2 is:
    I_b＝ I_C / h_FE = 10 mA / 20 = 0.5 mA
For the sake of simplicity, assume that the internal base voltage (V_be) is equal to the commonly used value of 0.7 V. Then, I_R2 flowing through R₂ is calculated as follows:
    I_R2 = 0.7 V / 10 kΩ = 0.07 mA
Therefore, the center value of the BRT’s base current (I_B) is:
    I_B = I_b + I_R2 = 0.57 mA
The input voltage (V_I) that provides an I_B of 0.57 mA is calculated as follows:
    V_I = R₁ * I_B + V_be = 10 kΩ * 0.57 mA + 0.7 = 6.4 V

It can therefore be considered that a V_CE of 0.2 V and an I_C of 10 mA can be obtained by applying an input voltage of 6.4 V or higher, although it is necessary to take device variations and temperature characteristics into consideration.

In practice, an input voltage (V_I) higher than 6.4 V should be applied to the B terminal. The following paragraphs describe the operation of the BRT when V_I > 6.4 V.

It is OK to apply a voltage higher than 6.4 V to the B terminal. Increasing the input voltage only brings the BRT into deeper saturation (i.e., decreases V_CE(sat)). This causes the base-emitter voltage of the internal transistor to increase slightly, but not significantly. (For your reference, Figure 4 shows the V_BE(sat)-I_C curve (h_FE=10) of the 2SC2712 that is equivalent to the internal transistor of the RN1402.) The base-emitter voltage varies only by several hundreds of millivolts in the collector current (I_C) range in which BRTs are commonly used.
Since the base voltage hardly changes, an increase in the input voltage causes the voltage across R₁ to increase, leading to an increase in the base current (i_b). Saturation collector current, which is determined by load resistance and other external factors, hardly changes. Since only the base current increases, the h_FE of the internal transistor decreases, causing the saturation level to increase (i.e., V_CE(sat) to decrease).
Note that an increase in the saturation level might cause the time required to turn off the transistor to increase.
The maximum value is specified for the input voltage (V_I). Also see the following FAQ entry.

• FAQ :What is the maximum voltage that can be applied to the base of a bias resistor built-in transistor (BRT)?