BRTs are generally used as switches. Ideally, their on-state voltage (collector-emitter voltage, VCE) should be as close to zero as possible.
The base current should be considered to achieve this condition.
To minimize VCE, BRTs are used in the saturation region shown in Figure 1. Typically, bipolar transistors are considered to provide an hFE of 100. However, since BRTs operate in the saturation region, it is necessary to assume that they provide an hFE 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 VCE of 0.2 V and an IC of 10 mA. Here, the RN1402 (with R1=R2=10 kΩ) is used as a BRT. Look at the VCE(sat) – IC curves shown in the RN1402 datasheet (Figure 3). When hFE＝IC / IB＝20, VCE(sat) can be read as 0.05 V to 0.06 V at IC=10 mA. So, let’s assume these conditions.
The internal base current (Ib) of the basic BRT circuit shown in Figure 2 is:
Ib＝ IC / hFE = 10 mA / 20 = 0.5 mA
For the sake of simplicity, assume that the internal base voltage (Vbe) is equal to the commonly used value of 0.7 V. Then, IR2 flowing through R2 is calculated as follows:
IR2 = 0.7 V / 10 kΩ = 0.07 mA
Therefore, the center value of the BRT’s base current (IB) is:
IB = Ib + IR2 = 0.57 mA
The input voltage (VI) that provides an IB of 0.57 mA is calculated as follows:
VI = R1 * IB + Vbe = 10 kΩ * 0.57 mA + 0.7 = 6.4 V
It can therefore be considered that a VCE of 0.2 V and an IC 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 (VI) higher than 6.4 V should be applied to the B terminal. The following paragraphs describe the operation of the BRT when VI > 6.4 V.