BJT High Side Driver Analysis

High side driver refers to an electronic switch that is configured to switch a supply voltage. It is commonly used in applications wherein the power source of a circuit is electronically controlled on and off by other circuits like a microcontroller. High side switch could be implemented by using BJT or MOSFET. IGBT could be an option as well but not as popular as BJT and MOSFET.

The circuit below is made up of BJTs. The high side driver is Q2 while Q1 serves as a pre-driver so that Q2 can be driven by a small signal coming from a microcontroller.

How the Circuit Works

When the microcontroller gives 3.3V, Q1 will conduct. This will effectively connect R2 to ground. With the proper sizing of R2 and R3, Q2 will saturate thus switching 12V to the collector of Q2.

When the microcontroller output is 0 volt, Q1 will turn off. This makes the other side of R2 to be in series to R1. Effectively, there will be no current flows to the base since R1 is also connected to the same voltage source in the emitter of Q2. Thus, Q2 will turn off.

Why not connect microcontroller pin to R2 directly?

The role of Q1 is to connect R2 to ground so that Q2 will function. Therefore, it is possible for a microcontroller pin to be connected to R2 directly. This time, the logic is reversed such as a low signal from a microcontroller will make the Q2 circuit work while a high signal (3.3V) will turn off Q2.

But examine carefully. When the microcontroller output is 3.3V, wherein it supposed to turn off Q2, there is still current flow from V1 down to R2 because the voltage at the end of R2 is only 3.3V. This is very much lower than the 12V level of V1. In short, the microcontroller cannot be able to turn off Q2. Thus, Q1 is important.

What is the role of R3 and R4?

Basically, both can be omitted. The only time they are critically needed is when the input to the base can float. When it floats, R3 will tie the base to the emitter thus still ensuring Q2 will turn off. In the same manner, R4 will pull the base of Q1 low to ensure turn off.

How to Ensure Q2 will go Hard Saturation

Turning on Q2 is not enough. It must go into hard saturation. Entering hard saturation means the voltage drop of Q2 is negligible. If Q2 cannot enter hard saturation, the voltage drop is significant. This is not what we wanted for a high side driver.

Check the Applied VBE

The applied VBE that we are talking about here is the voltage across the base and emitter junction considering the BJT is off. In a BJT circuit, when the BJT is on, the voltage across the base-emitter junction is fixed (diode voltage drop) to either 0.7V or 0.3V for Silicon and Germanium materials respectively. However, this will not happen if the applied VBE does not surpass the diode drop level. To check, let us do below.

VBE BJT off = Voltage base BJT off – Voltage emitter BJT off

Voltage base BJT off = V1 X R2 / (R2+R3) = 12V X 1K / (1K + 10K) = 1.09V

Above analysis considers that Q1 is at hard saturation as well.

Voltage emitter BJT off = 12V

VBE BJT off = Voltage base BJT off – Voltage emitter BJT off = 1.09V – 12V = -10.91V

The applied VBE when BJT is off is -10.91V. This is more than the -0.3V or 0.7V diode drop. The negative sign of the voltage denotes a PNP BJT transistor. Specifically, for BC807 BJT, the VBE requirement is maximum of -1.2V.

Since the applied VBE when the transistor is off is much higher than the requirement, then the transistor will turn on. Thus, the voltage across the base and emitter junction is clamped to a diode level.  

Set the Bias to Attain Hard Saturation

Hard saturation means the device is no longer amplifying. In practice, a maximum calculated Gain of very much less than the minimum gain specified in the device datasheet considers hard saturation.

Gain is collector current (Ic) divided by base current (Ib).

Gain (beta or HFE), maximum = Ic maximum / Ib minimum

In solving Ic maximum, the voltage drop across the collector-emitter is can be neglected, making the calculation straight forward as

Ic maximum = VEE / Rc

Where VEE is the supply connected to emitter (which is V1 in the above circuit) while Rc is the resistor in the collector.

In the circuit above, the collector current is given already as 200mA. This is the collector current to be used in the gain equation directly.

The base current can be computed as

Ib = IR2 – IR3

IR3 = VBE / R3

IR2 = (V1 – VBE) / R2

Ib = (V1 – VBE) / R2 – (VBE / R3)

Ib = (12V – 1.2V) / 1K – (1.2V/10K) = 0.01068A

So, the computed maximum gain is

Gain (beta or HFE), maximum = 0.200A / 0.01068A = 18.73

According to BC807-40, the minimum device gain is 250.

The computed maximum gain, which is 18.73, is very much less than the device minimum gain. Therefore, the BJT will saturate.

For many design engineers, a maximum computed gain equals to 10 or below is what considered hard saturation (rule of thumb). If we are going to adopt this value to the above circuit, we need to increase the value of R2.

Consider halving the value of R2,

 Ib = (V1 – VBE) / R2 – (VBE / R3)

Ib = (12 – 1.2V) / 500 – (1.2V/10K) = 0.02148A

The new maximum computed gain is

Gain (beta or HFE), maximum = 0.200A / 0.02148 = 9.31

The new maximum computed gain is less than 10. There is no doubt the BJT will enter hard saturation.

The principle applied to Q2 on how to ensure it will turn on and go hard saturation will be applied the same to Q1.