# Bjt base current calculation

In a bjt transistor that is used as a switch (maximum saturation, Vce minimum as possible), what is the minimum base current that is required for a corresponding desirable collector current?

• That would depend upon the current gain of a particular BJT. It would vary even within the same type of BJT e.g. 2N2222. Commented Jun 16, 2017 at 7:39

The transistor going into saturation isn't a property of the transistor itself, but instead a property of the circuit surrounding the transistor and the transistor, as part of it.

A question about Vce of an NPN BJT in saturation region

For this circuit with ideal transistor (current controlled current source CCCS) any base current large than:

$\Large I_B > \frac{\frac{V_{cc}}{R_c}}{\beta }$ will saturate the BJT.

But in real life, ideal transistor don't exist. For any real world transistor, the β is not constant. β varies with Ic, Vce, temperature. And what is worse, every single transistor will have different beta value and beta will changes for different operating conditions also. Also in saturation $I_C = I_B * β$ do not hold anymore.

So to overcome this problem with beta and saturation we are forced to use "overdrive factor" or "Forced Beta" trick.

We simply increase the base current well beyond $I_B > \frac{\frac{V_{cc}}{R_C}}{\beta }$.

We do this to make sure that we have enough base current to put the transistor well into saturation for every condition we have in our circuit.

Additional most BJT's vendors define saturation region when Ic/Ib = 10 (called Forced Beta). And the most data-sheet show Vce_sat for Ic/Ib = 10

So, to be one hundred percent sure that your BJT will be in saturation region you must use this so-called forced beta technique when choosing base resistor value.

$$\frac{I_C}{I_B} = 10$$

$$R_B = \frac{V_{IN} - V_{BE}}{0.1*I_C}$$

$$R_C = \frac{V_{CC} - V_{CE_{sat}}}{I_C}$$

Or we can use KVL and solve for $R_B$

$$I_B=\frac{V_{IN} - V_{BE}}{R_B}$$

$$V_{CE} = V_{CC} - I_C*R_C = V_{CC} - \beta*I_B*R_C = V_{CC} - \beta \frac{V_{IN} - V_{BE}}{R_B} * R_C$$ Solving for $R_B$

$$R_B\leqslant \frac{V_{IN_{min}} - V_{BE}}{V_{CC} - V_{CE_{sat}}}*\frac{\beta_{min}}{K}*R_C$$

And K = 3...10 - overdrive factor

• I wanted to drive a BJT as a low side switch. But the base of the transistor will be driven by an mcu. Then if I want a Ic close to 1A, if the Hfe is low enought I don't want to force the pin of mcu to source so much (10mA or much more for example). So, my concern was if the transistor will saturate when I limit the base current too much. Commented Jun 16, 2017 at 17:44
• By saturation I meant that I want to force the transistor to maintain a Vce below to 1V or as low as possible. But I do want to get a Ic=100mA. The problem is I don't know the actual load's value. It would be a led or a DC lamp or a relay (None of my loads would need current much more than 100mA, that's why I put that limitation). And my Vcc is also not a constant but it varies vrom few volts up to 30V. I could choose a current source instead of simple BJT but I need something very cheap and simple with few BOM. Commented Jun 16, 2017 at 18:17
• @MrBit In this case you can use MOSFET for example IRLZ44n
– G36
Commented Jun 17, 2017 at 13:33
• Could you give an example of a BJT with VCEsat = 9V ? I am curious about these freaks. Commented Jun 9, 2021 at 10:07
• @kellogs But the example does not show the transistor in the saturation region
– G36
Commented Jun 9, 2021 at 18:52

There is one formula you need to know to calculate the base curent : $H_{fe}=\frac{I_c}{I_b}$. You know from $I_c$ from you circuit and $H_{fe}$ from the datasheet. That lead to $I_b = \frac{I_c}{H_{fe}}$. I suggest you to take $I_b$ higher ($I_b >> \frac{I_c}{H_{fe}}$).

To fix that base current, you have to remember that $V_{BE}$ will affect the base voltage.

simulate this circuit – Schematic created using CircuitLab

In this exemple, $I_{R1} = \frac{V_{cmd} - V_{BE}}{R_1}$

• Sometimes Hfe is mentioned for a different Vce voltage than the voltage that is required. What happens then? Commented Jun 16, 2017 at 8:00
• @MrBit For different Vce? Are you sure it's not for different part? Hfe may change for different Ice most of the time. Commented Jun 16, 2017 at 8:03
• @MrBit Hfe may vary by a scale of 2 or 3. To be sure, set Ib 5 time greater than the origine calculation Commented Jun 16, 2017 at 8:05
• I suggest you to take Ib a little higher If you want a small Vce I would suggest making Ib a lot larger than Ic/Hfe, for example 10 times larger. Commented Jun 16, 2017 at 8:05
• Hfe may change for different Ice most of the time Note that is only true in saturation mode. But Hfe is usually specified when the BJT is not in saturation, then Hfe is constant over a very large range of Ic. Commented Jun 16, 2017 at 8:07

For saturation to occur over all temperature ranges take the hFE(min) and use the equation Ib_minSat = Ic / hFE(min) to find the minimum base current for saturation to occur. You select the desired Ic. Note that you want to make sure your Beta_Forced (i.e. Ic / Ib) is less than the hFE(min) in the datasheet.

If you're confused, see the example in Sedra Smith, Microelectronic Circuits 6th Edition, page 383:

Since βforced is less than the minimum specified value of β, the transistor is indeed saturated. We should emphasize here that in testing for saturation the minimum value of β should be used. By the same token, if we are designing a circuit in which a transistor is to be saturated, the design should be based on the minimum specified β. Obviously, if a transistor with this minimum β is saturated, then transistors with higher values of β will also be saturated.

Sedra Smith, Microelectronic Circuits 6th Edition, page 383

(Image source: ON Semiconductor P2N2222A datasheet)