# How do I really calculate these resistors?

I would like to focus on this circuit I am trying to calculate R10 and R11.

I want Ic = 10 mA. Transistor is 2N2222A.

This is an amplifier. If I understood how amplifiers work, I want Vc to be in the middle of the rail, so the amplifier will have maximum output swing. The circuit is powered by a 9V battery, so the middle of the rail = Vc = 4.5V.

So, I have calculated Rc like:

$R_C = \frac{V_{CC} - V_{CE}}{I_C}$

$R_C = \frac{9 - 4,5}{10 \times 10^{-3}}$

$R_C = 450$ ohms.

I have calculated Ib like this:

$I_C = \beta I_B$

$\beta = 225$ for 10 mA

so,

$I_{B} = 44,44 \thinspace \mu A$

Hence,

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

$R_B = \frac{9 - 0,7}{44,4444 \times 10^{-6}}$

$R_B = 186,750$ ohms.

The problem is: when I put that on the simulator, it gives me a Vc equal to 4.24V, almost in the middle of the rail but not exactly and the worst part is that the simulator gives me a Vbe = 0.562V.

As far as I know, this silicon transistor will need 0.7V to work and the value from the simulator is not showing the whole thing will work as expected.

How are these calculations really done, taking into consideration differences from theory to the real world?

• This is normal because the Vbe voltage is not constant, it will change with the Ib current. Also, the beta value will also change with Ic current. Also, every single BJT will have different beta value for a given current. The beta value will also change with temperature and with the Vce voltage. So to achieve Vc voltage equal exactly 4.5V you need to tweak Rb resistor value on the benchtop. – G36 Sep 19 '18 at 14:52
• SpaceDog - The configuration in Dan Mill's answer is the transistor amplifier to start learning on. R3/4 voltage divider sets the base voltage. This sets the emitter voltage as a vbe less. This sets the emitter current via R5. This sets the collector current (ignore base current as negligible) which sets the collector voltage via R1. Your common emitter amplifier is not used in practice until it becomes unavoidable, and then it's only used by experienced EEs. Treat the common emitter configuration as a switch, vbe > 0.7v on, vbe < 0.7v off. – Neil_UK Sep 20 '18 at 8:34

You've done the calculations correctly, but the entire basis for the calculations is just a rough approximation of transistor behavior. As you see, $V_{BE}$ isn't really exactly 0.7V (most of the time) and the $\beta$ isn't exactly 225 (most of the time).

This kind of biasing is very sensitive to variations in transistor parameters so it isn't used much in practice. A better biasing scheme uses four resistors, with a voltage divider for the base bias voltage and a resistor from the emitter to ground (for a little negative feedback).

• Thanks. I know the problems this kind of circuit has. I am just following through all kinds of biasing. One question is: will this circuit work in practice, with the values I have calculated? – SpaceDog Sep 19 '18 at 14:45
• @SpaceDog Try it and you will see. But, yes it will work the active region for sure. I have one question where you can buy a resistor with a value 186.750kΩ? In real-world electronics due to component tolerance, you do not need to calculate the component values with too many significant places and precision. – G36 Sep 19 '18 at 15:15
• ok. That was just the exact calculated value. Thanks – SpaceDog Sep 19 '18 at 15:20
• "will this circuit work in practice" - if your "practice" means 25C ambient temperature and if you select the transistor with b=225, then yes. If you will "practice" this circuit in Alaska or in Texas, likely not as well as you wish. – Ale..chenski Sep 19 '18 at 16:35
• @G36 take a stock resistor and start shaving ;-) – Stian Yttervik Sep 19 '18 at 18:13

You have done the calculations correctly.

However, your biassing scheme will only work well with one set of transistor parameters, it's very sensitive to variations in beta. A better biassing scheme will maintain a good bias point even with transistor variations.

A more stable one is shown below. simulate this circuit – Schematic created using CircuitLab

The 450 ohm resistor is calculated as you have done, as is the base current. I've then taken roughly 10x Ib, a nice round 500uA, and calculated resistors to give me 700mV and 3.8v drop at that current. That standing current swamps variations in Ib with beta.

As the transistor will want to try to keep around 0.7v on the base, it will act as an amplifier to try to keep the collector at 4.5v. R3 can be broken into 2 parts with the midpoint decoupled to ground to restore the AC gain.

You'll notice I've not attempted to allow for the base current being sourced from the R2/3 divider. This is deliberate. It shows that when you have a more stable biassing scheme, you can get away with significant errors, and still have a working amplifier. As transistor beta increases, the base current drops. If you have a scheme that will work well with zero base current, it means your design can be changed to use a really good transistor, and it will still work.

An even more stable circuit is shown in Dan's answer. You might well bypass R5, or a portion of it, with a capacitor, if you want more gain than Rc/Re.

You don't use that circuit in the real world....

Beta is horribly badly specified in real bipolar transistors (For example a random BC548 datasheet I just looked at give Hfe as 110 (min), 800(max), and it will vary with temperature and from device to device, so biasing the way you are trying to do it will give nothing good as a result.

Far more common is to use a scheme like the following: simulate this circuit – Schematic created using CircuitLab

Here the emitter resistance provides negative feedback (Its IR drop is subtracted from the base voltage) so Beta becomes (provided it is sufficiently large) mostly irrelevant, generally real circuits make extensive use of negative feedback to remove the dependency on poorly controlled things like device Beta. The other nice thing about this circuit is that the gain is (to a first approximation) the ratio of Rc to Re.

• For a linear amplifier, you are correct. For a simple switch, this bias network is perfectly acceptable, as long as you remember to consider worst-case beta. – John R. Strohm Sep 19 '18 at 15:39
• @JohnR.Strohm For a switch however you usually want the device in saturation, under which circumstances you want WAY more base current then Ie/beta would imply, typically to fully saturate a BJT you are looking at an effective beta of about 10! – Dan Mills Sep 19 '18 at 15:42