# Formula for R1 and R2 of a voltage divider bias

Please I'm asking if the stated procedure for R1 and R2 of a voltage divider is correct i.e if its valid way to come up with their respective values. If so, how do interpret the two statements mathematically. Thank you

• It would be helpful to draw a schematic indicating which is R1 and which is R2 and where they are connected. You can use the Circuitlab function. Also there's an implicit assumption in those formulae. Commented Sep 18, 2023 at 21:39
• Using V=iR, validating both of these statements is very trivial. Commented Sep 18, 2023 at 21:51
• Please do not insert images of text. There are several reasons, not least of which is they can't be read by the software used by visually impaired users. Also, it's'slightly out of focus which strains the eyes of those who can see. Commented Sep 19, 2023 at 0:01
• imran, I think you need to hire a tutor. Just saying. This isn't rocket science. And if you are dealing with BJT circuits, then you may already be far out of your depth given your question. Consider finding someone to help you out. Commented Sep 19, 2023 at 4:28

# Some history

Before determining the resistances R1 and R2, it would be good to understand why these resistors are needed at all. We can do this best if we mentally go back two centuries to when Ohm conducted his famous experiment and reinvented the famous voltage divider. I used this approach 15 years ago when my students and I conducted a series of real computerized experiments in the lab and described them in a Wikibooks story.

Now I do not have real students (who are more interested in their grade than in understanding circuit phenomena); however, I have these really inquisitive OP's and this magical virtual CircuitLab with the help of which we can work miracles.

# Understanding through simulations

For the purposes of intuitively understanding circuits, you can use the CircuitLab simulations below as follows:

1. The device parameters have initial values ​​that set the quiescent output voltage (current). They are adjusted so that the quiescent voltages (currents) have "exact" values. You can then explore the circuit in the following ways.

2. Open (by double-clicking on the device) the parameters window and start setting different values ​​while watching the measurement instruments. If the window covers the meter, click on a blank part of the window, and then Control-drag the schematic to the side.

3. If there is a graph under the schematic, enter Simulate > DC Sweep and then click Run DC Sweep. Change the device parameters and run again the DC Sweep.

# Reinventing the voltage divider

## The problem

Even then, the need arose to produce a voltage Vout lower than the supply voltage Vcc. So they came up with the idea of ​​somehow ​​removing part of the supply voltage and using the rest. How can we do it nowadays?

simulate this circuit – Schematic created using CircuitLab

## Dropping resistor

Constant dropping resistor: The first idea that comes to mind is to connect a resistor R in series with the power supply. "A resistor is something that gets in the way", we think, so it will reduce the voltage. What is our surprise, however, when we see that Vout does not change!

simulate this circuit

Varying dropping resistor: Maybe the resistance is not high enough? Then let's increase it (at least 10x)...or even make it variable.

simulate this circuit

In vain! No change...

## Adding a current sink

Gradually we begin to realize the problem... A resistor prevents (resists to) the current, but here no current flows. So there is nothing to get in the way... Simply speaking, the resistor is not a resistor when there is no current flowing... but it transfers all the voltage.

The remedy is to somehow draw current I from the resistor through some sort of "pull-down" element. This current will "create" a voltage drop VR = I.R that will be subtracted from Vcc. For example, we can do it with a constant current source I.

simulate this circuit

But then the output voltage will follow Vcc with a constant downward offset VR = I.R.

This technique is used to "shift" voltage variations, e.g., in op-amp internal structures.

## Adding a resistor sink

However, here we want to make a "divider" that proportionally decreases the input voltage. Then let's connect a simple ohmic resistor that will proportionally reduce the current depending on the voltage. Thus we get the classic 19th century voltage divider.

simulate this circuit

As you can see, since R1 = R2, the divider's gain is 0.5 (see the other answers for deriving the expression; I am just explaining the idea here).

## Op-amp biasing

The voltage divider we "invented" above is unloaded (open circuit). We can use it to bias, for example, an op-amp follower that has an extremely high (linear) input resistance...

simulate this circuit

## Emitter follower biasing

... or an emitter follower (common-collector stage) that has high (linear) input resistance (I will not explain why here due to lack of space).

simulate this circuit

## Loaded voltage divider

Let's now load our voltage divider with another resistor RL = 1k to see how it will behave.

simulate this circuit

The result is that its gain (output voltage) decreases.

But this is not a big trouble because we can foresee it and take measures.

## Base-emitter non-linear load

It is bad when the load does not have constant resistance but changes it when the voltage across it changes (as they say, it has non-linear or dynamic resistance). The base-emitter input of a transistor represents such a non-linear load.

The difference between this common-emitter stage and the common-collector stage above is that here the divider's voltage is directly applied to the base-emitter junction while above there is an emitter resistor (the so-called "emitter degeneration").

simulate this circuit

Let's adjust R2 so that a collector current of 1 mA flows. We get R2 = 708.8 Ω and Vbe = 657.7 mV.

## Equivalent circuit

To understand what the transistor input does, let's simulate the behavior of such a non-linear load through a variable resistor Rbe. For this purpose, we set R2 = 708.8 Ω as above and start changing Rbe until the voltmeter shows again Vbe = 657.7 mV. So, for this voltage, the base-emitter junction has a (static) resistance of Rbe = 78.16 Ω. If we change Vbe (R2), Rbe will change.

simulate this circuit

• That's alot.. will try to digest it carefully. Thanks for the help really Commented Sep 24, 2023 at 20:51

simulate this circuit – Schematic created using CircuitLab

First thing to do is draw a schematic - which you didn't do.

To solve for a standard voltage divider...
The current flowing through R1 is the same as R2 since the resistors are in series if you assume that whatever is connected to Vb requires an insignificant amount of current. Vb sounds like the base of a transistor which may be significant in the analysis. However, for this exercise assume that there is no additional loading on Vb which seems to be the case for the text quoted. Outside of textbooks (in real life) the current going in to the base of the transistor generally needs to be considered for biasing circuits.

The current flowing through R1 & R2 is: $$I = {Vcc \over {R1+R2}}$$

Vb is: $$Vb = I \; R2 = Vcc - (I \; R1)$$

This is basic Ohm's Law being applied to a very basic concept, namely, a voltage divider.
This is the steps I use solve a voltage divider circuit (find the current, solve for voltages across the resistors). I don't memorize the voltage divider equation since it uses excess neurons which I have in short supply.

• Although these types of divider are set up so that I >> Ib, in some circumstances, ignoring the base current can move the bias point away from that desired. Commented Sep 19, 2023 at 0:04
• The comment from @PeterJ is true. Circumstances matter. The base presents a load to the divider. If the load is negligible, then all works fine. But if that load isn't negligible, then it must be accounted in the computation. Often, with BJT biasing, the base load is about 10% of $I$ as shown in your answer. That may be significant enough to worry over (I usually include it.) But it's far away enough that some would argue otherwise. So no harm no foul, either way. But it still remains that the choice of $I$ as compared to the base load may significantly matter in some circumstances. Commented Sep 19, 2023 at 4:25
• However, I think it does NOT matter significantly if sufficient negative DC feedback is applied (resistor RE). By the way - this is the most important reason for implemeting such a negative feedback.
– LvW
Commented Sep 24, 2023 at 15:07