simulate this circuit – Schematic created using CircuitLab

The voltage divider is used in many circuits such as biasing the base of a transistor, etc.

Why is the voltage in the circuit on the left considered more stable than in the circuit on the right, if the voltage and current in the load is the same in both circuits?

EDIT: I have edited the circuit to clarify

My question refers to ..., if the load consumption changes, its resistance will change

So why is the voltage on the load more stable with a voltage divider than without a voltage divider?

Thank you

  • \$\begingroup\$ What value of load resistance will produce the same load voltage for both circuits? \$\endgroup\$
    – Andy aka
    Commented Jan 10, 2021 at 10:08
  • \$\begingroup\$ I think 50 ohms, sorry if I'm wrong \$\endgroup\$
    – Mario
    Commented Jan 10, 2021 at 10:12
  • \$\begingroup\$ So, do the potential divider calculations and assure me that the two scenarios are identical in load voltage. Don't guess. If you are to make a comparison, you need to make an objective comparison. \$\endgroup\$
    – Andy aka
    Commented Jan 10, 2021 at 10:14
  • \$\begingroup\$ You don't need to guess @Andyaka for this question of WHy is left more stable \$\endgroup\$ Commented Jan 10, 2021 at 11:18
  • 3
    \$\begingroup\$ Tony I have no idea what you are talking about (seriously). \$\endgroup\$
    – Andy aka
    Commented Jan 10, 2021 at 11:37

4 Answers 4


The load voltage is dependant of the "bottom" resistor in the voltage divider, thus it will be dependant on the parallel between the load equivalent resistor and R2 in left circuit, and the load equivalent resistor alone in the right one.

That said, if the load resistor changes also the load voltage will change, and this is a pretty undesiderable thing in a voltage reference. Paralleling the (probably high) resistor of the load to a (much smaller) resistor will make the load voltage more stable because the high load resistor contribution in the parallel resistor computation is more and more negligible as the load resistor get higher and R2 gets lower.

Anyway, this circuit is not the best for voltage stability. I would rather use the circuit proposed by Transistor in the post above.

  • \$\begingroup\$ Thanks a lot, That was my doubt, in many publications it is indicated that a resistance in parallel stabilizes the voltage but nobody explains why \$\endgroup\$
    – Mario
    Commented Jan 10, 2021 at 10:26
  • \$\begingroup\$ Happy to help! You're welcome \$\endgroup\$
    – Sixaxix9
    Commented Jan 10, 2021 at 10:31


simulate this circuit – Schematic created using CircuitLab

Figure 1. Equivalent circuit and example of operating point.

If your load is resistive then the two circuits can be made equivalent (but you'd need to recalculate your R5 value). In general you'll see the arrangement of 1a in transistor biasing circuits, etc. As shown in 1b the equivalent circuit has a better controlled maximum voltage with the output impedance being the parallel combination of R1 and R2.

1c's operating point will depend on the gain of Q1 and will vary from device to device - even in the same batch of components. Using 1a to bias the transistor will result in a much better defined operating point much less reliant on the gain of the transistor.

  • \$\begingroup\$ >> As shown in 1b the equivalent circuit has a better controlled maximum voltage with the output impedance being the parallel combination of R1 and R2 Does circuit 1b have better control of the voltage over the load? \$\endgroup\$
    – Mario
    Commented Jan 10, 2021 at 10:21
  • \$\begingroup\$ 1b is the equivalent of 1a. 1c is an emitter-follower. If you use 1a's biasing arrangement on 1c's Q1 the base will be held close to 4 V and the emitter will be held 0.7 V below that at 3.3 V. In 1c you don't know what voltage will be applied to RL. \$\endgroup\$
    – Transistor
    Commented Jan 10, 2021 at 10:35

Given any two of \$V_\text{LOAD}\$ and \$I_\text{LOAD}\$ and \$R_\text{LOAD}\$, you can always find the third. That's a complete specification for your theoretical load.

You also have a voltage supply, \$V_\text{SUPPLY}\$, and have provided two different proposals for providing the correct operating conditions for the load:

  1. Dropping resistor applied in series with the load, \$R_\text{DROP}\$.
  2. Resistor divider for the load, \$R_1\$ and \$R_2\$.

And now want to know how to compare the stability of the two cases.

But since the resistor divider can always be replaced by a Thevenin equivalent, \$V_\text{TH}\$ and \$R_\text{TH}\$, we really only have one case -- not two. Either way, it boils down to a voltage source and a series resistance. The only difference is that in the second case it's possible to have some control over the voltage source, too.

So there's only one case. Each case maps to just the same general statement:

  1. \$V_\text{TH}=V_\text{SUPPLY}\$ and \$R_\text{TH}=R_\text{DROP}\$
  2. \$V_\text{TH}=V_\text{SUPPLY}\cdot \frac{R_2}{R_1+R_2}\$ and \$R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$

\$V_\text{TH}\$ supplies \$R_\text{TH}\$ in series with \$R_\text{LOAD}\$.


Then the voltage on the load is just:


Stability is based upon the sensitivity equation. Using it (unshown here, but easily looked up on the web), we can find the following fact:

$$\begin{align*} \%\,V_\text{LOAD} &= \left[\frac{1}{1+\frac{R_\text{LOAD}}{R_\text{TH}}}\right]\cdot \%\,R_\text{LOAD}\\\\&=\left[\frac{V_\text{TH} -V_\text{LOAD}}{V_\text{TH}}\right]\cdot \%\,R_\text{LOAD} \end{align*}$$

The smaller you can make \$R_\text{TH}\$ the better, with respect to regulation versus load variation. Put another way, the closer \$V_\text{TH}\$ is to the desired \$V_\text{LOAD}\$, the better. And the resistor divider always has a closer \$V_\text{TH}\$, almost by definition.

This additional fact arrives from applying the sensitivity equation:

$$\begin{align*} \%\,V_\text{LOAD} &= \%\,V_\text{TH} \end{align*}$$

In case 1, \$V_\text{TH}\$ is just the supply voltage itself. So a 10% variation in the supply voltage will mean a 10% change in the load voltage, as well.

In case 2, \$V_\text{TH}\$ is a fractional portion of the supply voltage. So a 10% variation in the supply voltage will mean something less than a 10% variation in the load voltage, depending on the divider design.

So here too is a win for the resistor divider.

  • \$\begingroup\$ How did you derive the second equation for the sensitivity (in terms of Vth and Vload)? \$\endgroup\$ Commented Oct 29, 2023 at 18:33

You can see with the left Thevenin is 4V @ 60 Ohms and the right as 10V @ 200 Ohms

Rule of Law

% Load regulation is defined by some "rated" load Ohms or Amps by the % of change on the observed voltage. Thus a 1% change occurs if the load is 99 times higher than the source or you could say the source is ~ 1% of the load.

Thus regardless if the current is rated matched on both sides or short circuited, the critical factor is the source impedance and thus the left side has the obvious advantage for "Load Regulation error" with 60 Ohm source impedance.

Which side is more stable with loads and why?

Lower % load regulation error = lower source/load impedance ratio.

Actually, load regulation error is defined as Rs/(Rs+RL)*100% for rated Voltage and Amps and is measured by this test. But if you remember it as Rs/RL that's close enough for 1% ranges or less.

Now let's say I happen to have figured out you get the same voltage with 100 Ohms/3 for the load that yields 1.428 V @ 42.9mA.

Does that matter what the value is this question WHICH IS MORE STABLE??

You answer.
and Why?

Bonus question

Which side can supply the most power to any load of your choice?
What's the simple answer?


Your Answer

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