0
\$\begingroup\$

What's the formula for calculating generators internal resistance in a circuit with a load?

I am given the load resistance and the voltage sources voltage. What's next?

schematic

simulate this circuit – Schematic created using CircuitLab

I have tried number of ways in calculating the internal resistance of my power supply and I always get different values at different voltages.

For example:

First I measured open loop voltage of PS (Ug) then I measured voltage on output with load applied (Ul). The voltage drop on internal resistance should then be equal to: Urg = Ug - Ul , then I measured the current drawn from PS and put it in equation: Rg = Urg / I

But it doesn't seem to be the right formula, because when I apply different voltage I get different internal resistance, which should be the same all time...

\$\endgroup\$
5
  • \$\begingroup\$ Show what you have tried. \$\endgroup\$ Commented Jan 31, 2017 at 16:12
  • 2
    \$\begingroup\$ Your model might not be applicable to your power supply \$\endgroup\$
    – user28910
    Commented Jan 31, 2017 at 16:26
  • \$\begingroup\$ @StainlessSteelRat I just did. \$\endgroup\$
    – lucenzo97
    Commented Jan 31, 2017 at 16:38
  • \$\begingroup\$ @user28910 Then please edit it if you think that. \$\endgroup\$
    – lucenzo97
    Commented Jan 31, 2017 at 18:00
  • \$\begingroup\$ @Keno - I don't know anything about the power supply you are trying to characterize. Just saying that an idea voltage source with a resistor in series may not be an accurate model. Also, you didn't say what kind of results you are getting. Are the numbers very different from what you are expecting? Have you thought about how measurement errors might affect your results? \$\endgroup\$
    – user28910
    Commented Jan 31, 2017 at 19:20

2 Answers 2

1
\$\begingroup\$

Your experimental approach is valid. Your concern that the power supply internal resistance should be constant may be invalid, subject to these potential problems:

  • Your power supply has an internal current-limiter.
  • Contact resistance and/or wire resistance affects your measurements.
  • Your power supply output voltage is affected by temperature.
  • Your power supply output has "ripple" due to charging/discharging capacitors

Some voltage-regulated supplies have such low internal resistance that they cannot provide current to a very small-resistance load. Until current reaches a set-point, it acts as an almost ideal voltage source, with very low internal resistance. Your measurements may be affected by the first three potential problems.

The last problem arises because many unregulated supplies have non-linear elements inside (like diodes for example) that charge up an output capacitor to a DC voltage. Your experimental approach assumes that your voltage source is a linear circuit. This kind of DC supply is not linear:

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
1
\$\begingroup\$

So forget it is a generator. You have a series circuit with two resistors.

You know the open source voltage, which is \$U_G\$.

You know the Terminal Voltage \$U_T\$, which is also the load voltage \$U_L\$.

You know the load current I.

Kirchhoff's Voltage Law for a series circuit says the sum of voltages in a series circuit equals the applied voltage. \$U_G\$ is the applied voltage and \$U_L\$ is one of the two voltages.

Finally, you know the current, so Ohm's Law will give you \$R_G\$.

Edit your question if you still need help.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.