On my battery eliminator, if I measure source voltage without any load, it gives the correct voltage.

But when I add a load, the voltage drops.

And if I short-circuit the 2 output-leads with a very low resistance, say a piece of conductor wire, the voltage comes down to Zero.

voltage drop

The above image shows what is the diagram for connection.

Now I want to know , is there any formula or rule (mathematical expression) to calculate such voltage-drop?

What would be the calculation for this voltage-drop for other-kind of electric-source, such as battery, AC-source, etc. ?

Edited: intuitively it seems the voltage drop should occur even in ideal voltage source with no internal resistance; because the leakage through the parallel (bypass) path would drop the electrical pressure .

  • \$\begingroup\$ Does that battery eliminator has a datasheet? \$\endgroup\$
    – CL.
    Jun 1 '16 at 7:28
  • \$\begingroup\$ @ CL. no datasheet... the eliminator box I've made as project for my training-class. It contains a stepdown transformer (many output points), bridge rectifier, and a filter-circuit made up of 2 Electrolytic capacitors (each 1000 mu-F, 25 v) and a choke-coil. However the same phenomenon seems to occur in case of any-other kind of source (i've tested the short-circuit event in battery (cell) , but yet didn't tested impact of other-resistor in case of battery). \$\endgroup\$ Jun 1 '16 at 7:37
  • \$\begingroup\$ Any mathematical equation? \$\endgroup\$ Jun 1 '16 at 8:22
  • \$\begingroup\$ What is the saturation current of the transformer? Anyway, in practice, one would compute the output impedance from the measured voltage drop … \$\endgroup\$
    – CL.
    Jun 1 '16 at 9:16
  • 1
    \$\begingroup\$ V = I*R, or in this case you'd want R = V/I, which can be modified to: R = (V1 - V2)/(I1 - I2) this gives you the change in voltage in relation to the change in current where V1, V2, I1, I2 are the current and voltages measured with two different loads (note: No load doesn't give you any useful information in this instance). When you have R then you can plug it into V = I*R to see what voltage you get under any load (with a known current) \$\endgroup\$
    – Sam
    Jun 1 '16 at 9:48


simulate this circuit – Schematic created using CircuitLab

The reason why the voltage of the voltage source drops when you connect a load to it is that the voltage source is not an ideal voltage source. It has an output impedance. The output impedance is unavoidable during the construction of a power source. In theory you can split your voltage source into two components, the ideal voltage source (V1) and the output impedance (R1).

If you now measure your voltage source with a voltmeter you get the voltage of your ideal voltage source (V1), since no current is flowing through the resistor R1. (In reality some tiny amount of current flows through the voltmeter since it is not ideal either but we ignore this since it is negligible)

In the second case where you load down you voltage source the measured voltage drops because the current flowing through the load (R2) will also flow through the output impedance (R1). What you got here is a voltage divider.

$$VM1 = \frac{R2}{R1 + R2} V1$$

To calculate what amount of voltage your voltage source will drop if you connect a load to it you need to know the value of the output impedance. Then you can just use ohms law to calculate the voltage drop for a known load current.

$$U(R1) = R1 \cdot I$$

A problem with batteries as a voltage source is that the voltage source (V1) is dependent on the capacity left in the battery. Once the battery is empty it will rapidly drop, but because of battery chemistry this will only be visible when you load down the battery.

Addendum: Your "Battery eliminator" probably has a schematic similar to this:


simulate this circuit Since you have no regulation your voltage/current relation is mainly dictated by the transformer. Giving you an exact mathematical relationship is rather complicated since it involves both AC and DC. If you want to know the specifics you can read further here (requires knowledge of AC circuit analysis)

  • \$\begingroup\$ Yes. correct diagram. Just 2 e-capacitors (1000muF, 25V), a choke coil and a resistor (150 ohms, 1W) , very-similar to this-one. 4.bp.blogspot.com/-xsF1odnFELE/VGoz8muWr0I/AAAAAAAAAXs/… . I did-not mentioned the circuit diagram because i tried to keep the question size small and keep the focus on main point. \$\endgroup\$ Jun 1 '16 at 10:14

Voltage divider

The BEC has an output impedance. The load has an impedance. The two impedances form a divider that reduces the open-circuit voltage of the BEC proportionally.

  • \$\begingroup\$ please be slightly more specific. proportionally means? proportional with whom? \$\endgroup\$ Jun 1 '16 at 7:56
  • 1
    \$\begingroup\$ With the two impedances. Just as the article describes. \$\endgroup\$ Jun 1 '16 at 8:03
  • \$\begingroup\$ Only for battery-eliminator? not for other-kind of sources? \$\endgroup\$ Jun 1 '16 at 8:04
  • 1
    \$\begingroup\$ For every kind of everything. That's how physics works. \$\endgroup\$ Jun 1 '16 at 8:04
  • \$\begingroup\$ any mathematical equation? \$\endgroup\$ Jun 1 '16 at 8:23

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