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# What is the drop in load voltage and output voltage of the power supply during loading؟

#### electronium

##### Member
greeting to forum
A question has been occupying my mind right now, what is the voltage drop of the power supply output and load input when loading with a certain ampere?
for exmple:
source 12vdc
drop voltage terminal load؟ How is it calculated?
load= lamp 12v 21w

Last edited:
That's the Load or Output voltage regulation figure; typically the percentage drop in output voltage from no or minimum load current up to maximum load current, or a range of current.

If it's not in the specifications, you have to measure it yourself.

The other factor is Line, or Input voltage regulation for DC input - the change in load voltage as the input voltage varies from minimum to maximum, or though a given range.

For such as a voltage regulator IC, those figures should be in the data sheet.

See page three of this for examples with a 7805:

That's the Load or Output voltage regulation figure; typically the percentage drop in output voltage from no or minimum load current up to maximum load current, or a range of current.

If it's not in the specifications, you have to measure it yourself.

The other factor is Line, or Input voltage regulation for DC input - the change in load voltage as the input voltage varies from minimum to maximum, or though a given range.

For such as a voltage regulator IC, those figures should be in the data sheet.

See page three of this for examples with a 7805:
greeting enginner
I did the experiment, and surprisingly, the two terminal voltmeter or the load (the lamp) was voltage dropped and 10.5V reached 10.5V
How was it calculated?

Power supplies have a feedback loop that compares a reference V
to output V and drives the supply to match the Vout to the Vref. Vref
is what you ask the supply to provide.

In the above case. very typical, the Vout is divided down to match the
Vref. But circuit gain will move the Power stage to output the correct
Vout.

Regards, Dana.

Power supplies have a feedback loop that compares a reference V
to output V and drives the supply to match the Vout to the Vref. Vref
is what you ask the supply to provide.

View attachment 146187

In the above case. very typical, the Vout is divided down to match the
Vref. But circuit gain will move the Power stage to output the correct
Vout.

Regards, Dana.
Engineer,
if the subject is abstract, what is its formula? Why are the two ends of the lamp or load not 12 volts? It is 10.5 volts
Practical tests are very strange but real, how? From which formula is 10.5 volts obtained?

if the subject is abstract, what is its formula? Why are the two ends of the lamp or load not 12 volts? It is 10.5 volts
It's not abstract.
Other than the drop due to the (usually) small regulation error in the power supply, the drop is due to wire connection resistance in the circuit.
Using Ohm's law this is simply I x R were I is the current and R is the connection resistance.

The output volage of an unregulated (or poorly regulated) power supply will drop as the load current increases, because of the voltage drop across internal resistance in the supply.

Take a look at this for basic linear regulator operation :

Regards, Dana.

A regulated power supply provides the regulated voltage at the output terminals.* The current is carried by wires to the load; these wires have resistance, so there is a voltage drop across the cable of I × R [current x load]. If the load isn't too large, and the cable large enough diameter (less resistance), the voltage drop will be insignificant. If the load current is large, the cable too small and too long, the voltage drop could be substantial.

*Some power supplies can sense the voltage at the load using separate wires. This type of supply will compensate for the I×R drop in the cable.

Thank you all
I did the test without the power supply wire, the result was that the voltage on both ends of the load does not show a voltage drop and a constant 12 V voltage falls on both ends of the lamp.

Thank you all
I did the test without the power supply wire, the result was that the voltage on both ends of the load does not show a voltage drop and a constant 12 V voltage falls on both ends of the lamp.
Welcome to Ohm's law.

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