# A general question about voltage drop

#### raziell122

Joined Mar 28, 2023
52
Hello all!
In my project I built two different circuits, one of them gets 15V supply from lab power supply, and the other one is designed to work from 5V.
I wanted to use only one power supply in this project, so at first I tried to do a voltage divider by using 2 regular resistors to get 5V from the 15V and send this 5V to the second circuit, but when I connected this 5Volts to the second circuit, the voltage has dropped to around 1-2V.
I tried using an OpAmp as a buffer between the two circuits but the problem did not fix.
The next solution for me was to replace the buffer with a voltage regulator LM7805 and the problem is fix now.
I wanted to ask what may be the reason for this voltage drop? is it because the consumed current from the buffer that was higher than it can supply? and if so, why does it happen?
*Sorry for any English mistakes*

#### seanstevens

Joined Sep 22, 2009
213
If the current/power consumed by your 5V circuit is higher than what the voltage divider or your opamp buffer can supply, then you will have a voltage drop.
If the 5V is a critical requirement for its operation, I would not use a resistor voltage divider, its best to use a voltage regulator or depending on the power required, a correctly rated Zener diode with an appropriate resistor can also do the job.

#### raziell122

Joined Mar 28, 2023
52
If the current/power consumed by your 5V circuit is higher than what the voltage divider or your opamp buffer can supply, then you will have a voltage drop.
If the 5V is a critical requirement for its operation, I would not use a resistor voltage divider, its best to use a voltage regulator or depending on the power required, a correctly rated Zener diode with an appropriate resistor can also do the job.
But what causes such voltage drop? I would like to know the physical cause/explanation to understand it better.

#### Ya’akov

Joined Jan 27, 2019
8,171
AAC- ברוך הבא ל

Your thread has been moved to the Homework Help forum. The rules of AAC limit the sort of help that can be given with coursework-related questions. For that reason, all such questions must be posted in the Homework Help forum to make the nature of them clear.

(English is the language of AAC, my greeting was personal, to a חבר יהודי. In any case, I am not conversational in Ivrit so I wouldn't do very well trying to go further.)

#### Ya’akov

Joined Jan 27, 2019
8,171
Components and subsystems are able to deliver only a certain amount of power. So if something is rated for say, 5V@1A that would make it a 5W device. ($$\mathsf{W = E \times I}$$).

If you try to draw more current than the device can manage, the voltage drops to keep that total power dissipated within the limits of the device. Unless the current limiting is a deliberate feature, it will also get hot and possibly release the magic smoke making the device just another victim of the education process.

#### MrChips

Joined Oct 2, 2009
29,246
Voltage Divider

Vout = Vin x R2 / (R1 + R2)

When a load is connected across R2, i.e. from Vout to GND, the effective resistance of R2 is altered. Hence Vout is lowered.

A rule of thumb is the resistance of the load should be ten times greater than that of R2 in order to minimize the loading effect.

#### MrChips

Joined Oct 2, 2009
29,246
Voltage Regulator using Zener Diode

A simple and effective voltage regulator can be created with a zener diode and series resistor Rs.
When the zener diode is conducting, the current through Rs is given by Ohm's Law:

I = (Vin - Vz) / Rs

In other words, Rs is calculated for a given maximum load current:
Rs = (Vin - Vz) / Imax

The current Imax is shared by the zener diode and the load. Whatever current is not drawn by the load, i.e. Imax - IL, will flow through the zener diode.

The power rating of resistor Rs and the zener diode should be selected to be at minimum 20% greater than the power dissipated by each individual component.

#### raziell122

Joined Mar 28, 2023
52
Voltage Regulator using Zener Diode

View attachment 295356

A simple and effective voltage regulator can be created with a zener diode and series resistor Rs.
When the zener diode is conducting, the current through Rs is given by Ohm's Law:

I = (Vin - Vz) / Rs

In other words, Rs is calculated for a given maximum load current:
Rs = (Vin - Vz) / Imax

The current Imax is shared by the zener diode and the load. Whatever current is not drawn by the load, i.e. Imax - IL, will flow through the zener diode.

The power rating of resistor Rs and the zener diode should be selected to be at minimum 20% greater than the power dissipated by each individual component.
Voltage Divider

View attachment 295355

Vout = Vin x R2 / (R1 + R2)

When a load is connected across R2, i.e. from Vout to GND, the effective resistance of R2 is altered. Hence Vout is lowered.

A rule of thumb is the resistance of the load should be ten times greater than that of R2 in order to minimize the loading effect.
Components and subsystems are able to deliver only a certain amount of power. So if something is rated for say, 5V@1A that would make it a 5W device. ($$\mathsf{W = E \times I}$$).

If you try to draw more current than the device can manage, the voltage drops to keep that total power dissipated within the limits of the device. Unless the current limiting is a deliberate feature, it will also get hot and possibly release the magic smoke making the device just another victim of the education process.
Thank you all!
That was very informative