Voltage Divider Circuit Current Problem

I am using 7808 to turn 12v into 8 voltage which is working fine, but when i try to drop the voltage to 7.2 by using voltage divider circuit, it is giving voltage (7.2 something) but it is not turning anything on like my servo ( which is suppose to be running on 7.2)
I think there is some problem regarding the current capacity of resistors that i am using which is the normal resistor size that we usually use ( sorry for being dumb but i am really new in electronics and stuff)

Voltage dividers are for signals (very low current). Ohm's law will tell you why you can't use a voltage divider for power. As soon as you connect anything to the voltage divider output you change the resistances and therefore the output voltage.

From the Sparkfun voltage divider tutorial

Application Dont's

As tempting as it may be to use a voltage divider to step down, say, a 12V power supply to 5V, voltage dividers should not be used to supply power to a load.

Any current that the load requires is also going to have to run through R1. The current and voltage across R1 produce power, which is dissipated in the form of heat. If that power exceeds the rating of the resistor (usually between ⅛W and 1W), the heat begins to become a major problem, potentially melting the poor resistor.

That doesn't even mention how inefficient a voltage-divider-power-supply would be. Basically, don't use a voltage divider as a voltage supply for anything that requires even a modest amount of power. If you need to drop down a voltage to use it as a power supply, look into voltage regulators or switching supplies.

The mid-terminal of the 'voltage divider' circuit should not be used as a 'power' supply in general.

The reason is because the voltage at the mid-terminal is dependent on the amount of current flowing out from your actual voltage supply (and on the resistance of the upper resistor etc).

The higher the amount of current that flows through the upper resistor...... the bigger the voltage will be across that upper resistor. This means that the divider voltage may not be what you designed for (once you connect something EXTRA to that mid-terminal, such as your load).

The upper resistor is also going to limit the current to the rest of the your circuit as well.

This just means ..... if using a 'regulated' voltage supply, then use a voltage supply that generates the desired regulated voltage to start with. A voltage divider circuit shouldn't be used as a 'power' supply - in general.

Southpark:
The mid-terminal of the 'voltage divider' circuit should not be used as a voltage supply in general.

The reason is because the voltage at the mid-terminal is dependent on the amount of current flowing out from your actual voltage supply (and on the resistance of the upper resistor etc).

The higher the amount of current that flows through the upper resistor...... the bigger the voltage will be across that upper resistor. This means that the divider voltage may not be what you designed for (once you connect something EXTRA to that mid-terminal, such as your load).

The upper resistor is also going to limit the current to the rest of the your circuit as well.

This just means ..... if using a power supply (voltage supply), then use a voltage supply that generates the desired regulated voltage to start with. A voltage divider circuit shouldn't be used as a voltage supply - in general.

Should i use any type of regulator then ?

Mark248:
Should i use any type of regulator then ?

You can use any suitable regulator. One which allows you to make adjustments to provide your desired regulated output voltage ..... of 7.2 volt. Voltage regulator circuits often have components .... eg a resistor of a particular value, for setting the output voltage.

There are relatively cheap 'switching power supply' circuits that can allow you to efficiently convert higher DC voltages to lower DC voltages.

Can use a common diode to drop 0.7volt.
Leo..

A diode is an easy fix. But... 8V is only 10% over the nominal voltage and it will probably be OK.

You might check the servo specs to see if there is a maximum voltage rating. Like everything else in the "real analog world" voltages are never perfect so there has to be some tolerance. (But, they may not always specify or publish the tolerances/limits.)

Don't forget the thermal and current limitations of a linear 7808 regulator.
With a dropout voltage of 12-8= 4volt, anything over 250mA will need a heatsink.
And the 1Amp limit means one small (SG90) servo, nothing bigger.
Leo..

@Mark248

Perhaps this will help....

Consider you have an 8 foot platform.

You have two bungee cords, a heavy one and a lighter one. They are arranged so the heavy one is connected to the platform
The weaker one connected to the floor.

They are tied together and the connection point happens to be at 7 feet.

The connection point will stay at 7 Ft until you connect a weight to that point (i.e. a load) then the connection will start to drop. The heavier the load the more it will drop.

This is essentially what you have in your circuit.

Thank you everyone, you guys made it easy for me to understand what is happening, i have decided to use LM317 voltage regulator to get my 7 voltage and also it is 1.5 A so servo can work fine with it but can you please tell me does the size of heat sink matters ? as according to my calculation heat dissipation is around 6 watts which is alot but i am confused about heat sink size.

Mark248:
Thank you everyone, you guys made it easy for me to understand what is happening, i have decided to use LM317 voltage regulator to get my 7 voltage and also it is 1.5 A so servo can work fine with it but can you please tell me does the size of heat sink matters ? as according to my calculation heat dissipation is around 6 watts which is alot but i am confused about heat sink size.

Medium to large heatsink for 6W - this is something you calculate from the thermal data for the heatsink
and the maximum temperature rise you are prepared to tolerate.

Hi,
Can you post a link to your servo data/specs please?

Thanks.. Tom.. :slight_smile:

If you want to provide an amp or so at 7.2V for a servo from a 12V supply, don’t use a linear
regulator, that’s just going to waste power and generate lots of annoying heat you have to
get rid of.

A simple cheap LM2596 DC-DC converter module off eBay will give much more efficient power conversion,
you can set the output voltage to whatever you want, and no heatsinking required.

This thread is classic xyproblem - rather than asking how to power a servo from 12V,
you asked a question about the wrong way to do it, and then the thread proceeded to
discuss other wrong ways to do it, despite Southpark’s correct solution in #4

And I admit I got drawn in too! Sorry I didn’t just reiterate #4 when I should have.

Hi All,
There is still a choice to make. With the switcher, you do not
need a heat sink but, you have to use an inductor and a special
diode. As noted in other posts, circuit layout may be difficult
to optimize. I vote for the heat sink and extra diode approach.
But, what is the servo current? That may be the deciding factor.
Herb