Voltage Divider give different values [duplicate]

I have made a simulation of voltage divider, and that give me the same value like on this calculator: http://www.ohmslawcalculator.com/voltage-divider-calculator

But when I soldered my board I have measured voltage and It was around 4.5 V, but with that calculator should be around 12 V.

My two resistors have value 2000 Ohm (2 kOhm).

I was want to make 24 V to around 12 V with voltage divider, and then 12 V pass to LM7805 so my Atmega328 will work on 5 V.

Here is my PCB how works, really I'm not sure how is it possible to voltage to be 4.5 V but should be 12 V.

• Please add a schematic as well. – Bence Kaulics Jul 21 '16 at 14:17
• I have schematic in Fritzing but is really mess... because i started direct in PCB, but is pretty simple. – bosko Jul 21 '16 at 14:22
• Also an LM7805 can handle 24 V input voltage. It won't be more efficient if you try to lower the voltage with a divider because then you will waste power on the resistor in the divider. Maybe useful question. If you have such a high supply voltage available, I suggest you to use a buck converter. – Bence Kaulics Jul 21 '16 at 14:25
• Voltage dividers are not a suitable way of powering anything that consumes more than a few miliamps of power. – pjc50 Jul 21 '16 at 14:30
• You shouldn't draw current from a resistor divider. Don't do it. Please, don't do it! No! ${\sf \style{font-size:150%}{\text{No! Oh no}}\lower{1pt}\style{color:#303030;font-size:135%}{\text{o}}\lower{2pt}\style{c‌​olor:#606060;font-size:120%}{\text{o}}\lower{5pt}\style{color:#909090;font-size:1‌​??05%}{\text{o}}\lower{8pt}\style{color:#B0B0B0;font-size:90%}{\text{o}}}$ – dim Jul 21 '16 at 14:49

The problem is not your divider, the problem is this is not where you should be using one.

The LM7805 (with a load) in your case can be modelled as a resistance that draws current from your divider. This means the total resistance of the bottom half of your divider has changed, and hence the output you previously calculated has now changed.

Also as mentioned I think the LM7805 is capable of taking a 24V input voltage. Remove the voltage divider and you should not have any problem.

• If heat is a problem, use a buck converter as previously mentioned to drop the voltage - this is an efficient alternative – Nick B Jul 21 '16 at 14:31
• Good answer! And even without a load, the standby (quiescent) current may be enough to alter the voltage coming out of the divider. – Big6 Jul 21 '16 at 15:20
• Yes as Sixto says, even an LM7805 with nothing connected to it's output still loads your divider. Please only use a divider for very high impedance inputs! (If at all) – Nick B Jul 22 '16 at 9:20

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) What you appear to have done. (b) What you should have done.

You have made the mistake of thinking that you can connect a load to the potential divider without disturbing the voltage. Since you only want to reduce voltage to the regulator you just add a series resistor.

$$R3 = \frac {V_+ - V_{MIN}}{I_{MAX}}$$

where $V_{MIN}$ is the minimum acceptable input for the 7805.

Power calculations

For Figure 1b power in the resistor and regulator can be calculated as follows:

$$P_{R3} = I^2 R$$

$$P_{7805} = (V_+ - IR_3 - 5) I$$

• In this case there will not be problem with heating and voltage ? – bosko Jul 21 '16 at 14:40
• See the update. Create a spreadsheet and run the calculations across the range of currents you expect. Size to suit. Voltage wouldn't be a problem - the 7805 can take 24 V. – Transistor Jul 21 '16 at 14:48
• what values should be R3, C3, C4 if i want V+ to be between 24v and 20v ? – bosko Jul 21 '16 at 14:49
• C3 and C4 don't affect the voltage - they prevent oscillation. The values will be given in the datasheet examples. You need to calculate R3 for the range of voltages and currents you intend to use. – Transistor Jul 21 '16 at 14:52