Unexplained voltage loss in a simple voltage divider

[lukeex]

hey guys,

I am currently trying to implement a voltage divider to step down from 9V DC for a digital output.

I am using two resistors, 3.3M ohm & 2.7M ohm in series.

When I measure the voltage across both of them using a multimeter I get 9.5V

Across the 3.3M ohm resistor I get 2.2V and across the 2.7M ohm resistor 1.8V.

I don't have any idea where I have lost the 5V and there is no other resistances in series, it is connected straight to a 9V battery, using a breadboard.

If anyone has any suggestions at all it would be much appreciated.

Thanks

 

[vk6kro]

Your multimeter has a resistance of about 1 megohm, and it is upsetting the voltage divider when it is in circuit.

For example, put a resistance of 1.188 Meg across the 3.3 Meg resistor. The parallel combination would be 873 K

Now use this as a voltage divider and you get 873 / (873 + 2700) times 9 or 2.2 volts.

Your multimeter is behaving like that 1.188 Meg resistor.

You can get better multimeters that have a resistance of 10 Megs but they cost a bit more.

 

[sophiecentaur]

Or you could just compensate for the meter resistance and calculate the 'real', unloaded value. That would be a cheaper solution.

 

[Fish4Fun]

lukex,

You are likely to run into other problems using 3.3M & 2.7M voltage divider to power anything! The nominal current through your voltage divider is 1.5uA ( 9V/6M = 1.5uA). Assuming you have the 2.7M resistor connected to V+ and the 3.3M connected to ground, you would have a nominal 4.95V where the two resistors are connected; however, as soon as you put any sort of load in parallel with the 3.3M resistor (that is, attempt to use the 4.95V), the maximum current that can flow is limited by the 2.7M resistor to 3.3uA. There are very few circuits that are going to be useful with a maximum of 3.3uA.

You would likely be far better off to simply use a 7805 voltage regulator (or similar) to obtain your 5V if you need any current at all.

Fish