#### Llama16

##### 100+ Head-Fier

- Joined
- Mar 22, 2009

- Posts
- 326

- Likes
- 0

I didn't really know where to ask this but I thought that doing it in the members lounge would be a safe thing to do.

I'm reading alot about electronics lately and atm I'm at the voltage divider.

I fully understand the principal, and understand the formula's behind it.

Though I can't seem to get the effect of a load on a voltage divider.

I am reading everything from Welcome to Play-Hookey! very nice site.

The one thing we haven't accounted for as yet is the current drawn by the load. This will necessarily upset the resistance balance, since any load current will flow through R1, but not through R2. As a result, the load will reduce output voltage of the voltage divider by some amount. Appropriately, this effect is called loading.

Quote:

To calculate the effect of loading and its extent in any given instance, we must realize that the voltage divider circuit behaves in exactly the same way as a battery of voltage VOUT with a series resistor whose value is equal to the parallel combination of R1 and R2. The figure to the right shows the equivalent circuit for our example voltage divider. Now, we noted earlier that our example load draws 3.5mA at 5 volts. In accordance with Ohm's Law, this current will drop a voltage of 2.33333 volts across that 667 ohm resistor. Thus, our example voltage divider will not be able to provide +5 volts to this load. |

BTW: I tried alot of explenations, I found this one the best (allthought still don't really understand it, Why parallel??)

Would it mind if I post (after searching and trying to get things on other sites) such questions in here when they're on my mind?