Jun 5, 2009 at 3:52 PM Thread Starter

#### Llama16

Hey all

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?

Jun 5, 2009 at 5:08 PM
Quote:

 Originally Posted by Llama16 /img/forum/go_quote.gif Hey all 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. 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?

Probably a better question for the DIY forum.

Anyway, to visualize the effect of the load on the voltage divider, draw the load (Rload) as a resistor to ground at the node between R1 and R2. It's now apparent that R2 and Rload are in parallel. This changes the divider ratio, and hence the voltage delivered at the divider output.

Neat trick: If you need to supply an appreciable amount of current (mA) to the load, you can use an opamp buffer at the R1/R2 junction. The opamp supplies the current, and since the input of an opamp is very high impedance, it doesn't load the divider.

Jun 5, 2009 at 7:23 PM