Deciding factor with potential dividers

How much current does your load need to draw at 2V?

fred

You just can not say you want 2 volts out. You must specify the load resistance. If the 2 volt load is very high in resistance you usually will use a high value of resistance. If the load is low in resistance , you will have to use low values of resistance. It will also depend on some extent of the source to supply the needed current. If for example you want to get

2.5 volts from a simple two resistor devider into a high resistance load and you use a pair of 5 ohm reisitors in series across the load , the source must be able to supply 1/2 of an amp. This is usually more wasted power than you want. If you keep the resistors so they are more than 100 times the load resistance the error is usually small enough not to be noticed. As with most electronic circuits , there is a trade off in the components used with the power and price of them.

Looking to pull 250mA Max

2Vdc

To reduce a 5 volt source to a 250 ma load you should not be looking at resistors but another way. Maybe 6 silicon 1 amp diodes in series with the load. The reisitors will not hold the voltage constant under a changing load.

You won't really be able to use a resistor divider to provide 250mA, that is *way* too much current. You are better off using say a LM317 voltage regulator to drop the 5V input to give you 2V output. Then it will be 2V output regardless of the load you put on. A resistor divider voltage output will change with the current your load draws.

Dave :)

If by "equal rated" you mean equal valued, as in ohms, no. That would provide 2.5V, not 2V.

I'm assuming you imagine this:

: +5V : ,----------, : | | : | | : | \\ : | / R1 : | \\ ??? : | / : | | : --- | +2V : - V1 +--------, : --- 5V | | : - | | : | \\ \\ : | / R2 / Rload : | \\ ??? \\ 2V @ 250mA : | / / : | | | : | | | : '----------+--------'

In this case, your Rload resistance is as little as 2V/.25 or 8 ohms. But let's assume that it can draw a lot less current (for example, 1mA would suggest 2000 ohms, 1uA would suggest 2M ohm, etc.) So one of the first key questions to ask yourself is:

What is the range of voltage I want to accept at the +2V node?

I think you can see that if the answer is 2.0000 to 2.0000, it's going to be practically impossible to achieve. No tolerance of a voltage variation is impossible, if the load will itself vary it's loading. You have to have some finite, non-zero range you accept.

So let's say it is from 2.2V to 1.9V, with 2.2V being the voltage when Rload isn't even connected (no loading) and 1.9V being the case when Rload requires 250mA. First off, with no load, you just have R1 and R2 and they must develop 2.2V: V1*R2/(R1+R2) = 2.2V, V1 = 5V.

With Rload drawing current of 250mA, it's easier to imagine that the circuit has been transformed into this Thevenin equivalent:

: +2.2V : ,----------, : | | : | | : | \\ : | / Rth : | \\ ??? : | / : | | : --- | : - Vth +-- +1.9V : --- | : - | : | \\ : | / Rload : | \\ 1.9V @ 250mA : | / : | | : | | : '----------'

I'm not sure if you have already read about Thevenin, but the basic idea is that the more complex combination of V1, R1 and R2 can be replaced by a new, simpler Vth and Rth, which is what your Rload "sees" looking back into the V1, R1, and R2 circuit. If you haven't, this all may seem a bit strange and I'd recommend doing some reading and thinking on the idea on your own. It's not hard to follow, but I don't want to side-track the discussion right now with it. So just accept the above equivalent.

Now Rth is just R1 in parallel to R2, so we have Rth = R1*R2/(R1+R2). The difference in the 1.9V node and the 2.2V value for Vth is 0.3V. And we already know what current Rload is drawing in this case, 250mA. So that means Rth must be .3V/.25A or 1.2 ohms total. So we then know that: R1*R2/(R1+R2) = 1.2 ohms.

This gives us:

(1) 5*R2/(R1+R2) = 2.2 --or-- R2/(R1+R2) = 2.2/5

--and also that-- (2) R1*R2/(R1+R2) = 1.2

So, substituting (1) into (2), we get:

(2) R1*2.2/5 = 1.2

Which works out as R1 = 1.2 * 5 / 2.2, or 30/11 or about 2.7 ohms.

R2 is then simply R2/(30/11+R2) = 2.2/5, or R1*(2.2/5)/(1-2.2/5), which works out to 15/7 or about 2.1 ohms.

This would mean that your divider would be normally consuming about 1A of current. In other words, about 5 watts of power would be dissipated in R1 and R2, without ever hooking up your load. With the load in place, more than 5.5 watts total.

This is usually impractical, which is why other suggestions were offered.

Jon

To put it another way, and more briefly:

Suppose you want to regulate the voltage to within, say, 10% of the 2V. That means choosing divider resistors that will draw TEN TIMES the max load current: 10 x 0.25A or 2.5A.

This means using a power supply that is considerably costlier than what you'd need by using a different method.

Best to use a voltage regulator, so that the power source need only supply slighly more current than the load's 250 mA. AND the regulation will be way better than 10%!.

Regards,

Mark

Ian Tedridge wrote:

Suppose you transform this, using an NPN transistor:

: +5V : ,----------+-------, : | | | : | | | almost 250mA : | \\ | | : | / R1 | | : | \\ ??? | v : | / | : | | | : --- | |/ Q1 : - V1 +-----| NPN : --- 5V | |>-, : - | |