pcb - Basic voltage divider issue in my amateur circuit XBee

Once again I need your help in my project. I have a 4,5v powered circuit. I put a voltage divider with 2 resistors in series to reduce my initial voltage of 4.5 V to 3.4 V (to feed an XBee module).

The issue is that if I do not connect the xbee, the voltage divider reduces the input voltage to the desired voltage(3.4 V), but when I connect the XBee module, the resulting voltage divider voltage drops to a value very low (0.85 V), making the XBee malfunction. Surely it is a basic mistake, I am asking something that is very clear, but I'm newbies.

Basic Circuit:

Vin(4.5 V)-----R2(10 kΩ)---------R1(33 kΩ)--------GROUND
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                 Vout desired (3.4 V)
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                      XBee 

Answer

The basic mistake you are making is that you cannot use a voltage divider to power a circuit.

Let's do the maths.

For the sake of the maths I'll say the XBee needs 100mA to run. I don't know what the actual value is, but that seems a reasonable value to me for an RF transceiver.

When unloaded you have a simple divider with current flowing through two resistors. Let's calculate that current:

$$ R_T = 10,000 + 33,000 = 43,000\Omega $$ The current through that at 4.5V is $$$ I=\frac{V}{R} = \frac{4.5}{43000} = ~105 \mu A $$. Ok, so with that current through 10KΩ it drops $$ V=R \times I = 10000 \times 0.000105 = 1.05V $$ so the output voltage would be \$4.5-1.05=3.45V\$

All well and good.

Now let's add the XBee. That's going to draw 100mA through the output. So we have 105µA plus 100mA, or 100.105mA flowing through the 10KΩ resistor now. $$ V=R \times I = 10000 \times 0.100105 = 1001.05V $$ A thousand volts dropped over the 10KΩ resistor?! But, (you say), That can't be! There's only 4.5V available in the first place!

Quite right. So the entire (or certainly most of) the 4.5V gets dropped across that 10KΩ resistor leaving nothing left for the XBee to run on.

So if you want to draw more current than the voltage divider itself consumes, you're on to a loser right away. A divider like this is only good for small signals where you will be drawing minuscule amounts of current. Ideally the resistance of what you're connecting it to should be no less than 10 times that of the lower (33KΩ) resistor in the divider. Perfect for connecting to input pins on microcontrollers, ADC inputs, op-amp inputs, etc, where the input resistance (impedance) is typically in the hundreds of thousands, or even millions, of ohms. Not so good for powering something.

So what do you do then?

Well, the answer is very simple:

Use a low-dropout linear voltage regulator. They're ten-a-penny and usually just need a couple of capacitors with them. Very simple to use.