# 3V power rail from 5V

I've mistakenly ordered some 3V PIC18 L F14K22 instead of the 5V 18F14K22. 3V output is fine for the designed circuit but I need to make a 3V power rail from the available 5V rail.

Since only a couple of MCUs need the 3V power supply I've thought on using a simple resistor divider. Reading the wikipedia article on voltage dividers says any resistor matching the ratio will give the desired voltage but I'm pretty sure current matters when deciding resistor values.

Which resistor values would be adequate for powering a couple of 3V microcontrollers?

• 78L33s are very cheap from the right sources. – Ignacio Vazquez-Abrams Sep 28 '13 at 6:30
• I need to make it work, I don't have a 78L33S at hand and shipping is painfully slow where I live. The divider will be replaced in a future with a regulator. – NeonMan Sep 28 '13 at 6:36
• Do you have a power NPN and a diode available? – Ignacio Vazquez-Abrams Sep 28 '13 at 6:40
• NeonMan: anything is better than a resistive divider. Try a divider whose bottom half is diodes adding up to about 3V. – Kaz Sep 28 '13 at 6:41
• Can you point me to a reference design? I do have some diodes (not zeners IIRC) and NPN transistors. – NeonMan Sep 28 '13 at 6:47

First of all: It is very bad practice to use a resistive divider for this. A transistor buffer would improve the design a lot.

With the calculations below I try to show why.

A resistor divider will only work if the current through the divider is much larger than the current through the load. If the current through the load is same order of magnitude then the voltage will change significantly.

Say that your MCU loads the resistive divider between 0mA and 50mA (you have to verify these numbers) and you allow your voltage to vary between 2.8V and 3.3V. This means that at 50mA your divider should be at 2.8V and at minimum load 3.3V. This means that your MCU's power supply varies with load and it varies half a volt.

Say that your PSU is at 5.0V stable across the whole range of currents it delivers.

The top resistor would be $R_1 = \dfrac{500}{33} \approx 15 \Omega$ and the lower resistor would be $R_2 = \dfrac{500}{17} \approx 29\Omega$. Mind you that you need resistors rated for at least 500mW already for this setup.

The new power rail would easily vary from 2.8 - 3.3V, depending on load and you'll need some massive decoupling cap.

If you want the variation of the MCU's power rail lower, the current through the divider needs to be increased a lot (9 and 18 ohm rated 1W for a 3 to 3.3V variation). simulate this circuit – Schematic created using CircuitLab

(kudos to mathomatic for below calculations)

1-> R2/Umax = (R1+R2)/Ubatt              # unloaded resistor divider

R2    (R1 + R2)
#1: ---- = ---------
Umax     Ubatt

1-> Umin/R2 + Imax = (Ubatt-Umin)/R1     # when loaded, apply Kirchhoff's Current Law

Umin          (Ubatt - Umin)
#2: ---- + Imax = --------------
R2                 R1

2-> eliminate R2
Solving equation #1 for R2 and substituting into the current equation...

Umin*(Ubatt - Umax)          (Ubatt - Umin)
#2: ------------------- + Imax = --------------
(R1*Umax)                     R1

2-> solve R1
Solve successful:

Ubatt*(Umax - Umin)
#2: R1 = -------------------
(Imax*Umax)


$$R_1 = \dfrac{5(3.3-2.8)}{0.05*3.3} \approx 15 \Omega$$

2-> eliminate R1
Solving equation #1 for R1 and substituting into the current equation...

R2*(Ubatt - Umax)   Ubatt*(Umax - Umin)
#2: ----------------- = -------------------
Umax              (Imax*Umax)

2-> solve R2
Solve successful:

Ubatt*(Umax - Umin)
#2: R2 = ---------------------
(Imax*(Ubatt - Umax))


$$R_2 = \dfrac{5(3.3-2.8)}{0.05*(5-3.3)} \approx 29 \Omega$$

The traditional Zener-regulated pass transistor power supply is nothing more than an emitter follower with a Zener diode replacing the lower resistor in the base bias network. The traditional variable regulated power supply replaces the Zener diode with a variable resistor. The load goes in the emitter leg. Tie the collector high.

Taking a look at the datasheet, it appears that the worst-case current draw is some 15 mA per microcontroller, so a plastic 2N2222A or 2N3904 should be fully capable of running several of them simultaneously.

Assuming you need to supply 100 mA to your microcontrollers, those transistors will have a beta of about 100 (certainly well over 40), so design your base voltage divider to draw about 10 mA from 5V. That will give you a stiff-enough base bias voltage.

The datasheet says the PIC18LF14K22 needs Vdd in the range 2.7-3.6V. Standard red LEDs drop 1.7V, yellow LEDs drop 2.2V. Assuming Vbe = 0.6V, a red and yellow LED in series, replacing the "bottom resistor", gives Vb = 3.9V. 3.9 - 0.6 gives Ve = 3.3 V, and Bob's your uncle. Use about 110 ohms for the top resistor (the value isn't THAT critical: I'd use 100 ohms).

Breadboard it and play with it a little.