# Designing 4.3 V voltage regulator

In my design (for a hand-held device) I need a regulated power supply of 4.3 V. I am having 5 v voltage regulators. How do I convert this 5 v regulated supply to 4.3 v. I understand I can use potential divider circuit, but it will not be very accurate. Is there any better way to achieve this?

• Use an adjustable voltage regulator. Commented Sep 17, 2013 at 9:22
• Just out of curiosity, why do you need 4.3 V? Commented Sep 17, 2013 at 9:35
• How much current do you need? Is there a higher (unregulated) voltage available?
– Tut
Commented Sep 17, 2013 at 10:49

Maybe a Si diode (forward voltage ca. 0.6-0.7V) is a good enough regulator. You can use it to reduce 5V to ca. 4.3V.

simulate this circuit – Schematic created using CircuitLab

You can not use an ordinary voltage regulator (e.q. LM317) to regulate from 5V to 4.3V because the voltage difference between input and output voltage is to low. If you want to use an voltage regulator IC it must be a Low Drop one.

If you need output 4.3v with current up to 1.5 A, use LDO such as LD29150PTR

$V_I = 5V$

$V_{REF} = 1.23V$

$V_O = 4.3V$

eg,

$R1 = 100k\Omega$

$R2 = ?$

$V_O = V_{REF} \cdot (1+\dfrac{R1}{R2})$

$4.3V = 1.23V \cdot (1+\dfrac{100k\Omega}{R2})$

$\dfrac{4.3V}{1.23V} = 1 + \dfrac{100k \Omega}{R2}$

$3.49 - 1 = \dfrac {100k \Omega}{R2}$

$2.49 \cdot R2 = 100k\Omega$

$R2 = \dfrac{100k \Omega}{2.49}$

$R2 = 40k \Omega$

The LM317 Hemal suggests is no good. Just like most common three-leg regulators (78xx) they need a few volts input-output difference, so you'd need something like 6.5V in.

The solution is an LDO regulator, for Low Drop-Out. They're used the same way as the 78xx, i.e. there's input-ground-output, but they only need a few hundreds of millivolts between input and output.

If you only need low current you could also use a diode for the 0.7V drop, like a 1N4001 would give you, but that voltage will increase if you have higher current, so your output voltage may decrease to as low as 4.0V.

Sounds like you're trying to charge a lithium ion battery, or trying to run a circuit directly from a lithium ion source. The answer from Curd won't cut it (excuse the terrible attempt at a pun) - at no-load there will be some current trickling through there and slowly charging your circuit up to 5V which will definitely damage such a cell.

The easiest way to go is just to use an adjustable version of your voltage regulator, for instance a 1117-adj. This part just requires a resistive divider on the output to get any output voltage. The only catch is: your input voltage needs to be at least ~1.5V higher than your output voltage.

However, there are also specialized chips for lithium ion charging, for instance the ubiquitous MCP7383x.

Basic concept.

simulate this circuit – Schematic created using CircuitLab

The resistor is there to give the 7805 (actually, a 78L05) a nice, stable load. That creates a stiff bias point for the transistor, at Vb = 5V. The emitter will be at Vb-Vbe, and will be as stiff as the bias point (which, considering that the 7805 can deliver LOTS of current at 5V, will be pretty stiff). For the 2N3904, Vbe is about 0.6V, giving Ve = 4.4V.