# Drop USB voltage to 4.5v

I have different 4.5v small LED circuits (from 100mA up to 350mA consumption, with some 555 ICs) that I'd like to power using a small regulated USB wall charger (4.95V output, 0.8A).

As per my understanding (I'm new to electronics...) I can calculate the proper resistor for each circuit knowing the current. However I'd like a solution that can sustain 4.5V regardless of the current (within the limits above).

I've been reading about voltage dividers and zener diodes, however I'm not sure about the best approach for this small drop. What would be the the best way?

Update: these are independent small lighting circuits that I've build for small models, powered by 3 1.5v batteries. The problem I'm experiencing now is that the voltage of the batteries drops in time, dependent on the type (rechargeable, etc.) and brand. For example, because 3x1.5v = 4.5v, I've used some LEDs with fV 2.1 in series with a small resistor, to make a more efficient use of the drained current. However in time the batteries V drops to 3.8v, below the minimum level the LEDs need, affecting their brightness (and a new set of batteries is 4.8V!). In hindsight, maybe I should have considered a regulated power supply from the start, instead of trying to compensate for the variable V in the batteries during their lifetime... I might post a new question about this topic, but feel free to comment on this too :)

• What are those 4.5 LED circuits? My guess is that they can run fine on 5V, but I'd like to check. Commented Jul 2, 2013 at 7:23
• If what you have are LEDs with a forward voltage of 4.5V, then this is the wrong question to ask. What you want to regulate for an LED is the current, not the voltage. See How can I efficiently drive an LED? Commented Jul 2, 2013 at 11:25
• You may want a current driver or some transistors. Something like this < 1A led driver Commented Jul 2, 2013 at 11:59
• @Wouder: each circuit has a mix of different LEDs and resistors (parallel, series combinations), one of them also with a flasher made with a 400 mifroF capacitor and a N555. Commented Jul 2, 2013 at 17:41
• @Phil: no, I don't have LEDs with 4.5fV. These are independent small lighting circuits that I build for small models. I'll add more details to the question. Commented Jul 2, 2013 at 17:43

However I'd like a solution that can sustain 4.5V regardless of the current (within the limits above)

If you are happy with a constant voltage of 4.5V then you should consider a low drop out voltage regulator like this: -

It can be setup to deliver up to 4.5V and at 1A to the load and will only need 4.645V (typically) on the input to sustain the output voltage. Note that it needs a minimum load for it work correctly (1mA) but this shouldn't be a problem given your requirements.

This is typical of a series of many voltage regulators with low drop-out voltage - I'm not saying use this one - I am saying be aware of what TI and other folk may be supplying. It's likely you can find one with current limit circuits built-in. You could also apply a bit of current limiting in this device by having a resistor in series with the voltage feed - this should be chosen to develop just enough voltage across it (at the required load current) to sustain the output. Should load current increase, output voltage will drop.

• This answer is good, thanks! I'll will investigate about this (I'm constantly learning...), also checking out what I can find in the TI store (which I've used before). I've improved my question, feel free to check it out and let me know if this answer might not be a good fit based on the new information. Commented Jul 2, 2013 at 18:01
• I've searched TI and DikiKey for a similar component but the through hole variation, but I couldn't find it. Is there any similar? Commented Jul 3, 2013 at 14:18
• @Sebastian There are plenty of low-dropout regulators that are adjustable output - keep looking. Can't assist much today because really busy. I found it on the web by a generic search for "LDO adjustable" Commented Jul 3, 2013 at 14:59
• Indeed I've found a lot but in a different packaging (and the through-hole packages have at least a 1V drop), but I'll keep looking. Thanks again for your time and help. Commented Jul 3, 2013 at 22:36

A silicon diode has roughly a 0.65V drop. Putting it in series with the 4.95V from USB will get you close to 4.5V:

simulate this circuit – Schematic created using CircuitLab

An LDO will also work, and will have better regulation (voltage will vary less with current) and could be closer to 4.5V. But, there is no real such thing as exactly 4.5V and perfect regulation, only progressively better methods. A good engineer is one that chooses a solution that is good enough, and for powering an LED, this is probably good enough. An LDO will be more expensive than a diode, and you will probably have to order it. A diode is dirt cheap and you probably already have one.

• Thanks, however a .65V drop will be too much. I've updated the question to give a better idea of my needs. Basically, I have a bunch of small circuits powered by 3x1.5V batteries (each one with different consuption) and I want to change that using a USB charger, to eliminate the V variance that I see during the batteries lifetime. I don't mind buying a LDO (I buy from DigiKey), however I will need to study more about it. Commented Jul 2, 2013 at 17:54
• @Sebastian it sounds like your attempts to be more "efficient" by reducing the voltage across the resistor in series with your LEDs has also made the current regulation of your circuit very poor. It really would have been better to ballast the LEDs with a current regulator, which would maintain the same brightness over a wide range of input voltages. Commented Jul 2, 2013 at 18:19
• yes... now I can see that (I learn by doing!), however I'd like to avoid redoing these circuits.. can you elaborate on the LDO solution? Commented Jul 2, 2013 at 23:56
• @Sebastian well, Andy Aka's answer is that. Commented Jul 3, 2013 at 0:42

Check the forward voltage at the current (Ic) you want for the LED. This value can be found on the datasheet. Let's name it Vf.

Then you know that the voltage across your LED is Vf when Ic Amps is flowing through your LED.

Your power supply voltage (Vp) is 4.95V. The resistor in series with the LED has a voltage drop. let's name it Vr.

Thus Vf + Vr = Vp

You Know Vf and Vp. You can then compute Vr.

And because you know the value of the voltage across the resistor (Vr) and the current that flow though it (Ic), you can compute the value of the resistor that matches those value together.

Resistors follow the Ohm law: U=R*I in our case Vr = R * Ic.

You know Vr and Ic, you can compute R.

That's it, in an ideal world....

Here the voltage drop across the resistor is very small (~0.5V). And this value is computed from the value of Vf of the LED. But, an LED is a semiconductor with a dependency of the parameters on the temperature as well as tolerances.

Now imagine that you design for the worst case. In order to guaranty that the LED is never above its current limit, you will take the minimum value of Vf for the computation according to the tolerances. let's say that this value is just 0.5V lower that the nominal Vf. When it comes to the computation of Vr, the result will be ~1V. Which means twice the Vr for a nominal computation !

This means that the current through your LED and therefore the emitted light power will highly depends on the tolerance on the LED.

• This isn't answering the question asked. It may be answering a question the OP should have asked - but that has already been answered elsewhere. Commented Jul 2, 2013 at 7:52