# How to reduce 5 volts to 4 volts 1.5Amp with minimum components used?

I am trying to reduce 5 volts high current source to 4 volts with at least 1.5 amperes. I searched the internet, but I could not find 3 pin voltage regulator, 5V(in) 4V(out)voltage regulator. I think voltage divider is an option, but it consumes lots of power, LMZ12002 is an option but it requires at least 9 components.

Is there other options?

• There are simple programmable voltage regulators such as the LM317 which you can set the output voltage with just two resistors making three components the minimum number though you probably want an input and output capacitor too making it 5 components. The simple voltage divider is not an option as the output voltage will be load dependant. – Warren Hill Nov 30 '14 at 13:15
• How accurate must the 4V be? How accurate is the 5V? – Brian Drummond Nov 30 '14 at 13:27
• ... but check the LM317 dropout voltage. It's not an LDO, so pretty sure it won't work. – Brian Drummond Nov 30 '14 at 13:28
• LM317 is not an option, because it requires Vin-Vout at least 5volts. I want maximum 0.4 volt dropout when 1.5 Amp is consumed. – electro103 Nov 30 '14 at 13:33
• The LM317 dropout is 1.25V, not 5 Volt. But that still disqualifies is. – Wouter van Ooijen Nov 30 '14 at 14:32

There are definitely 3 terminal regulators with low dropout that suit this requirement.
See eg www.digikey.com and enter search terms.
As an example only The Seiko S816 works with an external transistor and not including decoupling capacitors requires TWO components total. Dropout is limited by the external transistor's characteristics. While a bipolar would usually be used, this would drive a suitable MOSFET and dropout voltages as sensibly low as desired would be possible.

Using Digikey's search also found

MIC29302 in stock $2.86/1 3A, 250 mV dropout typical at 1.5A. Agh/whoops - I now see I've arrived at the same device as Spehro :-). Search Digikey using their excellent parameter driven search. This and more are there. Also Semtech SC1592 in stock$1.82/1.
260 mV dropout at 3A BUT uses a special dual input supply mode - power conversion is low dropout but Vspply_control needs to be Vout + 1.5V. May or may not be useful.

Consider connecting a diode in series with the load.

edit: Depending on the application, this might be enough, but the forward voltage drop of a diode (or two) does vary with the load current and the temperature of the diode(s). Also, some power (1 W per A) is wasted. Depending on the duty cycle, heat sinking for the diode(s) may need to be considered.

If you, for example, want to drop the voltage supplied to a DC motor to make it run a little slower, a diode in series is good.

edit by Russell McMahon: A 2A rated silicon diode or 2 x 1A diodes in series will typically have 0.6 to 1.0 Volts drop at near rated current. Actual value varies with type - see data sheet. Voltage drop increases with current. Whether this is precise enough depends on application (the O.P. needs to give more information about the application).

• Does the diode allow 1.5 A current pass? – electro103 Nov 30 '14 at 13:36
• A diode certainly wins for least parts count. @electro103, yeah there are power diodes. – George Herold Nov 30 '14 at 16:13
• Diode forward voltages change more than 0.4Volts at 1.5A, so I select Russell McMahon's answer. – electro103 Dec 1 '14 at 5:15

An adjustable (because 4.0V is not a commonly used voltage) LDO regulator is the easiest route to take. For example, the Micrel 29150/29300 regulators can handle 1.5A or 3A with a dropout of well under 1V guaranteed.

You need to read the datasheet, calculate two resistors to get the required voltage and to follow the datasheet recommendations for the capacitors (especially the output capacitor) to the letter. For example in the above-linked datasheet where it says low ESR capacitors 'may contribute to instability' that means that under no circumstances should you use a large ceramic capacitor as the output cap without a series resistor. There are other options for the chip, this is just one possibility. Whatever chip will be dissipating about 1.5W at 1.5A so a small heat sink is called for. If you're actually using it at 1.5A, the 3A-rated part would be a good idea.

Voltage dividers will work well for low currents and well defined constant loads. 1.5A is too high of a current for good regulation and reasonable heat dissipation with a voltage divider.

The choice between linear regulator and switching regulator depends on the application. Generally linear regulators are less noisy, less efficient, and dissipate more heat. Switching regulators are more efficient and cooler at higher load current. Switching regulators work well when they have a high differential (input to output) voltage compared to linear regulators.

The LMZ12002 module is a good part but requires selection of many external components. Also, looking at the datasheet, it is difficult to tell if it will work with a low (1V) input to output differential voltage. The graphs show the lowest differential voltage of 2.2V. The minimum input voltage would be 6.2V not 5V for 4V output.

The NQR002A0X4Z is another prebuilt module with a lower differential voltage requirement of about 0.7V. The input voltage must be at least 4.7V for it to regulate to 4V at 2A. And it is a little cheaper and it still has pins. There lots of surface mount options.

NQR002A0X4Z at Digikey

http://apps.geindustrial.com/publibrary/checkout/NQR002A0X?TNR=Data%20Sheets|NQR002A0X|generic

Minimal parts count is a single resistor, RTrim. However, it is best to add a 10uF or 22uF low ESR cap (ceramic) to the input and the output. Connect them close to the pins.

RTrim calculation: 12/(4-0.6) = 3.52941k closest value is 3.57K, 1%

The C-tune and R-tune can be left out for general purpose use. They are for speed.

Leave the enable pin open and the part will turn on. Or connect it to Vin through a 10k resistor to be sure.

• In this application, the linear regulator is 80% efficient. Pout/Pin = (4*1.5) / (5*1.5) = 0.8. So probably not worth the hassle and expense of a switcher. – markrages Dec 1 '14 at 8:47

Most linear (Low Drop Out - LDO) voltage regulators need about 2V difference between input and output. Given the fairly high current limit, I'd suggest a Switch Mode voltage regulator.
For a drop in voltage, you are looking at "buck" regulators. Analog.com have a wide range of regulators: given your specifications on voltage in & out, current limit, the lowest part count I found was the ADP2106, with 9 components and PCB area of 45 mm^2

I am sure other manufacturers have similar product, possibly even better. This was the quickest example I could find.

• By definition, if it needs 2V, then it isn't LDO. All LDOs are linear, but not all linears are LDO. – Dave Tweed Nov 30 '14 at 15:19
• most engineers consider an LDO as 1V or under drop out. Marketing material tho, I've seen 2V listed for high amperage drop outs, and switching regulators marketed as LDO as well – Passerby Nov 30 '14 at 17:28

I think you can use diodes for this purpose. Apparently, a silicon signal diode has a Forward voltage of 0.7V and the diode drops as much as voltage across it. So a single diode can drop 0.7V, and there by multiples of 0.7V for every diode connected in series. For example, 3 diodes connected in series can drop the voltage to 0.7 x 3 = 2.1V. Only thing to keep in mind is the wattage of the diodes according to the current applied.

A 2W diode might do the job.

Hope this makes sense. (Correct me if I am wrong)