# Making a voltage drop while providing high current - how does this work?

I have a USB power source at 5V which needs to power a chip at ideally 4V (4.4V max) - but I need to be able to supply 2A of current for the "spike" current draws for the chip. In general the chip will probably only be drawing 200mA.

A simple voltage divider can get me the 4V, however I am not quite sure how to calculate the resistor values when it could draw anywhere from 10uA to 2A - the current draws could have wildly different resistance values.

I might be overthinking this myself, but does this still work as a simple voltage divider equation? You'd need resistors rated to handle the current, but is that all there is to it?

-- Yes, I've considered other circuit options for this kind of thing, I'm specifically wondering whether this could be done with a simple voltage divider.

• a voltage divider will not work ... during heavy current draw, the chip will be a low resistance device. since it would be placed in parallel with the lower leg of a voltage divider, the supply voltage to the chip would drop – jsotola Feb 16 '18 at 17:42
• You could use a resistor and 3.9 volt zener diode. (and a big capacitor maybe). – Andy aka Feb 16 '18 at 18:03
• @Andyaka - although note that you'll need a pretty big resistor, too (2A * 1.1V drop = 2.2W). – Jules Feb 16 '18 at 18:10
• @Jules no, if the average current were such that replenishment of the capacitor was good (duty cycle considerations) then maybe 300 mA is the current.... 300 mA x 1 volt = 300 mW. – Andy aka Feb 16 '18 at 18:16

A voltage divider will not work.

If you dis want to do this with a voltage divider, then you would need to make sure that there was much more current through the divider than could ever be drawn by the chip. Otherwise the chip will short out the lower half of the voltage divider, and the voltage will drop. But that would mean having e.g. 20A flowing through the divider, and even if USB could do that (it can't) it would make things very hot.

Use a low-dropout linear regulator, or, if accuracy is not important, a diode.

A linear regulator is designed to take a higher voltage and output a lower one. It will adjust it's resistance to make that work. You can buy one with a fixed output voltage of 4V, or you could buy an adjustable one and set the output with a couple of resistors. A basic regulator will cost a few pence, if you want 2A you will need to spend more like £1.

A diode is simpler, and uses a little less space. But it is a little trickier to get right, it needs a minimum current at all times to stop the voltage drifting up. So a linear regulator is probably easier to get right first time.

You will also want a capacitor next to the chip, to help smooth out those spikes. The datasheet will have some suggestions - follow them. If the spikes are rare and short, a suitable capacitor or capacitors will also mean you cna get by with a smaller or cheaper linear regulator.

• He's not going to be able to supply 2A from the USB port. They're only designed for 500mA MAX. There is more to this problem than simply dropping the voltage. He'll need some sort of large cap to supply the 2A spike, along with the voltage regulator. A diode probably wouldn't cut it in this case – DerStrom8 Feb 16 '18 at 17:42
• @DerStrom8, the OP did not say USB port ... the large cap is a good idea though – jsotola Feb 16 '18 at 17:43
• @DerStrom8 I thought 2.1A USB wall warts were pretty common these days. I assumed that is what he was using. – Jack B Feb 16 '18 at 17:47
• Thanks for the discussion. The USB power is from a charger that can provide 2.4A, it's just delivered over a USB cable and port. Still - you make a very good point about power delivery and it's definitely something I'm taking into consideration. – JRB Feb 16 '18 at 18:03
• @DerStrom8 - 2A is quite common for tablet chargers these days. I have a 3A supply that I use for running single-board computers with USB power inputs. Also note that USB3 has increased the limit, so 900mA is available from host USB ports these days. – Jules Feb 16 '18 at 18:07