reducing 5V to 4v or 4.2v MAX

[john2k]

I have a 12v to 5V 3A convertor. What would be the best way to further reduce it by at least 0.75V? I was thinking of a resistor but that will be generating a lot of heat. Is there a better way to reduce the voltage and have it stable? The current draw will vary. In most cases it probably won't draw full 3A but can sometimes peak at that.

 

[Bluejets]

One or two schottky diode and one silicon diode in series.

 

[(*steve*)]

what are you doing? I hope its not "charging a lipo battery".

A resistor won't work, not because it will dissipate a lot of heat, but because the voltage drop across a resistor depends directly on the current through it. That means you'll only get the range of voltages you require for a small range of currents.

An easy option is a diode or two. These are a bit like resistors in that the voltage drop across them depends on the current through them, but differs in that the relationship is not linear. In the case of diodes, the voltage drop is essentially 0 at zero curry, but rises very rapidly to about 0.6V, and then far more slowly after that (up to 2 volts at very high currents). Series diodes are often a solution, but in your case it may not be.

Another option is a linear regulator. In your case you'd need a low dropout (LDO) regulator. This will also dissipate quite a bit of heat (essentially the same as a resistor) but has the benefit of producing the same output voltage under effectively all loads.

Another option is a switch mode regulator. These have the benefit of *not* departing so much Great. However they are significantly more complex, and in this case the difference in heat dissipation will not be great.

 

[(*steve*)]

Aaagh, so many autocorrect errors. Too hard to edit on my phone! Have fun trying to figure out what I meant :-(

 

[swagguy8]

i'd use a linear regulator, like LM317, that one should fit your needs, it is capable of 1.75v - 30v. i don't know hows the accuracy on that you'll have to check the datasheet.
good luck on your project!

 

[Martaine2005]

voltage drop is essentially 0 at zero curry,

Never a good idea to answer when hungry!

Martin

 

[john2k]

what are you doing? I hope its not "charging a lipo battery".

A resistor won't work, not because it will dissipate a lot of heat, but because the voltage drop across a resistor depends directly on the current through it. That means you'll only get the range of voltages you require for a small range of currents.

An easy option is a diode or two. These are a bit like resistors in that the voltage drop across them depends on the current through them, but differs in that the relationship is not linear. In the case of diodes, the voltage drop is essentially 0 at zero curry, but rises very rapidly to about 0.6V, and then far more slowly after that (up to 2 volts at very high currents). Series diodes are often a solution, but in your case it may not be.

Another option is a linear regulator. In your case you'd need a low dropout (LDO) regulator. This will also dissipate quite a bit of heat (essentially the same as a resistor) but has the benefit of producing the same output voltage under effectively all loads.

Another option is a switch mode regulator. These have the benefit of *not* departing so much Great. However they are significantly more complex, and in this case the difference in heat dissipation will not be great.

Thanks for that very informative post. I'm basically running a Android tablet directly from a car power source and getting rid of the battery of the tablet. Currently it works fine with the 5V but I don't want to risk frying it. Most people have said that the tablet can support upto 5.25V and the range is between 3.85v and 5:25v, but I would rather be somewhere in the safe range around 4:2ish

 

[john2k]

I have lots of 1N5822 Schottky Diodes in my toolbox, so might try using one. Plus I plan to put a diode on the 5V output feed anyway. Does anyone know what the typical voltage drop will be for the 1N5822 on lets say very low current draw to something like a 2A current draw?

If on high current draw the voltage drop is as high as almost 1V then having two diodes in series might put me at risk of having too much voltage drop and hence the device will not get enough power.

 

[dorke]

John,
1.If possible ,the best way is to set the voltage of the "12v to 5V 3A converter" internaly.
Many of them have that capability with a dedicated trimmer on them.
2.The voltage drop of a 1N5822 for 10mA-3A would be 0.25-0.525V.
If you take that approach use 2 diodes in series.

 

[john2k]

John,
1.If possible ,the best way is to set the voltage of the "12v to 5V 3A converter" internaly.
Many of them have that capability with a dedicated trimmer on them.
2.The voltage drop of a 1N5822 for 10mA-3A would be 0.25-0.525V.
If you take that approach use 2 diodes in series.

I don't think it will be possible to set the voltage on the converter I have as it is enclosed. Below is a pic of the converter I have. I think I will wire 2 1N5822 diodes in series and see how that goes. If the max voltage drop for each diode is around 0.525v then 2 in series will have a max drop of just over 1v, which is not bad as it will bring the voltage to the desired 4V. If however, the tablet is drawing less current during less intensive use and the lowest draw is 0.25, then 2 1N5822 diodes will bring that down to 0.5v and hence I should still get 4.5v which is much better than having 5v.

 

[duke37]

A single 3A silicon diode would probably do what you want.

1N5408

 

[john2k]

A single 3A silicon diode would probably do what you want. 1N5408

Thanks for that info. I take it this has a higher forward voltage drop? whats the typical voltage drop from a very law current draw to something like 3A current draw? Also, do these generate heat? I know the 1N5822 don't generate too much heat. In fact the 1N5822 seems to generate very little heat.

 

[duke37]

The 1N5408 is a common standard diode. You can probably find voltage drop on the internet which may be much faster than mine.

The heat generated will be the same for the same voltage drop. A drop of 1V at 3A will generate 3W whether it is with a pair of Schottky diodes, LDO regulator, a resistor or a single silicon diode.

 

[dorke]

Heat generation is power dissipation related.
Using 2 *1N5822 will make each diode less hot than a single 1N5408.

1N5408- @3A the max voltage drop is 1V (that is 3W on a single diode).
1N5822 -@3A the max voltage drop is 0.525V (that is 1.6W on a single diode).

I think the 2 1N5822 diodes way is better,
and since you already have them ,use them.