0
\$\begingroup\$

I need to convert from 4 to 3.3V. So I was thinking of a diode, or voltage regulator. Voltage regulators, I know that it uses about 5mA to do the job.

How much current a diode uses ? I cant find this information...

[sorry I deleted the former question, was wrong 3.3 not 4.4]

\$\endgroup\$
11
  • 2
    \$\begingroup\$ Could you explain more about what you are doing? What provides the 4 V? What uses the 3.3 V? Why are you concerned about 5 mA? What do you want to achieve here? I can't tell from this and I didn't read your prior, deleted question. \$\endgroup\$
    – jonk
    Oct 19, 2018 at 23:25
  • \$\begingroup\$ I'm asking how much current a diode absorb, so I can opt for the diode or a regulator. \$\endgroup\$
    – Jackt
    Oct 19, 2018 at 23:48
  • 1
    \$\begingroup\$ if you're simply putting the diode in series (forward-biased) and using the 0.6v or 0.7v drop of the diode to bring you down to 3.3v, then all of the current going to your load (is that about 5 mA ?) is going through the diode. with voltage drop and series current, you can easily calculate the power that the diode will dissipate. \$\endgroup\$
    – robert bristow-johnson
    Oct 20, 2018 at 0:19
  • 2
    \$\begingroup\$ I may be misunderstanding your question, but at this time I currently believe that your mental model of the situation leading to your posed dichotomy still needs some work. I don't think I can offer much help, for now. I wish I could. I just can't see how, for now. \$\endgroup\$
    – jonk
    Oct 20, 2018 at 0:20
  • 1
    \$\begingroup\$ you do not have a clear understanding of voltage, current and power .... you are totally on the wrong track with the idea of current dissipation ..... the diode does not dissipate current, it dissipates a small amount of power in the form of heat .... the "zero" that you keep talking about is vapor \$\endgroup\$
    – jsotola
    Oct 20, 2018 at 2:32

3 Answers 3

Reset to default
1
\$\begingroup\$

I need to convert from 4 to 3.3V. So I was thinking of a diode, ...

Presumably you are considering using the forward voltage drop, Vf of a silicon diode placed in series with the load to drop about 0.7 V to bring your voltage down.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Using a diode to reduce supply voltage.

This works but you need to be aware of a few things:

  • It is not a regulator. It just drops voltage and the voltage drop depends on the current. At more than a few mA it typically drops about 0.7 V but at lower currents the voltage drop will decrease. The I-V curve for the diode will help you calculate this.
  • If V1 increases or decreases the the load voltage will increase or decrease with it. Again, it's not a regulator.
  • The fact that you're looking for 3.3 V suggests that you want to power a micro-controller. You need a regulator.

... or voltage regulator. Voltage regulators, I know that it uses about 5 mA to do the job.

  • Regulators do require some current to operate.
  • Regulators also require some "headroom" - the difference between the input and output voltage - to operate. If you are thinking of a 78xx series regulator then these need about 2 V or so. You don't have that much headroom so you would be looking for a "low drop-out" (LDO) regulator.

How much current a diode uses?

A diode doesn't "use" current. It passes current but reduces the voltage. The power dissipated in the diode can be calculated from \$ P = VI = 0.7 \times I \$.

\$\endgroup\$
6
  • \$\begingroup\$ Thanks for the usefull information ! So zero ! Thanks at you and robert. \$\endgroup\$
    – Jackt
    Oct 20, 2018 at 1:35
  • \$\begingroup\$ A question. Supose that i'm using a battery, will the batt discharge sooner (cos of this voltage/power loss) with the diode than without ? \$\endgroup\$
    – Jackt
    Oct 20, 2018 at 2:36
  • 2
    \$\begingroup\$ @Jackt, how can information be useful when it appear that you do not understand it .... if you did, then you would not have said so zero. \$\endgroup\$
    – jsotola
    Oct 20, 2018 at 2:37
  • \$\begingroup\$ Enlight me jsotola then \$\endgroup\$
    – Jackt
    Oct 20, 2018 at 2:39
  • \$\begingroup\$ @jsotola: I think the OP means that this circuit doesn't require any additional current, "so zero". \$\endgroup\$
    – Transistor
    Oct 20, 2018 at 8:44
1
\$\begingroup\$

The diode may be a good choice if your load current and input voltage are fairly stable and known.

Keep in mind that the nominal 0.7V or so drop from the diode will be less at very low current and you'll pretty much get the full input voltage with a very light load. If whatever your load is goes into a sleep mode and the current drops to about zero then you'll see that voltage increase to maybe 3.9V. If that is within specs, then you are okay, but then maybe you don't need the diode at all.

There are various regulators. The LP5912-3.3DRVT is one. It can output 500mA with only 0.25V drop (typically less than 50mV at 200mA) and typically draws only about 30uA for its own purposes. It can handle up to 5.5V input.

If you choose to use an LDO regulator, be sure to carefully read the section on capacitors and follow it to the letter. In this case, 1uF ceramic capacitors are suitable on both input and output.

\$\endgroup\$
0
0
\$\begingroup\$

Jackt, the answer you are looking for is you can do it either way, the useable life of the battery will be comparable, 99% deteremined by the "load" you have on it. One small difference you might want to consider, if the battery voltage drops below 4 V, then the same voltage drop will appear after the diode. If you need a consistent 3.3 V, a regulator would be the better option.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.