I just got a voltage regulator and I was playing with it in a simple DC circuit. My Source is 3.25V, the OUT pin on the voltage regulator (L7809CV) shows 2.25. My confusion starts with the 1V drop in the energy. I'm not sure I understand how this works very well. If my voltage regulator is a 9V isn't it suppose to take in a lower voltage and increase it up to 9V ? I know the L7809CV is a linear regulator. Do all linear regulators drop voltage? Would I need a booster to increase the voltage instead, or this very one can be set up to work as a booster? Another question, what happens when the voltage in my source drops to (say) 200mV, would the voltage regulator be able to maintain the same 2.5V output until the battery is almost dead? I am not sure how to make sense of the information presented in the data sheet. Thanks much

  • \$\begingroup\$ Volts is not a unit of energy. \$\endgroup\$ – Olin Lathrop Jan 23 '14 at 12:56

78xx series are linear regulators, the input should be about 2v - 2.5v higher than the output voltage (please read below). For 7809 the input should be about 11v - 11.5v (again read below)

The dropout is given in the datasheet of 78xx, it shows a typical voltage drop of 2v and a max drop of 2.5v (note that this is from a 7805) enter image description here

The specs for 7809A are the following:

enter image description here

And for 7809C the following

enter image description here

As you can see in the red rectangle there is only the typical value of 2v shown (no max voltage drop), but I think we can assume the same 2.5v as for 7805. Furthermore the voltage range (green rectangular) and yellow rectangle for which the datasheet shows the output voltage spec (blue rectangular) uses a range with 10.6 or 11.5v minimum input depending on the "A" or "C" type.
I think we should design circuits based on the worse case scenario so it is better to use the max drop of 2.5v rather than the typical of 2v, so we should feed the regulator with 11.5v minimum just to stay cleat of troubles.

To boost the voltage you'll need a switching regulator used in a boost configuration. The minimum input voltage specs depend on the specific implementations, as long as the circuit works withing proper input range the 200mv drop will not affect the output

  • \$\begingroup\$ The dataheet snippet you show clearly says that 2.5 V headroom is needed, yet you tell the OP it only needs "at least 2 V". That is wrong. \$\endgroup\$ – Olin Lathrop Jan 23 '14 at 12:58
  • \$\begingroup\$ @OlinLathrop The snipper shows a typical value of 2v and a max value of 2.5v (the headers were too far to the top and were cropped originally). You are correct in that we should always use the worse case scenario (2.5v in the shown image), but the image is actually from the 7805 table rather than the 7809 table where only a typical value of 2v is shown. I have updated my reply with additional info and taking into consideration your point. \$\endgroup\$ – alexan_e Jan 23 '14 at 15:12
  • \$\begingroup\$ @alexan_e, thank you. I hope it won't be too long until I will be able to read datasheets. I was looking at this one instead. It's a LTC1871 booster ideal for batteries. What do you think ? \$\endgroup\$ – Nactus Jan 23 '14 at 15:52
  • \$\begingroup\$ wow this actually makes sense (when you present it clearly like that) \$\endgroup\$ – Nactus Jan 23 '14 at 16:00
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    \$\begingroup\$ @Nactus I guess reading the datasheet takes some practice but I'm sure you'll soon learn what to look for. The device that you show is a boost regulator and will suite you (assuming you want output>input) but seems rather expensive, You can probably find cheaper ones in ebay. You may also want to check Recommendations for converting 2~7V to 5V? , maybe you'll find something that suits you in the replies. \$\endgroup\$ – alexan_e Jan 23 '14 at 16:35

You got it another way around, they drop voltages not boost

Linear regulators regulate voltage by chopping off extra voltage as heat


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