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I am trying to use a STMicroelectronics L7805CV to increase voltage from 7.5v to about 12v, but I can't get it work. I've placed 470 ohm resistors between out and common and between common and ground, which should give me about 10v, but instead it only gives about 7.5v. It's like the 7805 can't handle any higher, when I put a second 7805 in parallel, the voltage rises to 8v.

I've made the following observations:

  • 100 ohm resistor between out and common, commond connected to ground; The voltage is a nice 5.0v.
  • 470 ohm resistor between out and commond, common connected to ground; the voltage drops to around 3v, this I find very strange. The load is lower so it should be easy to maintain 5v?
  • 100 ohm resistor between out and common, and a 100 ohm resistor between common and ground; the voltage is around 8v.
  • 470 ohm resistor between out and common, and a 470 ohm resistor between common and ground, the voltage is around 7.5v.

My setup is similar to http://www.rason.org/Projects/regulator/regulator.htm

Any ideas on why this isn't working? According to the datasheet, you can use the device this way.

Thanks in advance!

edit: If you're reading this and are trying the same. This is not the way, try this: Step up DC/DC 5V -> 6V

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    \$\begingroup\$ The 7805 is a 5V output linear regulator. It's never going to boost a voltage, and you seem to know that. What are you talking about trying to increase the voltage to 12V? \$\endgroup\$
    – Matt Young
    Commented Jun 22, 2014 at 21:56
  • \$\begingroup\$ Various sources indicate that it is possible to use a fixed regulator to get a higher voltage, including the datasheet. You do this by placing a resistor between out and common, because the voltage is fixed, a fixed current will flow from out to common. When you place a resistor from common to ground, the fixed current will create a voltage across this resistor, increasing the voltage of OUT, with respect to ground. See the link in OP for more explanation, or this topic: electronics.stackexchange.com/a/57234/46076. \$\endgroup\$
    – Jim Daniël Teunis
    Commented Jun 22, 2014 at 22:02
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    \$\begingroup\$ Yes, you can create an adjustable regulator from it, but you can't take a lower voltage as an input and make it higher. \$\endgroup\$
    – Matt Young
    Commented Jun 22, 2014 at 22:05
  • \$\begingroup\$ What is your input voltage? \$\endgroup\$
    – markrages
    Commented Jun 22, 2014 at 22:20
  • \$\begingroup\$ Jim - the answer by Oli and the data sheet are correct, BUT you have read into them information which is not actually present - easy to do when you come across something new. The output can be higher than the regulator's set value BUT in ALL cases a linear regulator works by effectively creating a dynamic resistor which the load current flows through and results in a voltage drop in the process. Vout must ALWAYS be less than Vin - and there will be an internal drop as well - typically around 2V for a 7805 type device. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jun 22, 2014 at 22:47

2 Answers 2

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The link recommends a fairly moronic way to increase the voltage from a 7805, and incorrectly equates the 78xx parts with the LM317.

One of the differences is that the internal currents (a few mA) that the regulator uses goes out the output of an LM317 and out the GND of a 7805. The ADJ terminal of an LM317 typically sources 50uA of current, not 5-8mA, so a resistor divider is reasonable. The LM317, OTOH, has a requirement for a minimum output current to keep the output in regulation. The 7805, since it dumps the internal current out the GND pin works fine down to no load.

If, for some reason, you want to increase the output voltage of a 7805, use a zener diode between the GND terminal and ground, 6.2V will give you 11.2V out.

The dropout voltage of a 7805 is around 2V typical at 1A (somewhat less at lower current, can be worse over temperature), so if you want 12V out, you need to give it 14 or 15VDC in.

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  • \$\begingroup\$ Clear answer, thank you. If I buy a LM317, can I use that to get higher voltages than the input? If not, what is the most common/cheapest way to go for example from 5v to 12v? Efficiency is not important here. \$\endgroup\$
    – Jim Daniël Teunis
    Commented Jun 22, 2014 at 22:44
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    \$\begingroup\$ 1. You cannot. 2. Boost converter. \$\endgroup\$
    – EM Fields
    Commented Jun 23, 2014 at 0:01
  • \$\begingroup\$ I concur with EM. You could also buy a DC-DC converter such as the P7805-Q24-S12-S: cui.com/product/resource/digikeypdf/p7805-s.pdf \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jun 23, 2014 at 0:17
  • \$\begingroup\$ Can you 'unaccept' this answer, as it is clear wrong (7805 / 317 are not boost convertors) \$\endgroup\$
    – RJR
    Commented Jun 23, 2014 at 3:05
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    \$\begingroup\$ Oops, sorry - I missed the last sentence. \$\endgroup\$
    – RJR
    Commented Jun 23, 2014 at 4:00
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A boost regulator is needed for this application. A boost regulator is a switch mode regulator that will boost a lower voltage to a higher voltage.

The amount a current needed is not specified which will have an impact on the specific components needed.

A device such as the On Semi NCP1403 would be useful. The datasheet is

here -> [http://www.onsemi.com/pub_link/Collateral/NCP1403-D.PDF][1]

The datasheet shows some typical boost regulator designs that can be modified for your application.

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