How do I 12v to 6v (not 5v) using 7805?

Discussion in 'General Electronics Chat' started by Grayham, May 18, 2010.

  1. Grayham

    Grayham Thread Starter Member

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    Hi,
    I have been searching for this for a while but found nothing.

    I have made the circuit from 12v to 5v using the circuit from this datasheet
    http://www.amplebiz.com/spec/L7805.pdf
    On page 18 (figure 19), but without the resistors R1 and R2.

    But I want 6v instead of the stock 5v this regulator outputs.
    I am new to electronics and cannot understand the formula for R1 and R2 to make this 6v.

    Can someone please explain to me this formula:
    VO = Vxx (1+R2/R1)+Id*R2

    What is the 'Id' variable otherwise I might be able to work it out?
  2. rjenkins

    rjenkins AAC Fanatic!

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    Id is the chip's supply current, between positive input and it's ground or common terminal.

    It's in Table 4 of the data sheet.

    Unlike for a regulator that is designed to be variable, the Id current is not fully specified. The datasheet just says it will not more than 6mA, so you may have a bit of trial and error.

    Try a 100 Ohm resistor between the reg common and your circuit 0V. That will need 10mA to drop the one volt you need.

    Then add a 470 Ohm resistor in series with a 470 Ohm preset between the regulator output and common, that will allow you to adjust the reference current between roughly 5 and 10 mA. Together with the actual Id of the chip, you should be able to adjust the preset for 6V out.
  3. beenthere

    beenthere AAC Fanatic!

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    You can use an LM317 variable regulator or get an LM7806 in place of the 7805.
  4. Bill_Marsden

    Bill_Marsden Moderator Staff Member

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    You can also use a similar configuration the LM317 on the 7805 to bump its voltage up. Personally I'd buy a new regulator.

    If you don't understand what I'm talking about, ask.
  5. SgtWookie

    SgtWookie Expert

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    That's close.

    Id can vary due to temp, Vin and Iout.

    If you figure that Id will be around 5mA +/- 5% or so, that works to get a "ballpark" output voltage with just a resistor from the GND terminal to GND, so roughly 200 Ohms per volt increase on the output, +/-10%.

    However, if you use a 1K resistor from Vout to the ground terminal, then you have ID= 5mA +/-10%, plus 5mA +/- the tolerance of the regulator, which is roughly 4.8% up to around 750mA out - so you increase the precision somewhat - because the regulator's output voltage needs a 5mA load before regulation is guaranteed.

    With the 1k resistor from VOUT to the GND terminal, then adding a 100 Ohm resistor to GND will give you roughly 6v out.

    If any degree of precision is required, I'd go with an LM317 instead.
  6. Bill_Marsden

    Bill_Marsden Moderator Staff Member

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    You could also put 1 or 2 diodes on the ground lead of the 7805, which would raise the voltage by .6 to 1.2V.
  7. SgtWookie

    SgtWookie Expert

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    A 1N4148/1N914 will have a Vf of about 0.7v @ 5mA current at room temperature.
    A 1N4002 will have a Vf of about 6.5v @ 5mA current at room temp.

    The trouble is that changes over temp will result in more variation than if a resistor were used; and you only have adjustments in rather coarse steps.
  8. Grayham

    Grayham Thread Starter Member

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    I tried that and it does work Vout is +6v :)

    But I also tried ONLY a 200ohm resistor from common to GND and that produced the same results.

    What is the different between using 2 resistors (like your example) and a single resistor just on IC's common to GND?
  9. SgtWookie

    SgtWookie Expert

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    Assuming 1k from the OUT terminal to the GND terminal, and then 100 Ohms from the GND terminal to actual ground.

    That's a very good question. I'm afraid that I cannot give you a very good answer (exception below), as the data in the datasheets does not indicate that you will receive any better result from using a single resistor from GND to ground, than using two resistors (for example, 1k and 100 Ohm vs a single 200 Ohm resistor).

    The only exception is, that with a 1k resistor to the GND terminal and a 100 Ohm resistor from the GND terminal to ground, is that the minimum 5mA load is satisfied for guaranteed regulation. If your load was less than 5mA, then using the two-resistor method would be beneficial.

    What the datasheet was attempting to imply is that the output voltage regulation was better than the GND terminal current regulation. Were that the case, it would be beneficial. However, the output voltage regulation is no better than the GND terminal current regulation, so it's a moot point.

    Reading and interpreting datasheets can be difficult, and requires a lot of practice. If it's any consolation to you, I make my fair share of mistakes trying to read and interpret them properly.
    Grayham likes this.
  10. Grayham

    Grayham Thread Starter Member

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    StgWookie thanks I will stick with the proper datasheet method of 2 resistors that being said.

    Would you mind explaining (I still can't work it out) how you came up with the 1K and 100ohm resistor values?
    What was your formula? (my 200ohm was just trial and error on Multisim).
  11. SgtWookie

    SgtWookie Expert

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    Sure.
    The datasheet says that the output load must be >= 5mA for guaranteed regulation.

    R = E/I = 5v/5mA = 1000 Ohms from OUT to ADJ

    I used 5mA for the GND terminal current, for a total current of 10mA.

    (6v-5v)/10mA = 100 Ohms.
  12. Grayham

    Grayham Thread Starter Member

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    One more thing, this may be obvious, but wouldn't having a 1k resistor as you mentioned from pin2 (GND) to pin3 (VOUT), while making the vreg. more stable at very low currents, it would also product more wasted heat in my circuit.

    Meaning that the 7805 would be working harder for the extra 5ma plus the resistor would be doing some dissipating too?

    Is it common practice to ignore small amounts of wasted heat overhead like this?
  13. SgtWookie

    SgtWookie Expert

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    25mW power will be dissipated in the 1k resistor.
    The 7805 will dissipate an additional (Vin-Vout)*.005A = power.

    Yes.

    If you will always have at least 5mA load on the output, you won't need the extra 1k resistor.

    If you will have less than 5mA current through your load at any time, then the power is not wasted; it's for a good cause - proper regulation.

    If you're worried about wasted power, then you shouldn't be using linear regulators in the first place, as with a Vin of 12v and Vout of 6v, you are using more than half the power in heating up the regulator instead of powering the load.
  14. Grayham

    Grayham Thread Starter Member

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    Yeah I'm making a kind of generic 6v power supply somewhat so I guess it could be anything I'm pulling form it. I'll keep the 1k resistor there in this case :)


    Any recommendations to other IC's that would be better for creating a 6v power supply out of a 12v source?
  15. SgtWookie

    SgtWookie Expert

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    A DC-DC converter is what you need.

    You can either buy a finished module for around $15-30, or build a switcher using a dedicated IC, or try to "roll your own". Switching supplies use switched inductors to control the supply output. It takes awhile to understand switching supplies.

    This site: http://www.smps.us/
    has lots of tutorials and ideas.
  16. SgtWookie

    SgtWookie Expert

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    I realize that parts are expensive and selection is not great there in Australia.

    You might wish to peruse Roman Black's pages, in particular this two-transistor switching power supply:
    http://www.romanblack.com/smps/a02.htm

    As far as inductors, Ronald Dekker has a great resource page here:
    http://www.dos4ever.com/flyback/flyback.html
    If you have access to an oscilloscope, you can build Ronald's simple "Inductor Test Bench" to verify the performance of inductors that you make from salvaged toroids.

    Junked computer power supplies are a great source for free toroids; it just takes a few moments to snip them out. Your local computer repair shops may have a number of them for the asking.
  17. Norfindel

    Norfindel Active Member

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    You can also use a 7806, you know?
  18. SgtWookie

    SgtWookie Expert

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    A 7806 will be just as inefficient as a 7805 with a 200 Ohm resistor from the GND terminal to ground.

    But, this whole exercise can be worthwhile to people everywhere, as often one can't simply walk into a store and purchase a 78xx regulator in the voltage you want. However, many vendors carry the 7805 and 7812. A 7805 is more generally useful, as you can easily add a single resistor from GND to ground to get just about any Vout >=5v that you require, within the limits of the regulator of course.

    If you call the resistor from GND to ground R1, then:
    R1 = 200*(Desired_output_voltage - Vreg_out)
    So, if you wanted to get 7.5v out of a 7805 regulator, you'd calculate:
    R1 = 200*(7.5-5) = 200*2.5 = 500 Ohms.

    It's not an exact formula, but it will certainly get you "in the ballpark".
    Last edited: May 28, 2010
  19. Norfindel

    Norfindel Active Member

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    Of course, that was just a response to the original post.
  20. Grayham

    Grayham Thread Starter Member

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    Yeah I have looked at this page a while ago, might check it out again.

    This calculation does work well for me :)

    Now one more thing I need to query on for this circuit, here is the proper datasheet circuit for the 7805 for 6v (with 5ma minimum load always):
    [​IMG]



    Where here is the circuit you are referring to in your last post (R3=200*1)=+1v+5v=6v
    [​IMG]



    Now here is my own circuit that uses the same calculation but has constant 5ma current load on the IC to give it above minimum load (no matter what R3 is, U5 stays above 5ma):
    [​IMG]


    What I don't understand is what is the difference between the 1st image and the 3rd image. Isn't the 3rd one better because it ensures the load is above 5ma?
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