9V to 3V DC converter

Discussion in 'General Electronics Chat' started by hazim, Feb 10, 2010.

  1. hazim

    hazim Thread Starter Active Member

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    Hi all

    I want to build a small dc-dc converter to convert 9v from a 9v battery to a circuit that requires 3v with current about 15mA only. A simple one would be like the attached circuit. I built a circuit with same connections as in the attached circuit using BC109 instead of 2N3055, R=100Ω and a 3V zener diode.

    The resistor is going hot (not very hot), the zener is in connected in reverse, it's connected right.. The output voltage is dropping to about 2.3V and the circuit isn't working good using this conversion method as I see; even if the voltage is 2V the circuit should work normally.

    The converter using 2n3055..
    [​IMG]

    The circuit
    [​IMG]

    I don't want to use a voltage regulator IC such as LM317 because I want only about 15mA 3V for the circuit and the size is small...

    Any ideas? it's not a must to use the converter as that I mentioned above, like the first circuit

    Hazim

    Attached Files:

  2. hazim

    hazim Thread Starter Active Member

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    I tried the circuit with 2x1.5V batteries. It gives very high noise on some batteries and it works great on others!. also it works great using a variable power supply I have built previously..
    Last edited: Feb 10, 2010
  3. mik3

    mik3 Senior Member

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    The output voltage is about 2.3V because of about 0.7V voltage drop across the base-emitter of the transistor. You need to use a 3.7V zener diode to take a 3V output.

    What do you mean when you say it should work even with 2V?

    Using a voltage regulator IC requires less space rather than building a regulator with discrete components. IC regulators for 100mA outputs are just 1/4 the size of the LM317.
  4. hazim

    hazim Thread Starter Active Member

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    Yes that's right. How come I missed out such thing!

    The circuit works fine even with 1.5V, I tried it with my variable power supply.

    Actually I didn't know about that.. Anyway know I'm very far from any electronic parts store...

    The problem still exist here, can't I make a simple converter (9V to 3V ~20mA) using only resistor, zener and transistor??
  5. thyristor

    thyristor Member

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    The following LM317 circuit will do what you require and the LM317 itself is in a tiny TO-92 package.

    If you make R2 = 390Ω and R1 = 270Ω, the output will be 3v.
    ie: [1.25 x (1 + R2/R1)] volts

    You can ignore the "Iadj" term in the equation shown on the attachment as it will only make a few millivolts difference to the output.

    Attached Files:

  6. hazim

    hazim Thread Starter Active Member

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    thyristor:) I have LM317 voltage regulators here 30cm far from me. As I said before, I prefer to do it using a resistor and zener and a transistor if required... I can do it simply using these parts but there is a problem with the circuit when taking voltage from the converter..
  7. SgtWookie

    SgtWookie Expert

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    The Zener diode needs current flow to regulate properly. If the current is too low, it's voltage will drop. You will need to look at a datasheet for your particular Zener diode to find out what current they used to determine the Zener's breakdown voltage. Many typical Zeners require at least 10mA current to properly regulate.

    Then, subtract the Zener voltage from the input voltage, and divide the result by the current required for the Zener to regulate to determine the resistor needed.

    For example, if the 3.6v Zeners' breakdown voltage is rated at 15mA, and you are using 12v for a supply, then:
    R = (12v-3.6v)/15mA
    R = 8.4 / 0.015
    R = 560 Ohms
    Don't forget to calculate the wattage requirement for the resistor.
    Since P=EI, and E=8.4 and I=15mA, actual power requirement is 126mW, double for reliability = 252mW. You could use a 1/4W resistor in this case.
  8. Audioguru

    Audioguru New Member

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    Why not use two inexpensive AA cells that will last a lot longer than an expensive 9V battery??
  9. mik3

    mik3 Senior Member

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    What exactly do you mean?

    You put 1.5V input voltage and you get 2.3V output?
  10. Audioguru

    Audioguru New Member

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    No.
    He powered the lousy circuit with 1.5V instead of 3V and he thinks it works fine.

    The horrible circuit is a nightmare. The output transistors are biased the same way that we were taught never to use. If their gain is high then they are saturated. If their gain is low then they are cutoff.
    It is bad to feed DC to headphones.
  11. hazim

    hazim Thread Starter Active Member

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    SgtWookie, your calculated resistance is the maximum value of R that can be used, and so no output can be taken from between the resistor and the zener. For example, I need an output of 15mA at 3V to feed the circuit and 15mA for the zener. Using a 9V source I can calculate R = (9v-3v)/(15mA+15mA)
    R = 6/0.03 = 200 Ohms
    I have 0.03^2 * 200 = 0.18W or 180mW for the resistor and the circuit will consume 3v * 0.015A = 0.045 or 45mW only.the resistor alone will consume 4 times more than the circuit. This is all without using a transistor off course. If I used a transistor as the connection in the first attached circuit (the converter) then the resistor will consume about half the power calculated above (only 15mA will pass through the resistor -neglecting Ib<1mA for Ic=15mA)
    But still there is a high power consumption comparing to the circuit consumption...

    Audioguru, I'm concerned in the size but actually I now see that using two AAA batteries will fit my need...

    mik3, as Audioguru said; and to Audioguru too, the circuit really worked fine with 1.5V, I tried it and not only thinking... Yes I know it's bad to feed DC to headphones, they look as if they will damage at high noise...

    I'll use two AAA batteries as I said before and if anybody is interested in the circuit, actually it gives a very high gain with very clear sound.. It can detect very remote sounds and you could hear them clearly and they will seem to be near you..

    Anyway, thank you all:)
  12. thyristor

    thyristor Member

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    Other than an LM317 and the couple of resistors needed are probably 1/10th of the size of 2xAAA batteries :(
  13. SgtWookie

    SgtWookie Expert

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    An LM317L is available in a TO92 package; output is 5mA to 100mA at whatever you set via R1/R2.
  14. hazim

    hazim Thread Starter Active Member

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    No, you missed the 9V battery with the LM317 and its resistors.
    2 AAA batteries have approximately the same size of the 9V battery. Also the 2 AAA batteries doesn't require voltage regulation and so the overall power consumption is reduced at least 2 times..;)
  15. hazim

    hazim Thread Starter Active Member

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    I'll not use it this time but I will buy some of it later. These ones are very useful in many circuits and applications. Thanks for the information.
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