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-   -   9V to 3V DC converter (http://forum.allaboutcircuits.com/showthread.php?t=34149)

hazim 02-10-2010 08:36 AM

9V to 3V DC converter
 
2 Attachment(s)
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..
http://forum.allaboutcircuits.com/ca...1&d=1265794240

The circuit
http://forum.allaboutcircuits.com/ca...1&d=1265794240

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

hazim 02-10-2010 09:40 AM

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..

mik3 02-10-2010 10:13 AM

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.

hazim 02-10-2010 10:57 AM

Quote:

Originally Posted by mik3 (Post 212915)
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.

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

Quote:

Originally Posted by mik3 (Post 212915)
What do you mean when you say it should work even with 2V?

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

Quote:

Originally Posted by mik3 (Post 212915)
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.

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??

thyristor 02-10-2010 11:04 AM

1 Attachment(s)
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.

hazim 02-10-2010 11:15 AM

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..

SgtWookie 02-10-2010 04:29 PM

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.

Audioguru 02-10-2010 04:42 PM

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

mik3 02-10-2010 05:10 PM

Quote:

Originally Posted by hazim (Post 212923)
The circuit works fine even with 1.5V, I tried it with my variable power supply.

What exactly do you mean?

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

Audioguru 02-10-2010 05:45 PM

Quote:

Originally Posted by mik3 (Post 213037)
What exactly do you mean?

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

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.


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