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Old 19th December 2006, 04:47 PM   (permalink)
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Post Reduce 5 volts to 3 volts, suggestions?

I am new electronics. I built a very simple 9 volts (battery) power supply to 5 volts using a 7805 voltage regulator. How do I to reduce the 5 volts to 3 volts, suggestions?
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Old 19th December 2006, 05:03 PM   (permalink)
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use an adjustable linear regulator. What are the current requirements?
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Old 19th December 2006, 05:34 PM   (permalink)
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..Thanks "OutToLunch"...but let say that I still want to used a regular 7805 (which works great) and want to get 3 volts. Do I used resitors (which should I used?) or diodes (again which should I used?)...this is for a music synthesizer I am prototyping. (on 3 volts the tuning is good..more than 3volts it gets out of tune)
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Old 19th December 2006, 05:41 PM   (permalink)
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the 7805 is a 5 volt regulator, it doesn't make any sense to use it for 3v. If you want to get 3 volts from the output of the 5 volt regulator, resistors aren't going to cut it... you could use a zener diode regulator but if you're going to go out and buy a 3v zener, you might as well buy an LM317 adjustable voltage regulator and just use it instead of the 7805 in the first place. It's used almost exactly the same way, but with a couple of resistors to choose the output voltage.
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Old 19th December 2006, 05:46 PM   (permalink)
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You can't use a 7805 to create 3V. Resistors are not a good option unless the load is always constant - even then, though, they are not a good option. If you want to use diodes, then the selection depends on the load current - if the diode cannot handle the amount of current required, then put them in parallel. Diodes have a forward drop of around 0.6V, so three in series would get you to about 3V.

Or you could go with the LM317, using either 12V or 5V as the input voltage (5V input wastes less power)
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Old 19th December 2006, 05:50 PM   (permalink)
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Thanks evandude.... , let say what I have available is the 7805 (which is exactly true) ....so with a zener diode I can drop the voltage to 3 volts.?
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Old 19th December 2006, 05:53 PM   (permalink)
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try this circuit...the ground pin is made floating...

the output will be 5*(1+R2/R1) volts....this is assuming R2 is chosen small enough so that drop across it due to current thru ground terminal can be neglected...

you can adjust output voltage from around 1.5 volts to 30volts,assuming proper heat sink is provided...
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Old 19th December 2006, 07:15 PM   (permalink)
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the circuit shown has a bottom limit of 5V - unless you can somehow create a negative impedance.
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Old 19th December 2006, 07:21 PM   (permalink)
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If your output is 5*(1+R2/R1), how do you figure you can get 1.5 volts output? that would require R2/R1 to be a negative value, ie - you'd need a negative resistor! Your circuit can only regulate to voltages HIGHER than the 5v that the regulator is meant for. Your circuit is exactly the same as the way you'd use an LM317 adjustable regulator, however that device has a much lower regulation voltage (1.25v) which allows it to achieve lower output voltage.

*edit: outtolunch beat me to it
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Old 19th December 2006, 10:38 PM   (permalink)
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An LM317 will work but only at lower currents; above a certain load the droput voltage will become a significant factor.
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Old 19th December 2006, 10:44 PM   (permalink)
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Quote:
Originally Posted by Hero999
An LM317 will work but only at lower currents; above a certain load the droput voltage will become a significant factor.
also depends on whether you use the LM317 to regulate the 5v output of the 7805 regulator to 3v, or to regulate the original 9v to 3v. Either way, I would assume output currents will be low since you don't generally run anything high-current off a wimpy 9v battery.
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Old 20th December 2006, 12:28 AM   (permalink)
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You never did say what you're driving with the 3 volt load.
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Old 20th December 2006, 01:41 AM   (permalink)
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Learn LDO regulators quickly you should. Micrel you should try. Good stuff they make.
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Old 20th December 2006, 04:51 AM   (permalink)
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You could do the extremely elcheapo thing of putting 3 Silicon Diodes in series on the +5V Rail... It will drop 2.1V and your 5V will theoretically become +2.9V

1N4004 will do the job, 400V at 1A, the 7805 only outputs 1A anyway... use the 1N5004 if you want to be safe...

Again I stress the elcheapo-ness of this method...
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Old 20th December 2006, 05:19 AM   (permalink)
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Series diodes don't have a really consistent voltage drop. Temp and current levels will change the voltage significantly, plus you have few options in how much voltage to drop because silicon diodes only come in 0.7v.

The zener shunt regulator- using a resistor in series with a zener diode- is generally not efficient because the shunt arrangement must always draw the peak current required by the device. If you need a peak of 25mA, then a properly sized zener shunt reg will draw 25mA even when the device is doing nothing. Actually it's a bit worse because you probably want to size the resistor to work as the battery runs down to 7.5v or so, so at 9v (or more for a warm, fresh battery) it will draw even more than 25mA.

There are proper 3 terminal regs which put out 3v, but they're uncommon. 3.3v is common though generally not from Radio Shack. An adjustable reg is common and will allow you to set 3v. You can feed the 3v reg off of the 5v line or the 9v line. At 5v there may be a voltage dropout issue dependng on the parameters of the reg and circuit. It also means the 5v reg dissipates most of the heat for both of them. Putting it on the 9v source means no dropout probs and each device handles the heat from its own current.
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Last edited by Oznog; 20th December 2006 at 05:22 AM.
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