voltage regulator to get 1.5 volts from 9 volts?

Started by Solidhex, May 31, 2009, 05:15:21 PM

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Solidhex

Yo

  Been reading up on threads about voltage regulators. I wanted to use a 9 volt battery to power some old school fuzz circuits that run on 1.5 volts with a led. What would you guys suggest as being the best compromise between simplicity and stability? I'm hoping to get something together that doesn't involve op amps. I've seen the circuits that use a voltage divider with a transistor. I'd like to figure something out that maintains its 1.5 Volts even as the 9 volt dies...

--Brad

Gus

Easy way is to use an opamp like a TL061

Wire the opamp as a buffer out to - input

Make a voltage divider from two resistors so the center node is 1.5VDC maybe 15K at the bottom and 68K at the top(should be close about 1.6VDC you could add a trim pot to the top of the 68K if you want to be an exact 1.5VDC) place a cap at the center to ground ADD a series resistor from the two resistor and cap node to the + input, maybe 1K(very important to do to save the protect the opamp input pair).

power the fuzz off the output of the opamp  Old school fuzzes at 1.5VDC were low current so the opamp output current should be well in spec.

Work out the voltage division at different supply voltage 9VDC, 8.5VDC, 8VDC and calculate the output of the buffered divider.

supply voltage/total series resistance = current in the string.  that current x 15K is the voltage.

or find a 1.5VDC drop diode like a certain type of red LED and use in place of the 15K


Ripthorn

Cheapest solution is a 1.8V zener diode (I've never seen a 1.5V one), costs maybe .08 and will maintain a relatively tight tolerance for any voltage above 1.8V.  Of course, depending on the current going through it, you need the appropriate power rating.  Next best solution, voltage divider, then an LM317 in my opinion.
Exact science is not an exact science - Nikola Tesla in The Prestige
https://scientificguitarist.wixsite.com/home

Paul Marossy

#3
I would be inclined to use an adjustable voltage regulator IC chip like the LM317. The support circuitry for the LM317 is extremely simple.

brett

Yes, the LM317 can go all the way down to 1.2 V.
Actually, buy a couple of LM317s.  They can be used for all sorts of things, including dead simple constant current sources.
Brett Robinson
Let a hundred flowers bloom, let a hundred schools of thought contend. (Mao Zedong)

Paul Marossy

Quote from: brett on May 31, 2009, 10:27:40 PM
Yes, the LM317 can go all the way down to 1.2 V.
Actually, buy a couple of LM317s.  They can be used for all sorts of things, including dead simple constant current sources.

Yeah, LM317s are handy little devices to have around.

Gus

A 317 does not regulate well at under a few ma. page four 3.5 to 5ma.  You could load it with a resistor to increase the current but the battery will drain faster.

http://www.national.com/ds/LM/LM117.pdf

That is why I posted the TL061 idea people seem to overlook the 317 etc need to be loaded to work correctly. The selection of the TL061 was part of the design
http://focus.ti.com/lit/ds/symlink/tl061.pdf

  There is a red LED that the voltage drop is more constant at different ma and is about 1.5VDC.

One could also do this with an transistor with high Hfe set as an EF.  Voltage divider node set to .6VDC + 1.5VDC =about  2.1VDC connected to the base, collector to +9VDC power the circuit from the emitter(the circuit is the load in the emitter leg)

  What voltage drop do you get from two 1n4004 series connected? Resistor then node then two 1n4004s you might want to do a simple test with say a 47K to 100K "top" resistor


Paul Marossy

Quotepeople seem to overlook the 317 etc need to be loaded to work correctly.

Oh, yeah... good point.

Gus

an idea

http://www.aronnelson.com/gallery/main.php/v/gus/9to1_5.GIF.html

any questions?

You could add an R and C between the opamp and effect for more noise reduction

Kind of crude but should work OK.  One could make the LED a zener and adjust the resistor.

You could also remove the opamp and reduce the pot value and use a high Hfe NPN transistor C to 9VDC and B to the wiper E "feeds" the circuit

R.G.

Two silicon diodes make a good 1.2 - 1.4V "zener". 1N400(n) works fine. You absolutely must put a resistor between them and the +9V power.
Another one is to use an NPN transistor with a a resistor from base to emitter (R2) and from base to collector (R1) as the "zener". A transistor used this way makes the collector-emitter voltage be (R1+R2)/R2 times the transistor's base-emitter voltage, so it can be dialed in by making R1 and R2 variable. It's called a "Vbe multiplier".

However, there is a problem. Any linear regulator (transistor, opamp, voltage reg chip) will waste the extra (9V-1.5V) times the load current as heat. That means you take six times as much power out of the battery as the effect needs. If you use a "zener" solution, you must make the current limiting resistor to limit at the biggest necessary current, and it pulls that current all the time, making losses even worse.

You don't get to efficient regulators with low outputs or high input/output differential voltgaes until you start using switching regulators.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

JDoyle

Quote from: R.G. on June 03, 2009, 01:32:01 PMAnother one is to use an NPN transistor with a a resistor from base to emitter (R2) and from base to collector (R1) as the "zener". A transistor used this way makes the collector-emitter voltage be (R1+R2)/R2 times the transistor's base-emitter voltage, so it can be dialed in by making R1 and R2 variable. It's called a "Vbe multiplier".

R.G. - what about feeding the circuit directly from the emitter (or collector with a PNP) of the Vbe Multiplier by making the Vbe mult. drop 7.5V, rather than it's current source? The emitter would have to be well bypassed because the impedence of the Vbe mult. will change with the current required from the load circuit, but that way only the current needed will come out of the battery, not the current needed to bias a reference as well.

It would require a careful selection of R1/R2 with knowledge of what the circuit will draw, both peak and average.

This is seriously stream of consciousness on my part so I wouldn't do this without verification from another party...

Regards,

Jay Doyle

brett

Hi again
if the "old school" fuzzes are the FZ ones ( :icon_wink:), then the required current is tiny (and therefore Gus' point about loading the LM317 is very relevant.  The LM317 is also overkill for such a use).  A couple of mA at 1.5V can be had with minimal heat and battery wastage by using a voltage divider (1k and 150 ohms) or a the junction of couple of 1N400x diodes and a 1k resistor.   And we wouldn't be using enviro-unfriendly disposable batteries, anyway, would we? 

Of course, the 1.5V should be AC grounded with a large cap (22uF would be ok in this case).
cheers
Brett Robinson
Let a hundred flowers bloom, let a hundred schools of thought contend. (Mao Zedong)


R.G.

Quote from: zachary vex on June 04, 2009, 03:46:39 AM
1.5V battery.
Just like the technique of hiding things by throwing them way, high up, it only works for short periods of time.  :icon_lol:
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Paul Marossy

QuoteI'd like to figure something out that maintains its 1.5 Volts even as the 9 volt dies...

If you used a D size battery, it would last a very long time.

QuoteAnd we wouldn't be using enviro-unfriendly disposable batteries, anyway, would we? 

I rarely ever use batteries in any of my effects. They are expensive and go in a landfill somewhere after you throw them away. (insert emoticon for look of disapproval here)

Quote1.5V battery.

That was actually one of my first thoughts but I didn't say anything because of the how the original question was presented.

Lurco


Gus

#16
Two 1n400x might not make a good voltage ref used with just a resistor and powering the circuit from the resistor diode(s) node(except at low current changes in the circuit being powered).   But you can work with them if you understand why they might not be good Vrefs.

Look at the curves of the 1n400x diodes the voltage drop changes with current .6 to over 1VDC.   

There is a red led that is kind of cool it has fairly constant voltage drop at different led currents good with CC BJT circuits.

I look at the now easy to find specs of parts with things like a google search

So how do you work with this? You buffer the Diode(s) and run it at one current so what ever the diode voltage drop is it stays a bit constant and sense its voltage with a high input resistance circuit. For the circuit I linked  If the voltage from the diode is over what is wanted use the 1meg pot to the low power FET input TL061(look at the idle current draw in the link).  If the diode drop is the voltage wanted remove the 1 meg pot and connect the ref to the + input with say a 1K.

I tried to make the voltage constant even as the battery voltage goes down and be low powered.

The original post was about powering a old school 1.5VDC fuzz and still having the 9VDC to power the LED and to be low power If I understood it correctly.

At the currents an old school fuzz is run at even a switcher might use up more power than this buffer vref idea.  I have not done the math.  Designs are almost always a compromise.

There is a way to use just a 1.5VDC cell.  There are circuits on the web that power white leds with 1.5VDC cells the problem is they store energy in a coil and there will be switching noise to be aware of.

MikeH

Quote from: Lurco on June 04, 2009, 11:39:14 AM
Quote from: zachary vex on June 04, 2009, 03:46:39 AM
1.5V battery.

there are 6 of them in each 9V block!

So you could use one and feel only 1/6 the guilt of using a 9v battery.  :icon_lol:

I don't know about other cities/counties/states/countries, but where I live we can recycle batteries with our curb-side recycling. 
"Sounds like a Fab Metal to me." -DougH

Lurco

Quote from: MikeH on June 04, 2009, 12:29:42 PM
Quote from: Lurco on June 04, 2009, 11:39:14 AM
Quote from: zachary vex on June 04, 2009, 03:46:39 AM
1.5V battery.

there are 6 of them in each 9V block!

So you could use one and feel only 1/6 the guilt of using a 9v battery.  :icon_lol:

I don't know about other cities/counties/states/countries, but where I live we can recycle batteries with our curb-side recycling. 

dismantle 1 and use it 6 times!  :icon_eek: