adding a led to a sunn buzz?

[foxfire]

i just drew up a perf layout for a sunn buzz and i'm getting my build list together and got to thinking about adding a on/off led. this is my first non 9v project. so if i add a led will there be enough voltage for it, and or will it just drain the battery that much faster?

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[runmikeyrun]

If you have space, add 9v for LED, or maybe double battery clip for LED (3v) and single for circuit.  Or run them both off 9v w/ voltage dividers.  Or run both off  double AA clip.  I think it really depends on if you want to carry AA batteries with you to your shows.  I know i wouldn't.

[johngreene]

i just drew up a perf layout for a sunn buzz and i'm getting my build list together and got to thinking about adding a on/off led. this is my first non 9v project. so if i add a led will there be enough voltage for it, and or will it just drain the battery that much faster?

The short answer is 'no'.

LED's typically have a greater than 2V Vf.

http://www.kpsec.freeuk.com/components/led.htm

So if you want to add an LED indicator, you need more voltage as runmikeyrun mentioned.

--john

[foxfire]

i like the idea of using a single 9v for both with a voltage divider. only problem is i don't know how to make up a voltage divider. would you mind pointing me in the right direction? and thank you for the help thus far.

[foxfire]

would it be as simple as, 9v battery to a 4.7kR (for running the led) and from the 4.7k into another resistor that gets the voltage down to 1.5v to run the sunn buzz?

[yeeshkul]

here is what you need:
http://www.diystompboxes.com/analogalchemy/emh/emh.html

[yeeshkul]

there will also be a voltage drop on the LED, which is about 0.7V. So you need something like 20k resistor. I suppose the input resistance of the box is fairly big due to the emitter follower ... which is allright but the 20k will lower the imput impedance to something less than 20k .... so ...

[foxfire]

here is what you need:
http://www.diystompboxes.com/analogalchemy/emh/emh.html

thanks for the handy link. i'm just not sure how to apply a voltage divider to my project? if what they call Vout is the supply for the led than is what they call ground the supply for the sunn buzz circuit?

[yeeshkul]

Vin is 7.5.V (9V-1.5VLED voltage drop)
Vout is here 1.5V, held due to the 4.7k/17k divider (which looks quite strong because 4.7k is smaller than 17k)

sorry i was first counting with 0.7 voltage drop, but it is more in case of LEDs - lets say 1.5-2V
i also edited my second message ...

honestly i would omit the LED thing for this case , it just destroys the idea of the emitter follower placed in the input

[foxfire]

wait, wait, wait, i think i just caught up to everybody! the voltage divider is only for the sunn buzz. the led and buzz are separate but share the power supply? so it is battery, 4.7k pulldown resistor for the led then the voltage divider to the buzz?  i thnk i'll just have to build it, get it working then try tinkering with it.

[slacker]

yeah that' s right. For it to work you'd have to put a big capacitor from the voltage divider to ground, something like 100uF or so. Otherwise you'd probably get lots of oscillation, which might sound cool but isn't what you want. 

[yeeshkul]

Ok  . I guess my answer was hasty anyway .... I don't know what you mean by "voltage divider separately" but first you have to set the right current to open the LED. That may be 2mA, so you need 4.7k. And after that you can split this 4.7k to 2 separate resistors in serires to make the voltage divider and have the right current at the same time. I see 3.7k and 1k. It also means that 1k will be in parallel to the input which makes it really low. I am not sure if i got what you want though. hehehe

[yeeshkul]

or do you want the LED+4.7k branch parallelly to the divider? I see, that may be the way ...

[Sir H C]

What about an incandesent lamp bulb?  They make those that work off of 1.5 volts (for some of the small flashlights).  Might suck the current though.

[foxfire]

my goal is to be able to use a 9v battery and if possible add a led. so what i was thinking to pull juice for the led and then set up the power divider based on the left over voltage. but then again i'm a newb with a capital N so i don't fully know/understand how putting all of that in front of this pedal will affect the way the pedal works?

[yeeshkul]

1. If you wanna use the "leftover voltage" as you say:
then the circuit will be connected in parallel to the second divider resistor. This resistor must be small (1k) to maintain the 2mA for the LED (together with the LED resistor which must be 4.7k - 1k = 3.7k ). When you have two resistors in parallel (that 1k resistor and input resistance of the circuit), then the total resistance is smaller than the smaller of the two resistors - less than 1k. So the total input resistance will be smaller than 1k. And that is catastrophically low. It will suck a lot of signal from your guitar to the ground and for example Wah is not gonna work.

2. If you wanna connect the whole LED+4.7k branch in parallel to the divider then keep on mind that the circuit was designed for particular bias current (often 1mA) and you have to provide that current despite the divider. It however means you will have to use small resistors ... and we are back in the same troubles.

[yeeshkul]

here is a little illustration to the text above:

[Seljer]

in that 2nd example.... could you change R2 for two silicon diodes in series? or for the actual LED? as if you were doing a zener diode voltage regulator

[foxfire]

ok, so no led. i have to give you a big thanks Yeeshkul for taking the time to explain it all to me so here it is, THANKS. just one more question, can i do a voltage divider with r1 at 10K and r2 at 50K? please for give me if i'm just not getting it.

[yeeshkul]

Sure you can . 1-5/10-50/10k-50k/10M-50M ... will split the voltage the same way: 1/5(where bigger voltage sits on the bigger resistor V=R*I).

However: the bigger the resistors are - the smaller is the current through the divider (I=V/R). So to get the enough bias current, first you have to find the resistor R=R1+R2 to get the right current and then split this sume to the voltage divider to get the right voltage. I can see on my calculator something like 9k totall for 1mA through the main branch ... you may need more though.

Another thing about voltage dividers is their "strenght". The strong divider will keep the the chosen values of the splitted voltage without changes dependent on changes of the impedance of the circuit (which is frequency dependant so it will vary). That happens when we take the important voltage from the bigger of the two resistors (this is the same issue as the inner resistance of voltage sources - the bigger current it gives, the smaller is the inner resistance, the stronger the source is ). And that is another thing you cannot do here.

It works as follows: impedance of the circuit changes -> current to the circuit changes -> current through the divider changes -> VR1 changes => VR2 changes.

When R1 <<  R2 the voltage drop on R1 is small(V=R*I). When it is big, the voltage drop is big and pushes the voltage on R2 down (VR1+VR2 must be 9V).