powering an amplifier light with 9 volt dc

Discussion in 'The Projects Forum' started by sim84, Dec 12, 2014.

1. sim84 Thread Starter New Member

Dec 11, 2014
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Hi, I'm a total beginner at electronics (as in, sort of worked out the difference between volts and amps last week kind of beginner), but I am making a guitar pedal board project (most of the work is woodworking), and I want to be able to use one of the 9 volt dc outputs on the power supply which runs my pedals to power a small light bulb in the style of Fender guitar amps. Apparently Fender normally use a 6.3 volt bulb. When I google search for 9 volt bulb all I get is a bunch of stuff about powering a light bulb with a battery. Can anyone tell me if what I'm trying to do is possible? Here's a link to the power supply I have, in case that helps:

Cheers!

Simon

2. DickCappels Moderator

Aug 21, 2008
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Should not be difficult. You can put a resistor or a Zener diode in series with the bulb to bring the voltage down to about 6.3 volts.

The Zener will give you the most predictable results. If your bulb draws less than 150 milliamps (and it almost certainly does) you can use a 1N5223B Zener diode with the cathode (marked with a band) toward the postive power supply.

http://www.es.co.th/Schemetic/PDF/1N5221-67B.PDF

3. #12 Expert

Nov 30, 2010
16,026
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I tried looking for a bayonet base 9V bulb, but didn't find any. Please proceed as you were.
The current for 6.3V incandescent bulbs that I saw ranges from .15A to .25A

4. djsfantasi AAC Fanatic!

Apr 11, 2010
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The zener diode is the simplest method. Or if you're up to it, make a voltage regulator circuit to drop the 9vdc to 6.3vdc. It can be made with just a few parts and is based on the LM317 adjustable voltage regulator.

5. djsfantasi AAC Fanatic!

Apr 11, 2010
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OK, so maybe the LM317 circuit is overkill. Here's the zener circuit suggested by DickCappels:

The zener part number is from LTSpice, who indicates the manufacturer is On Semiconductor. The DigiKey part number is BZX84C6V2LT1GOSCT-ND

6. #12 Expert

Nov 30, 2010
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There is also an LM7806C chip (page 5).

ps, I like to run lamps and LEDs below their rated current. They last MUCH longer.

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7. GopherT AAC Fanatic!

Nov 23, 2012
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@djsfantasi
With ~200 mA draw for the bulb, the resistor in your Zener schematic should be more in the range of 12 ohms.

It is likely easiest to just use a 12 ohm, 1 watt resistor since there is a stable power supply and intensity of the bulb is not critical (just an indicator light).

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8. alfacliff Well-Known Member

Dec 13, 2013
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or if you want to draw less current, us a series connected 3 volt zener to drop 9 to 6 volts.
with .2 amp draw, as listed, the 1k resistor would drop about 200 volts, way too much for a 9 volt battery.

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9. djsfantasi AAC Fanatic!

Apr 11, 2010
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My oops in the resistor value. You guys nailed me on it! Thanks.

The current through the resistor is, by KVL and Ohm's law:

$\frac{(V_{in} - V_{z})}{Rs} = I_{load}$, or $R_s = (V_{in} - V_z)/I_{load}$l

This is the maximum current through the zener diode and so, RS is chosen to limit the current through the diode to some value such that the maximum power dissipated by the diode is below the maximum power rating.
Since I knew the Voltage source, the zener voltage and the load current - I can calculate the resistor needed,

10. sim84 Thread Starter New Member

Dec 11, 2014
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you guys are awesome. part numbers and everything, thanks! the equations are a little over my head, but can i ask what makes the first diagram overkill? i'm putting heaps of money in parts into this project so i don't mind doing a little more work or spending a little more in parts if it's a better way of doing it. cheers!

11. DickCappels Moderator

Aug 21, 2008
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Wait a moment...my suggestion was not to make a shunt regulator, but to put a 2.7 volt Zener is series with the bulb.

(ignore the voltages in the illustration above, its just a picture I found on the web).

Why waste power heating up the resistor?

Dec 11, 2014
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13. MrChips Moderator

Oct 2, 2009
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Either way you look at it, the resistor or Zener has to dissipate the excess power.

Put two 6V bulbs in series.

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14. MrChips Moderator

Oct 2, 2009
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My ebike needed a replacement bulb for the rear light powered from 36V.
Where do I find a 36V bulb? I wired three 12V bulbs in series. Problem solved.

15. takao21203 Distinguished Member

Apr 28, 2012
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there is 7806 its justt less common

16. DickCappels Moderator

Aug 21, 2008
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607
Yeah, MrChips, why not put that power to work?

I can't say for sure sim84 because we don't know the amount of current your bulb draws. The diode in your link is good for up to about 140 ma. Above that, you would be better off with a higher power rated diode.

The power, P, dissipated in the Zener diode is P = Zener Voltage x current.

Derating strategies vary but I don't like to push current ratings in diodes beyond 80%, even in non-critical applications.

17. GopherT AAC Fanatic!

Nov 23, 2012
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Why no
Why not a resistor? So much more common. So much easier for a newbie to understand and the same efficiency. No performance gain at all vs a series diode.

A 12 to 15 ohm resistor will be perfect - select a bit higher 22 to 33 if bulb current is below 200 mA.

18. sim84 Thread Starter New Member

Dec 11, 2014
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i've just realised that one of the pedals i'm getting needs its own power supply, so i'll actually need 2 lights that can be switched on and off independently as i switch on and off the different power supplies. i have the options of powering them with either a 9, 12 or 15 volt output from my main psu though. this is the bulb i was thinking of using:

https://www.amplifiedparts.com/products/P-47

the problem is, if i only switch on the 2nd psu, the light won't come on, since the 1st psu powering that light would be switched off. is there a way of using ac to power the lights? i'm in australia, so it's 230 volts. thanks!

19. #12 Expert

Nov 30, 2010
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Is there a way of having each power supply turn on its own light bulb?

20. sim84 Thread Starter New Member

Dec 11, 2014
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not as far as i know. not without buying an extra distributor unit for the 2nd psu which would cost \$70