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hot tubes
Silver Member
Canada
162 Posts |
 Posted - 08/20/2009 : 03:15:05
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i have a DUNLOP DC BRICK , that i have run out of 9v plugs for use ,
but i have 2 , 18 volt spaces left , could i tame down the voltage with a resistor and use one of those to run a 9v wah pedal ?? if so
how much resistor will i need ?
thanks big time !!! |
Edited by - hot tubes on 08/20/2009 03:15:53 |
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cctsim
Silver Member
United Kingdom
418 Posts |
Posted - 08/20/2009 : 05:47:27
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| I would use a voltage divider with two 10k resistors. The resistors are connected in series from 18V to ground. The middle point should have 9V depending on how closely matched the resistors are. |
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hot tubes
Silver Member
Canada
162 Posts |
Posted - 08/20/2009 : 07:24:11
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voltage divider ??
i'm not sure what that is , sorry .. |
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DeFrag
Moderator
USA
3409 Posts |
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hot tubes
Silver Member
Canada
162 Posts |
Posted - 08/20/2009 : 20:29:59
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| thanks for the great info guys !! |
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IM28
Copper Member
26 Posts |
Posted - 08/31/2009 : 21:48:38
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The actual voltage availble to your 9v pedal will depend on the loading of the voltage divider. As long as there is no current (power)going to the wah pedal the voltage from the 10k/10k divider will be 9 volts. Once power is deilvered to the load things start to change...
Lets assume that the pedal draws 10ma. Your pedal may draw more or less depending on signal levels, LED states etc.
With 9v at 10ma of current you have a 900 ohm load. With a 900 ohm load and a 10k/10k voltage divider the output of the divider is 1.4 volts...not what you are looking for. You do need to know the current load placed on the divider to calculate the voltage drop. If you have a meter you can take some measurements of the pedals current draw to determine the load resistance. Use this calculator to get the right divider values.
http://hyperphysics.phy-astr.gsu.edu/hbasees/electric/voldiv.html
The problem is the exact resistances you need may not be available in standard values.
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Laurie
Double Platinum Member
Canada
4854 Posts |
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cctsim
Silver Member
United Kingdom
418 Posts |
Posted - 08/31/2009 : 23:18:54
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A good power supply should be regulated and usually provides 9V independent of the load. Otherwise if you were to supply more than one pedal you will be in trouble.
I think the example in the link is oversimplifying RL only for DC and resistors only. In practice RL is distributed, includes semiconductors and depends also on the amplification settings. So you can not say that 9V/10mA=900 Ohm represents the load resistor.
The voltage divider with two equal resistors method is used in almost every boss pedal to derive the 4.5V bias from the 9V battery or power supply. If you want peace of mind, you could add another 500k to 1M Ohm resistor to make sure that the voltage divider is not overloaded.
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cctsim
Silver Member
United Kingdom
418 Posts |
Posted - 08/31/2009 : 23:29:07
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Something like this:
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IM28
Copper Member
26 Posts |
Posted - 08/31/2009 : 23:46:20
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I agree it is over simplified as the actual load is not known. the 900 ohms is an example.
"So you can not say that 9V/10mA=900 Ohm represents the load resistor."
A 10ma current draw at 9 volts DC is a 900 ohm load.
OpAmp bias currents are in the microamp (or lower FET types are a few picoamps) range so the voltage drop is negligible for practical purposes.
The voltage divider places a resistance in series with the load. Depending on the current draw of the pedal there will be a voltage drop across the series resistance. This effectively raises the impedance of the PS and limits the current. Some power supplys have this series resistance as a "sag" feature.
Just like powering an ACA pedal from a PSA supply= Voltage drop because of the series resistance in the ACA type pedals.
I am not saying you cant use a voltage divider in this application, but you do need to know the current load to make it work right.
The best solution is a VR as per Laurie
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Laurie
Double Platinum Member
Canada
4854 Posts |
Posted - 09/01/2009 : 02:14:34
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Have to stick with my earlier suggestion... use a voltage regulator  |
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cctsim
Silver Member
United Kingdom
418 Posts |
Posted - 09/01/2009 : 11:07:05
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| Yes, a voltage regulator (78L09) is the best solution. I can work off a 30V input voltage directly. |
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stinkfoot
Silver Member
Sweden
181 Posts |
Posted - 09/02/2009 : 15:23:39
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I might be over-obvious here , but what kind of wah is it? A normal Crybaby/Vox type pedal only draws a minute amount of power (less than 0.5mA for a GCB-95), that an alkaline will easily last a couple of years, as long as the input cable is unplugged when the pedal isn't being used.
/Andreas |
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paisleyfender
Bronze Member
Germany
70 Posts |
Posted - 09/02/2009 : 15:34:10
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But why at all the hassle with the voltage divider?
Each 9V outlet of the Dunlop DC brick can provide up to 55 mA, so couldn't he simply use a daisy chain with two plugs going from one outlet to one effect pedal and the wha pedal simultaneously? |
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IM28
Copper Member
26 Posts |
Posted - 09/02/2009 : 20:11:04
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quote:
Originally posted by paisleyfender
But why at all the hassle with the voltage divider?
Each 9V outlet of the Dunlop DC brick can provide up to 55 mA, so couldn't he simply use a daisy chain with two plugs going from one outlet to one effect pedal and the wha pedal simultaneously?
Best idea yet...no soldering, no parts, no math just plug it in |
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hot tubes
Silver Member
Canada
162 Posts |
Posted - 09/03/2009 : 15:34:10
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thanks for all the great thoughts here , in the end thats what i did,
i used a low amps prob from my autoshop , found that a few pedals were only drawing aound 20ma , so i just hooked into one of the other pedals. so far its working great !!
thanks again .. |
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Topic |
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voltage drop with resistor ??
