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Roland Cube 30 amp pedal


Jazzer2020

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The amp has two different switch type options. Both have LED's so I'm guessing the pedal you go was either a generic pedal or some budget pedal sold with the unit.

 

One is the which is a momentary "unlatch"-type footswitch.

 

The second is an which has two switches which combines a latch and momentary-type switching and each footswitch can be assigned latch- or momentary-type.

 

The switch can control channels and/or effects depending on how the jack is wires. If you have a single momentary switch and its wired to the sleeve and ring it will control the channels. If you have an FS6 with two switches the tip is used to control the effects. I'm guessing since its a momentary switch it may scroll through the effects as you tap the switch.

 

 

Without having a pedal in hand, schematic or the amp to check the voltage supplied I really cant advise you on how to add an LED. If its being used for channel switching it should be obvious without an LED which channel is on when you have the channels set for lead or clean. LED's often have a resistor in series with the LED to drop the voltage I'm guessing it might be a 100K resistor if the pedal receives 9V phantom from the amp, but you really need to check that voltage to be sure.

 

It could be low enough where no resistor is needed, It could be higher and need a larger resistor. If the resistor is the wrong value, you could damage the amp. Then you have to know if the switch is latching/momentary, normally open or normally closed and compatible to wiring the LED up. You may wind up having to wire in a different switch that costs at least $10 and have to wire in a stereo cable and jack to get the phantom power.

 

Because of these unknown factors My suggestion is just buy the right switch. The single switch is about $30 new and you can buy used ones for $15 which is about ho much you'd spend modding a generic pedal. The dual switch is about $50 new but you can probably find one for less used.

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  • 4 weeks later...
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The amp has two different switch type options. Both have LED's so I'm guessing the pedal you go was either a generic pedal or some budget pedal sold with the unit.

 

One is the which is a momentary "unlatch"-type footswitch.

 

... Because of these unknown factors My suggestion is just buy the right switch. The single switch is about $30 new and you can buy used ones for $15 which is about ho much you'd spend modding a generic pedal. The dual switch is about $50 new but you can probably find one for less used.

 

Thanks again for your reply WRG!

Same thing happened. I didn't notice your private reply until now.

 

Not long after I posted I discovered that the type of pedal I was using was wrong.

I knew I wasn't using the one that was designed for the amp (it doesn't come with it, it's an extra)

but I thought the one I was using was a similar type to the original.

 

It wasn't though. Mine is a Roland momentary, single pole that basically just switches from clean to solo/overdrive. That's because it doesn't have ring-tip/sleeve (stereo). It's a mono 1/4" plug.

 

Once I discovered that the proper pedal uses ring-tip/sleeve (stereo) a light went off in my head. :)

Then I realized I could also use the foot switch to turn effects on and off in the amp (something I never knew I could do with the pedal).

 

So I think I'll do as you suggest and just buy the proper switch for the amp instead of messing around with it. But I hope it will come with LED's, something I'm not sure it does.

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... Without having a pedal in hand' date=' schematic or the amp to check the voltage supplied I really cant advise you on how to add an LED. If its being used for channel switching it should be obvious without an LED which channel is on when you have the channels set for lead or clean. [b']LED's often have a resistor in series with the LED to drop the voltage[/b] I'm guessing it might be a 100K resistor if the pedal receives 9V phantom from the amp, but you really need to check that voltage to be sure...

 

Technically, the voltage does not change. The resistor is in place to limit the current flow and will always have the same voltage drop across it. If the source is 9V and the forward bias voltage of the LED is approximately 2V then that leaves 7V to be dropped across the resistor. The value of the resistance will be selected to provide enough current to adequately illuminate the LED while limiting and preventing excess drain on the power supply.

 

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Technically, the voltage does not change. The resistor is in place to limit the current flow and will always have the same voltage drop across it. If the source is 9V and the forward bias voltage of the LED is approximately 2V then that leaves 7V to be dropped across the resistor. The value of the resistance will be selected to provide enough current to adequately illuminate the LED while limiting and preventing excess drain on the power supply.

 

I think you misinterpreted my statement. I wasn't talking about individual voltage drops even though that occurs. I meant it as reduction in power to the LED. A resistor reduces the total current in a series circuit. The voltage drops occur across the individual components in that circuit.

 

If the supply voltage 9V, and you have an LED that requires 2V and needs 20mA (0.020A) to operate,

 

The resistor needs a 7 v drop. 7V divided by 0.02A = 350ohm. (or the nearest higher standard value which is greater). This will leave a voltage of 2V and .02A for the LED to operate properly.

 

Many use higher resistor values so thay aren't blinded by an ultra bright LED, but too high and it wont have enough power to light.

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