Help with an LED puzzle...
Hello all, can you please help me with this puzzle ?
This is a simplified part of my DIY ECC82/12AU7 overdrive pedal.
I have connected a red LED from between the two stages to ground (see picture), in order to achieve some soft clipping.
I can see the LED glowing every time the clipping occurs, as I was expecting it to do.
BUT : When I cut off the second stage (see switch S), the LED stops glowing !
I cannot imagine any reason for this behaviour.
I tried connecting a 1M resistor in parallel with the LED, reversing the LED, even connecting it to the +12V, but nothing happened.
The LED glows only when the second stage is connected !
Can someone help with a solution ?
This is not only a theoretical problem, as the glowing LED can be the base of a compressor circuit, if an LDR is connected to the input of the next stage.
Thank you very much in advance.
How much current does it take to light the LED? Well lets just guess at 20ma. Where is that current comming from? No DC current passing through the 100n cap, so it must be grid current. Sence the plate current of a 12AU7 is about 1 ma this is a whole bunch and is probally not a good pratice. How is the clipping working? Does it do what you want? Is the mechanism for clipping just cutting off the top of the AC signal peaks as is what it is usally used for, or are you causing the second tube to go into some kind of ugly state of operation? Not sure.
Hello firechief, and thank you for the very fast reply.
I think you just gave me the answer. Here is what I understand:
Yes, the LED is clipping by cutting off the top of the AC signal peaks, when the voltage is enough to make it contact, but the plate current is not enough to light it.
The LED lights because it's "stealing" some current from the grid of the next stage. That makes sense.
Thank you once again !
And it doesn't take a lot of current to light up a red LED. I'm currently (doh) getting a nice red glow with only .8mA in one of my circuits.
Hello all, I think I just found the solution:
The second part of the circuit is acting as a load for the first part, in order to keep it work.
When it is disconnected, the capacitor cannot charge/discharge properly, because the LED is not always conductive.
So, I added a resistor in parallel with the LED (47k is OK, see picture), and the LED started glowing again, with the peaks of the signal.
The glow is not too bright, of course, but it's enough to activate a sencitive LDR.
Why waste the power? Use two LEDS, wired in parallel, but with opposite polarities. That will work just like the classic diode clipper used in pedals.
Thank you Don, I will try it.
|All times are GMT. The time now is 05:29 AM.|
vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2016 DragonByte Technologies Ltd.
Copyright ©1999-2016 diyAudio