I've spent a ton of time looking at FAQs and searching forum posts for this, but I just can't get my head around it.
I've built a Tube Reamer with a pair of Si diodes. For bonus points, I want to add a switch and a pair of Germanium diodes and then switch between them. So, the original circuit looked like this:
(http://s10.postimage.org/6aqyrz0uh/Screen_shot_2011_12_03_at_11_02_22_AM.png)
Then, I added the DPDP (On/On) switch, like this:
(http://s12.postimage.org/sl7h5yf4t/Screen_shot_2011_12_03_at_11_05_27_AM.png)
The next effect of this was...nothing. It's as if the tone pot was completely disabled. I'm fairly sure that by wiring the diodes in the way I've done it, the open diode loop is acting as a short to render the other diode loop useless. I thought that perhaps putting one pair of diodes on the capacitor side of the switch and one pair on the IC side would do the trick, but I'd like to understand what I'm doing before tearing into the perfboard again.
Can someone show me the correct way to do this before I kill another pad of PostIt notes, trying to figure it out?
Thanks!
Here you go:
(http://img14.imageshack.us/img14/1849/diodeb.png)
I might not be reading this correctly, but that circuit appears to take the germaniums and simply switch them in and out of the circuit, not switch between the pairs.
To be clear, I want to switch between having
- the germaniums off, the silicons on and
- the silicons on, the germaniums off.
Thanks!
The germanium diodes have a lower forward voltage, so they will conduct first, and the silicon diodes will have no effect. No need to switch them out of the circuit
Hmm. I should have been more precise in my question. You may have given me the answer that only solves this specific issue.
It appears that in your diagram, the Silicon diodes are never out of the circuit. In fact, you could cut the leads on the Germaniums and it would be electrically identical.
So, the switch in your diagram only appears to switch the Germaniums in and out of the circuit. Are you saying that when in the circuit, they are chosen because they have a lower forward voltage and so, current will not pass through the Silicon diodes but instead, only through the Germanium diodes?
While this may be true of this particular component pair combination, I'm looking for the broader answer of how to correctly switch two, parallel sets of components. Assume, for example, that we're not talking about diodes but rather, two sets of "Anti-confabulating harmoniums" with identical forward voltages; what is the correct way to accomplish this?
Alert me if i bungled up .
(http://i44.tinypic.com/1cbit.jpg)
Suicufnoc is right, the silicon diodes don't really need to be switched out. The won't conduct much because the germanium diodes' forward voltage is much lower. His diagram will work fine.
If you insist on switching out the silicons, this is the easiest way. Like I said, you don't need it in this circuit. But it would be useful for modding. For example, if you decide to switch the germanium diodes with LEDs later, this switching method would make life easier.
(http://i44.tinypic.com/2ewk5le.png)
@DavenPaget
Please use dots for connections with three or more wires. I can't really tell what's going on there.
Quote from: CynicalMan on December 03, 2011, 06:21:47 PM
Suicufnoc is right, the silicon diodes don't really need to be switched out. The won't conduct much because the germanium diodes' forward voltage is much lower. His diagram will work fine.
If you insist on switching out the silicons, this is the easiest way. Like I said, you don't need it in this circuit. But it would be useful for modding. For example, if you decide to switch the germanium diodes with LEDs later, this switching method would make life easier.
(http://i44.tinypic.com/2ewk5le.png)
@DavenPaget
Please use dots for connections with three or more wires. I can't really tell what's going on there.
OOPS . Yeah my bad .
Note that you only need to connect one side of the diodes to the switch, like in CynicalMan's diagram. They won't conduct if one side is lifted. Thus, you only need a SPDT.
Quote from: Keppy on December 03, 2011, 06:45:19 PM
Note that you only need to connect one side of the diodes to the switch, like in CynicalMan's diagram. They won't conduct if one side is lifted. Thus, you only need a SPDT.
I drew a DPDT .
> The next effect of this was...nothing.
Sure. The one cap is across BOTH diode pairs.
This is what you wanted to do:
(http://i.imgur.com/zcvua.gif)
> in your diagram, the Silicon diodes are never out of the circuit.
True; but the Ge diodes conduct first.
Say you want to keep tall people out. Build your door 6 foot tall. Now say that sometimes you want only short people. Put a barrier 3 feet up. You can keep the 6' barrier, it "doesn't matter" because the 3' barrier won't pass anybody over 3'.
As mentioned: this can be simplified to a SPDT switch. Alex's plan does the right thing.
(http://i44.tinypic.com/2ewk5le.png)
It is possible to use reverse-logic:
(http://i.imgur.com/nbtKJ.gif)
A specific "advantage" is that the whole diode and cap network "can" be wired ON the switch terminals, avoiding lug-strips or added PCB pads. The disadvantage is nasty repair/modification.
DP> Alert me if i bungled up.
It appears the cap is not connected when switch is in the up position.
In switching, omit the stuff that is NOT switched. As I read Yeahno's request (and understand typical diode clippers), the cap is always across the clipper, whatever clipper is used.
Quote from: Keppy on December 03, 2011, 06:45:19 PM
Note that you only need to connect one side of the diodes to the switch, like in CynicalMan's diagram. They won't conduct if one side is lifted. Thus, you only need a SPDT.
A DPDT will work, but is more complicated than necessary. I was referring to CynicalMan's diagram, not yours. Sorry for the confusion.
All,
Thanks so much. Not only have you solved the problem (while the others are all "correct", Paul's solution is both electrically and conceptually where I'm trying to get to), you've taught me a ton in the process.
So, my thanks to everyone who took the time to think about this and reply. It's enormously encouraging and helpful.
Hope I can return the favour!
Yeesh.
Having just found the time to actually solder this up, I stumbled into a bigger mystery.
I worked up the diagram Paul presented ("This is what you wanted to do:"). In that diagram, it appears that C1 is directly connected to IC1's pins, with the addition of two leads to the switch attached to those traces. The diode pairs are then attached to the poles of the switch.
Doing this, and after much fumbling around, I discovered that the circuit, a Tube Reamer, is actually distorting without the use of any diodes at all. That is, if I land C1 on IC1 and omit the switch completely, I get a crunchy circuit, no diodes required!
After a brief WTF moment, I then disconnected C1 from IC1 and went back to having a circuit with gain (lots of gain).
Thinking I might be reading the drawing incorrectly, I tried Cynical Man's approach and got exactly the same results - connecting C1 directly to the IC gives me distortion, regardless of whether the diodes are inserted or not.
Stranger still: when I did insert the diode pairs, the silicon side sounded pretty much the same as the circuit with no diodes, while the germanium side has no distortion at all - clear signal with gain. Swapped out the germaniums with new parts, same result. Checked polarity with a meter, meters correctly.
Clearly, I am doing something wrong or not understanding the circuits presented.
Help?
I'm an idiot.
I think I've figured out that the TR circuit is supposed to distort, regardless of the presence of diodes. The diodes only affect how it distorts.
The Germaniums now distort rather nicely. The silicons distort, but with a longer decay, less "crunch", less spitting.
Or not. Found an open connection, closed it, and now the germaniums have stopped distorting.
Crap.
I moved the diodes to breadboard with two leads back between C1 and IC1.
When I use LEDS: distortion. LEDs light up.
When I use germanium diodes (1N60 or 1N34): I can just barely hear the faintest hint of distortion. Otherwise, it sounds like a completely clean signal.
This can't be right, can it?
If your multimeter has a diode function, test them out. You should get readings of 0.1-0.3V in one direction.
The diodes check out okay.
After breadboarding the diodes, it turns out:
- Any combination of Germaniums yields zero distortion.
- Any combination of anything else (LED, Si diodes, one Germanium diode) yields distortion, each combination yielding a different tone.
What am I missing about the nature of Germanium diodes?
Something is not wired right.
Go through the Debugging thread and let's get the basic unit working right before you go on to switching.
Shall do, Paul.
Thanks.
Quote1.What does it do, not do, and sound like?
Distortion pedal. Works, distorts. But when the original Si diodes (1N34) are swapped for Germanium, all I get is a a clean, loud signal. Paul suggests that something's not right with the circuit.
Quote2.Name of the circuit =
ROG Tube Reamer
Quote3.Source of the circuit (URL of schematic or project) =
http://www.runoffgroove.com/tubereamer.html
Quote4.Any modifications to the circuit? Y or N
Y. Power jack, LED added, following this diagram:
http://www.generalguitargadgets.com/diagrams/switch_lo_3pdt_tb_dcjack.gif?phpMyAdmin=78482479fd7e7fc3768044a841b3e85a
Quote5.Any parts substitutions? If yes, list them.
Yes.
100k-A pot swapped for a 250k-A
Quote6.Positive ground to negative ground conversion? Y or N
N.
Quote7.Turn your meter on, set it to the 10V or 20V scale. Remove the battery from the battery clip. Probe the battery terminals with the meter leads before putting it in the clip. What is the out of circuit battery voltage? =>
8.45
QuoteNow insert the battery into the clip. If your effect is wired so that a plug must be in the input or output jack to turn the battery power on, insert one end of a cord into that jack. Connect the negative/black meter lead to signal ground by clipping the negative/black lead to the outer sleeve of the input or output jack, whichever does not have a plug in it. With the negative lead on signal ground, measure the following:
Voltage at the circuit board end of the red battery lead =
3.97
QuoteVoltage at the circuit board end of the black battery lead =
3.67
QuoteNow, using the original schematic as a reference for which part is which (that is, which transistor is Q1, Q2, etc. and which IC is IC1, IC2, C1, and so on) measure and list the voltage on each pin of every transistor and IC. Just keep the black lead on ground, and touch the pointed end of the red probe to each one in turn. Report the voltages as follows:
UA1 (which is an NE5532):
P1 = 3.95
P2 = 3.94
P3 = 3.58
P4 = 0.00
P5 = 3.94
P6 = 3.94
P7 = 3.96
P8 = 8.34
D1
A (anode, the non-band end) = 3.93
K (cathode, the banded end) = 3.92
D2
A = 3.92
K = 3.94
QuoteFor extra credit, while you're waiting for someone to tell you what they see, probe the pins of each of the electrolytic caps, verifying that the voltage on the (+) pin is more positive than the voltage on the (-) pin. If it's not, that cap will eventually fail, whether it's the immediate cause of the thing not working or not.
The only polarised cap is reading = 0VDC at negative, 4.1V at positive
> Voltage at the circuit board end of the red battery lead = 3.97
Voltage at the circuit board end of the black battery lead = 3.67
From a 8.45V battery?? Something wrong there.
> UA1 P8 = 8.34
And how is there more voltage on the chip than on the battery lead??
However..... in this circuit, the diodes may not clip much. There's 1:6 gain after the diode stage, and the second stage will clip about 3V. So if diode voltage is greater than 0.5V, the output stage clips first and masks some of the diode clipping.
So it is a bit odd that it uses 0.6V Silicon diodes.
Just for curiosity: lift one end of the 100nF cap. Gain will drop, turn-up the output Volume. Now can you hear a difference Ge Si LED?
Quote> Voltage at the circuit board end of the red battery lead = 3.97
Voltage at the circuit board end of the black battery lead = 3.67
From a 8.45V battery?? Something wrong there.
Yes. I am an idiot.
It's 8.34V.
Quote> UA1 P8 = 8.34
And how is there more voltage on the chip than on the battery lead??
Now that I've metered it properly, there isn't.
Quote
However..... in this circuit, the diodes may not clip much. There's 1:6 gain after the diode stage, and the second stage will clip about 3V. So if diode voltage is greater than 0.5V, the output stage clips first and masks some of the diode clipping.
So it is a bit odd that it uses 0.6V Silicon diodes.
So you're saying that you'd want diodes with a lower voltage than .5v or they won't do anything? This would be the forward voltage, right?
QuoteJust for curiosity: lift one end of the 100nF cap. Gain will drop, turn-up the output Volume. Now can you hear a difference Ge Si LED?
I lifted the 100n. Gain dropped. I turned up the volume and was still able to hear the switch between diode pairs. But this wasn't at issue before.
Again, to be clear: I *do* hear a difference when I switch between diode combinations (including diode combinations that include one germanium diode). But combine two Ge diodes and the circuit reverts to the perfect, loud, crystaline sound described above - there is no distortion whatsoever. Any other combination (Ge/Si, Si/LED, LED/Ge) gives me lovely, crunchy distortion. It's only the Ge/Ge combination that, mystifyingly, doesn't.
Thanks for your patience with me as I fumble around.