Cap orientation help needed

[motterpaul]

In this Diezel Diefet - it calls for two 100n caps where all I have are electrolytics. Can someone please confirm the best polarity for these for me - thanks in advance...  100n caps top left. I am guessing both have anode going down. Let me know.

[Neil mcNasty]

Are you saying that you have 100nf electrolytic caps?
I have never seen that before... Are you shure that it's nf and not uf/mf that you've got?
As far as I can tell, a 100nf electrolytic is something I've never come across... (Electrolytics normally range from 0.22uf (220nf) and upwards..

Here is a alternative solution:
If you connect caps in paralell, you double the value, so if you got 2 x 47nf caps you can put them side by side and get 94nf (stick 2 of them in the same holes on the vero), that should be close enough...
There's normally a 5-10% error margin on componets, except potensiometers that has a 20% error margin, so you should get pretty good results using two 47nf caps in parallel, instead of a 100nf.

If you actually have a 100nf electrolytic, I would have the + side towards the input of the signal, but that requires some knowledge about how the audio signal travels trough your circuit, unless it is the input cap (then reverse it, as far as I know...)

Hopefully someone with a slight better knowledge about this will chip in an correct me, cause I might be wrong... But I could also be right... 😜

[Frank_NH]

I have some of those 0.1 uF electros (came in an assortment pack).  Recently I've gone the other way and have been collecting 1, 2,2, 3.3, 4.7 and 10 uF non-electros from Smallbear.

[motterpaul]

Yeah, thats what I have. I have non-polarized 1uf and I had 104s, but I just ran out so I need to use these.

Answer - help?

[Freppo]

Cap far left should have cathode to vol 3 (down on the layout) and the other should have cathode to treble 2 (up on the layout)

[motterpaul]

Thank you!

[induction]

The one on the left: negative leg downward, positive leg on Deep 2.
The one on the right: positive leg downward, negative leg on Treble 2.

Edit: Freppo types faster than me, but at least we agree.

[motterpaul]

Another question on this circuit. The 1n5817 diode pair - are those used for clipping? I think the clipping on this is a little harsh - would another diode sound less hairy?

NEVER MIND -  I see now those are used for the charge pump current.

[pnadora]

I had this problem as well on one of my builds. I looked for a polarised cap or a ground row so that I knew a point with negative polarity. Then I followed the way to the cap in question remenbering that polarity switches at caps and stays the same at resistors. Worked for me so far (although its probably a noob workaround) :P

[motterpaul]

Hey Pnadora...

In other words - what I thought was right was actually backwards; because caps switch polarity? I also looked for the ground connection and tried to point the cap's ground towards ground, but are you saying that is the backwards approach? Because the answers seemed to be the opposite of what I thought.

Wow, another thing I learned today. Incredible.

[induction]

Any phase issues (reversals or shifts) with caps are not relevant to this particular question. Phase shifts are an AC effect, and polarized cap orientation is only a question of DC bias.

For our purposes, you can just pretend that the cap is jumper for AC, and an open circuit for DC. Just figure out where the DC voltage is higher and put the positive leg there. There are a few common scenarios that you see over and over (I'm assuming a positive supply voltage for all of these):

1. Filter caps: These often go from V+ to ground or from Vref to ground.
Solution: Just put the negative leg on ground.

2. Coupling caps: These caps are in series with the signal and connect different stages of the circuit and allow us to introduce or remove a DC shift at the circuit input, output, or between subcircuits.
Solution: Pedal inputs and outputs should have a DC voltage of zero. Internal subcircuits are often centered around a reference voltage. The reference voltage will be positive for circuits with positive supply voltages, and negative for circuits with negative supply voltages. Figure out which side of the cap touches the part of the circuit with a DC shift and put the positive leg there. Both of the 100n caps in question on the Diefet are an example of this. The one connecting the Deep control to the Volume control is an output cap. The Deep control is connected to the final op-amp stage (DC biased to Vref) and the Volume control is connected to thecircuit output (ground). The other one connects to the input of the final op-amp stage (again, DC biased to Vref) and the other side connects to the tone stack. Notice that there is a DC path to ground on this side (through the TMB pots) but no DC path to the supply voltage, because the signal path from the previous jfet stage passes through other caps before it reaches the 100n. Thus the DC bias on this side should be zero.

3. Hi-pass filters: These are also in series with the signal, but there may not be a DC shift across the cap. Solution: Don't use polarized caps here.

4. Tone stacks/tone controls: These are connected to resistors and potentiometers in networks that can be simple or very complex.
Solution: Look for DC connections (through resistors or active components, but not blocked by other caps) to the supply voltage and/or to ground. This one requires some expertise or experience.

Needless to say, you must look at a schematic when you are working this out. Trying to figure out correct cap orientation from a layout is difficult and error-prone. This means you must find the schematic for the circuit, and you must be able to read schematics to figure it out yourself. What if you can't read schematics?
Solution: Learn to read schematics. Or ask someone who can read schematics to tell you the answer.

Also note that the answers will be reversed in circuits with negative supply voltages. With bipolar supplies, the correct orientation will depend on exactly where the cap is placed, but the analysis will be the same.

In very tough cases, you can just sim the circuit and calculate the DC voltage on both sides of the cap.

[strassercaster]

Induction i really appreciate your posts . You know your stuff and put in laymans terms thanks

[strassercaster]

socket them and deal with it later. there only two so only 4 combinations. find a schematic and post it so we can help😀. good luck hope it turns out well thats a complex build take your time .electrolytics will probably be fine but hype says otherwise if they sounded better they would be used in those smaller values more often .just saying never heard of this pedal ii am off to you tube for a listen (hopefully)

[motterpaul]

Yeah, I read that post and said "right on" - It is not that caps flip the ground, it is that caps turn AC into DC, thereby putting all the voltage on the plus side - with is different from swapping ground.

I also really appreciate all of the different contexts he gives us for how a cap will react. He has been a big help with my understanding of theory.

And in the long run it is the theory that we all need to get down. So we can visualize what is happening every step of the way. I really wish I had taken a class in this in college.

[strassercaster]

also if you have some  47n you could twist them together to make 94n that would be within tolerence of 6 percent electrolytics can be off by as much as 40percent sometimes they are really inaccurate.sounds like  your on the right path.

[strassercaster]

yes i wish  i would have taken some classes  as well😀. youth is wasted on the young so they say. Hey we are here doing it now and the internet is smarter than any single person and can be a wealth of information .

[Silver Blues]

motterpaul wrote

Yeah, I read that post and said "right on" - It is not that caps flip the ground, it is that caps turn AC into DC, thereby putting all the voltage on the plus side - with is different from swapping ground.

That's not entirely correct. A cap doesn't "turn AC into DC", that's what a rectifier is for. I'll refer you back to Induction's post:

"For our purposes, you can just pretend that the cap is jumper for AC, and an open circuit for DC."

It doesn't "turn" anything "into" anything else.

[motterpaul]

yes, silver, you are absolutely right, I was thinking diodes in a rectifier format - sorry about the error and thank you for correcting.

[motterpaul]

"Figure out which side of the cap touches the part of the circuit with a DC shift and put the positive leg there.  Both of the 100n caps in question on the Diefet are an example of this. The one connecting the Deep control to the Volume control is an output cap. The Deep control is connected to the final op-amp stage (DC biased to Vref) and the Volume control is connected to the circuit output (ground). The other one connects to the input of the final op-amp stage (again, DC biased to Vref) and the other side connects to the tone stack. Notice that there is a DC path to ground on this side (through the TMB pots) but no DC path to the supply voltage, because the signal path from the previous jfet stage passes through other caps before it reaches the 100n. Thus the DC bias on this side should be zero."

I am trying to understand the phrase "Figure out which side of the cap touches the part of the circuit with a DC shift and put the positive leg there."  Okay - so DC shift means there is a coupling cap which is changing the state of the DC (or eliminating it) for the coupling of sub-circuits within a larger circuit, because one side has different DC requirements from the other side. So, in this case both 100n caps have one side connected to Vref and the other side eventually going to ground. The DC bias on the side of the 100n caps that goes to ground should have 0 bias - is that correct?

diefet_schem.jpg

In other words, when you say "Figure out which side of the cap touches the part of the circuit with a DC shift and put the positive leg there" - that means to put the positive side on the side before the bias shift? and that the ground goes to the side after the bias shift? Meaning the "shift" is from a section with a DC+ bias containing the IC, to  audio subsections (the tone stack and the volume pot) where we really do not need DC. Am I on the right track, or completely off base?

(also - sorry about the earlier mistake, I don't know why I was thinking diodes when we were talking about caps).

I actually just noticed the original schematic is here:

http://techniguitare.com/forum/ressources/image/1663

[induction]

motterpaul wrote

I am trying to understand the phrase "Figure out which side of the cap touches the part of the circuit with a DC shift and put the positive leg there."  Okay - so DC shift means there is a coupling cap which is changing the state of the DC (or eliminating it) for the coupling of sub-circuits within a larger circuit, because one side has different DC requirements from the other side. So, in this case both 100n caps have one side connected to Vref and the other side eventually going to ground. The DC bias on the side of the 100n caps that goes to ground should have 0 bias - is that correct?

That's correct, in this case. There is no source of DC voltage on the side of the cap that touches ground, so the bias must be zero there. In some other cases, it might not be possible to say that the bias is necessarily zero on one side and Vref on the other, but it should be possible to say which side is higher in voltage.

In other words, when you say "Figure out which side of the cap touches the part of the circuit with a DC shift and put the positive leg there" - that means to put the positive side on the side before the bias shift? and that the ground goes to the side after the bias shift? Meaning the "shift" is from a section with a DC+ bias containing the IC, to  audio subsections (the tone stack and the volume pot) where we really do not need DC. Am I on the right track, or completely off base?

You can't really generalize to 'put the positive side on the side before the bias shift' without knowing which way the bias shifts. The general rule always applies: put the positive side where the DC voltage is higher. Beyond that, it's a question of figuring out how the DC voltages sit in the circuit.

I've just written an extensive post on how to read schematics, that touches on the cap orientation issue and may be of interest to you. I put it in a separate thread because it's really long.