schematic reading

[Belt]

I've been studying schematics for a bit.  I finally have a schematic and unpopulated PCB.  I have been tracing the circuit to learn more about them.  I have a question.  It might be simple... but there doesn't seem to be any rhyme or reason.  (not yet at least).



What's the difference with the resistors going to ground compared to the one's that go to 4.5v bias?  Where do the ones that go to bias connect to?  In my studying, the "bias" connection simply goes to another "bias" connection.  If that is the case, then how do you decide from a schematic layout which bias connection goes where?

Thanks

Don

[demonstar]

The points labelled bias need to go to a 4.5V supply. Hey, but you've got a 9V supply but not a 4.5V supply. Well what is usually done is a potential divider (voltage divider) is used to hold the non-inverting input (the one labelled '+') of the opamp at 4.5V. This means the opamp has maximum clean headroom swing in both positive and negative directions.  It can swing positive by as much as 4.5V before clipping and negative as much as 4.5V before clipping.

The formula for calculating V_out of a potential divider is R₂/(R₁+R₂)*V_in=V_out

V+
  \
  /
  \  R₁
  /
  l
  l---- V_out
  l
  \
  /
  \  R₂
  /
---- Ground
--
 
In your case I'm preety sure you need three separate bias potential dividers for each bias point. You also need to keep the resistors LARGE eg. *10⁶ ohms.

There is a way to work out how big which I can explain if you'd like but as long as you make them big you will be O.K.

Hope that helps. Any questions, just ask. 

[demonstar]

I forgot to say you need to put a capacitor across across V_out but that needs to be done too.

[frokost]

Some schematics don't list all parts. In this case, the power supply section is omitted. Check out the  complete SD-1 schematic at http://www.freeinfosociety.com/electronics/schemview.php?id=169 for the missing parts.

[Naz Nomad]

[Belt]

The schematic I am looking at is the BYOC shredmaster.  I believe the power section is in the schematic.  I can't find any rhyme or reason for the "biases" though.  I will continue to study demonstar's explanation.  I used the SD1 because I don't have an image of the shredmaster... just pdf.

[Belt]

Some schematics don't list all parts. In this case, the power supply section is omitted. Check out the  complete SD-1 schematic at http://www.freeinfosociety.com/electronics/schemview.php?id=169 for the missing parts.

This is a great example.

The +9v and +4.5v  ---  what component do they connect to.  And how do you tell?

I have read about learning schematics and I think I am missing something.

Thanks

[demonstar]

The simpler version...

If you look at this...

"http://www.tonepad.com/getFile.asp?id=115"

R₂,R₃ and C₁ make the voltage divider in the schematic (the one I've linked to (thanks tonepad!)). You need to make three of them (potential dividers) and connect the junction between R₂ and R₃ on each to the points that require to be connected to Vbias just like R₄ is in the schematic I linked to.

Hopefully that may clear things up a bit.

[demonstar]

Quote from: frokost on Today at 17:37:08
Some schematics don't list all parts. In this case, the power supply section is omitted. Check out the  complete SD-1 schematic at http://www.freeinfosociety.com/electronics/schemview.php?id=169 for the missing parts.


This is a great example.

The +9v and +4.5v  ---  what component do they connect to.  And how do you tell?

I have read about learning schematics and I think I am missing something.

Thanks

It really is as simple as it sounds. All you have to do is connect the places on the schematic that say 9V to 9V and the places that say 4.5V to 4.5V. You can create those voltages anyway you want but we use a PP3 (9V battery) so we already have a 9V source. So you would connect the points labelled 9V to the 9V supply which is the battery. The issue now is we need a 4.5V supply too but we don't have one and don't want to use any more batteries or another power source so we have to get it another way using what we have got. The simplest way of getting this 4.5V is using a potential divider.

Note: a potential divider can provide a voltage less than that of the the power source's voltage but not higher!

[Belt]

Hey demonstar, I will check out my schematic. 

Thanks so much for your help.

[ayayay!]

Hey Belt, those are good questions and we're glad you're not bashful. 

One way to look at it:  See in the SD-1 schem above, the black dot likes to show up right before a transistor or an opamp (amplfier = triangle.)  Or usually very nearby to it. 

So some schems have say, an open arrow which could be 9v, and a filled in arrow that means 4.5v or even another say grayed-out arrow that may mean I dunno 8v  ( A Boss BD-2 Blues driver uses this if I recall) 

In all these cases, the arrow, black dot or "+9V".... whatever it says, it's "looking" for voltage from somewhere.  All YOU have to do is figure out a way to help voltage get to that point.

[Belt]

That was the simplest explanation I could've asked for.

You guys are tremendous help.

thanks

Don

[aron]

Don,

The DIY FAQ also has a section on V.R., 1/2V+ etc... in it. Check it out if you haven't.

[Belt]

thanks Aron.  I'll check it out too.

[Belt]

"Basically if you see 4.5V or V.R. or 1/2V+ ( all the same), you find all the places on the schematic that reference the label and connect them together."  from the wiki page.



Using that info...  On the first schematic, the 470k - 100k - 10k all get connected together.

Then the 9v cathode leg goes to the 9v side of the power supply section.

Assuming that's all correct, where would the 4.5v be derived from?    The 4.5v would all connect to the power supply section.

Right?

[Naz Nomad]

Yup, like my pic up there ^^^

[Belt]

Thanks.  I appreciate the help.

Are the power supply schematics pretty universal?  What capacitor value should be used?

[demonstar]

"Basically if you see 4.5V or V.R. or 1/2V+ ( all the same), you find all the places on the schematic that reference the label and connect them together."  from the wiki page.

You find all the places and connect it to 4.5V or V.R. or 1/2V+ for the purpose of what we are doing here. However that does not always mean they are all connected together. This is one of those cases where everyone of the points cannot be connected together (I think but I could be wrong so someone please tell me if I am!). My reasoning for this is that you'd be joining places that AC could flow between that should not be joined.

Then the 9v cathode leg goes to the 9v side of the power supply section.

You've lost me. (Sorry) 

Assuming that's all correct, where would the 4.5v be derived from?    The 4.5v would all connect to the power supply section.

Right?

As I pointed out earlier it can come from anywhere but the most practical method, the one that is generally used and the one that should be used here is using a poetntial divider (voltage divider).  They are what I explained above and what Naz posted a picture of too.

What capacitor value should be used?

It will affect the input impedance of the pedal (so I believe) and it will also set the frequency cut-off acting as a filter in combination with other components (other resistances and capacitances in the circuit). The equation for calculating the frequency of a RC filter (either high pass or low pass) is f_c=1/(2piRC). The values of R and C are total resistances/capacitances of all the components affecting the filter.

I think generally in that situation I once heard if you just stick in a 47uF cap you'l be fine as long as it has the required operating voltage.



I'm pretty confident that what I've outlined is correct and I certainly don't pretend to be an expert so if anyone else can correct, fill in the gaps or just confirm it that would be great.