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Just realized I accidentally placed the +9V tap at the +4.5V tap. Anyone reading it should realize the error pretty quickly.

Here's the schem for anyone who wants to fix it. Not worth anymore effort though, IMO.

Pardon the newb question, and maybe this isn't the place to ask it, but can anyone explain the benefit of the buffered power supply? Is it less noisy?

Here are some reasons to use a follower for a reference supply:
-Its stiff. Really stiff, without sacrificing current consumption due to low value resistors.
-Higher value resistors can be used, since the follower is doing the heavy lifting. That can mean lower current consumption (a balance between what the follower draws and what the simple divider draws).
-Higher value resistors mean lower value capacitors, which in this case have been negated. Don't ask why. I didn't design it.
-Its isolated from the rails. This means you can slam it with (relatively) huge signals and currents without causing the same disruptions in other sections of the circuit. There are limitations here, of course...
-A closer approximation to linear operation can be achieved as a result of the isolation. Think about it. You'll see it.
-Its confusing! Makes people stop and ask why, and anytime that happens, I'd say you've got a winner on your hands.

Low impedance out too, so less susceptible to external noise interference, and less voltage drop under operation. I use them on occasion when I need a bias supply for several devices.

mmolteratx wrote:Just realized I accidentally placed the +9V tap at the +4.5V tap. Anyone reading it should realize the error pretty quickly.

So all "+9V" should connected to "V+" instead right? VB remain as 4.5V.

Jack Deville wrote:Here are some reasons to use a follower for a reference supply:
-Its stiff. Really stiff, without sacrificing current consumption due to low value resistors.
-Higher value resistors can be used, since the follower is doing the heavy lifting. That can mean lower current consumption (a balance between what the follower draws and what the simple divider draws).
-Higher value resistors mean lower value capacitors, which in this case have been negated. Don't ask why. I didn't design it.
-Its isolated from the rails. This means you can slam it with (relatively) huge signals and currents without causing the same disruptions in other sections of the circuit. There are limitations here, of course...
-A closer approximation to linear operation can be achieved as a result of the isolation. Think about it. You'll see it.
-Its confusing! Makes people stop and ask why, and anytime that happens, I'd say you've got a winner on your hands.

Thanks for the informative reply! Just out of curiosity, would the results be the same with a BJT or JFET buffer?

newly wrote:

mmolteratx wrote:Just realized I accidentally placed the +9V tap at the +4.5V tap. Anyone reading it should realize the error pretty quickly.

So all "+9V" should connected to "V+" instead right? VB remain as 4.5V.

Yes.

After being hit up by IVIark to see if I had done any more work on this, I cleaned up the schem and did a quick retrace. The original doesn't really do anything with the unused op amp half, but I included it tied to Vr, which is better practice. Result is attached.

Here's the vero




and one with a standard voltage divider for vref, doing away with the second opamp. I'm not convinced it will make a difference in something like this, and so would prefer to save the space.

Only got 1uf electros lying around. Can anyone help tell me which way round they should go until I get replacements?

hey Miles
You can make a quick and dirty 1u nonpolar wannabe capacitor by tying together the negatives of 2 1u electro
then solder positive leg to board

So I just have two together with the positives on the board?
+--+
That's great, thank you : )

yes

^^^^ 2 uf?

I think it would only be 2uf if they were wired in parallel : )

287m wrote:hey Miles
You can make a quick and dirty 1u nonpolar wannabe capacitor by tying together the negatives of 2 1u electro
then solder positive leg to board

No! Capacitances in series adds up like resistances do in parallel. 2*1µF in series would effectively give you 0.5µF.
To get 1µF you'll need two 2µF caps in series

C7 into the volume pot has a -3dB point of 16Hz! You could easily use 470n for the 1u caps in the audio path with no significant change in the (audible) frequency response. You could put two 470n caps in parallel for the "Bass" cap....

Blitz Krieg wrote:^^^^ 2 uf?

no, two 1 uF polarized (+-) wired in series like +--+ , still 1uF

grrrunge wrote:

287m wrote:hey Miles
You can make a quick and dirty 1u nonpolar wannabe capacitor by tying together the negatives of 2 1u electro
then solder positive leg to board

No! Capacitances in series adds up like resistances do in parallel. 2*1µF in series would effectively give you 0.5µF.
To get 1µF you'll need two 2µF caps in series

normally, you right.
normal elco actually conduct in the reverse direction, two 1uF polarized act like they each have a diode in parallel with them that conducts when the voltage is backwards for that one cap.

Where did you learn that? I have never seen any diodes in the equivalent diagram for electrolytics. Only capacitance, inductance and resistances.
As far as i know, the only reason electrolytics are polarised devices, is because improper DC-bias will break down the dielectric layer.