Walk me thru this circuit....

[swinginguitar]

Over the weekend I built the Ruby 386 based guitar amp (with the bassman mods listed on runoffgroove):



I'd like to understand a little more about what each component (particularly the various caps) contributes. Specifically:

- the 100uF from rail to rail
- the 3k9/47n combo after the buffer (47n changes to .1uF with the bassman mod)
- 100n from pin 7 to ground
- 10ohm/47n combo and 220u on the output
- what does the 220 across the volume pot (bassman mod) do....treble bleed as the volume turns down?
- this amp sounds a bit bassy/dark to me. which cap do i tweak to change that....the 220 output cap?
- i used a 100K volume pot (all i had lying around)...how does that affect the sound vs the specified 10k?

[petemoore]

- the 100uF from rail to rail
  Smooths the DC.
  - the 3k9/47n combo after the buffer (47n changes to .1uF with the bassman mod)
  the 3k9 sets up the Jfet bias/buffer output impedance, the cap is a DC blocker [which can', depending on if it's small enough, attenuate bass signals]- 100n from pin 7 to ground
- 10ohm/47n combo and 220u on the output
  This string is allowing certain HF's to be shunted to ground.
- what does the 220 across the volume pot (bassman mod) do....treble bleed as the volume turns down?
  There's no cap on the 'volume', the 220uf lets AC frequencies drive the speaker, blocks DC from speaker coil.
- this amp sounds a bit bassy/dark to me. which cap do i tweak to change that....the 220 output cap?
  That would load the output increasingly as frequencies drop [ie could reduce bass], before the amplifier is probably a better location for frequency control, the 47n looks like the only existing point for rolling bass back [smaller value there].
- i used a 100K volume pot (all i had lying around)...how does that affect the sound vs the specified 10k?
  10k is bigger load = slight volume drop.

[anchovie]

+1 on lowering the value of the cap to the 386 input - it's forming a high-pass filter with the parallel combination of the volume pot and chip input impedance. By increasing the volume pot value you're also letting a lot more bass through. Try 22n as you're aiming for the Bassman version and see how that sounds.

[swinginguitar]

- what does the 220 across the volume pot (bassman mod) do....treble bleed as the volume turns down?
  There's no cap on the 'volume', the 220uf lets AC frequencies drive the speaker, blocks DC from speaker coil.

The Bassman mod (not shown on the schematic) calls for a 220 across the volume pot....pretty sure it's a treble bleed

- i used a 100K volume pot (all i had lying around)...how does that affect the sound vs the specified 10k?
  10k is bigger load = slight volume drop.

10k is a bigger load than 100k?

What about the cap from pin 7 to ground?

@anchovie: the Bassman mod calls for a .1uF in place of the 47nF...how would your suggestion of 22n sound vs that? (I'm using the .1uF at the moment...guess that's a little more bassy huh)?

[anchovie]

@anchovie: the Bassman mod calls for a .1uF in place of the 47nF...how would your suggestion of 22n sound vs that? (I'm using the .1uF at the moment...guess that's a little more bassy huh)?

It's a rough estimate based on the idea of the volume pot resistance being in parallel with the chip's 50k input impedance.

10k||50k = 8.3k
100k||50k = 33.3k

So with the 100k pot the resistance is roughly 4 times bigger, therefore you should make the 100nF 'Bassman' cap roughly 4 times smaller to have the filter roll off at roundabout the same frequency.

The cap on the volume pot is indeed a treble bleed.

Look at the LM386 datasheet for info on the pin 7 cap.

[edvard]

10k is a bigger load than 100k?

Yes.
You're making the same mistake I did when I first started to learn electronics.
I thought when a resistance was high, it took more effort to push the electrons through the thing, therefore the driving force (pickups, preamp, etc.) had a higher load to push against.
Almost the exact opposite is true.
To give a very visual but probably inaccurate analogy, imagine the input resistance (impedance) as the "road" your signal is "walking" on.
The higher the impedance, the firmer the road, therefore the easier time your signal has of it; a smaller load.
The lower the impedance, the softer and muddier the road, which means a larger load.
Hope that helps...

[swinginguitar]

10k is a bigger load than 100k?

Yes.
You're making the same mistake I did when I first started to learn electronics.
I thought when a resistance was high, it took more effort to push the electrons through the thing, therefore the driving force (pickups, preamp, etc.) had a higher load to push against.
Almost the exact opposite is true.
To give a very visual but probably inaccurate analogy, imagine the input resistance (impedance) as the "road" your signal is "walking" on.
The higher the impedance, the firmer the road, therefore the easier time your signal has of it; a smaller load.
The lower the impedance, the softer and muddier the road, which means a larger load.
Hope that helps...

Ouch...OK...I'm trying....the analogy makes sense, but can you explain in a little more detail?

Maybe V=IR? ...increasing the resistance yields a higher voltage? 

[CynicalMan]

Maybe V=IR? ...increasing the resistance yields a higher voltage?  

V = IR. If voltage is constant, current and resistance are inversely proportional. That means that the lower the resistance is, the more current there is. So, if a source is outputting a signal in to a low resistance, it has to supply more current. If the same voltage is going into a higher resistance, it has to supply less current.

Another aspect of loading is that every source has an output impedance, which is similar to a resistance. This makes a voltage divider with the load. That means that the lower the load resistance is, the more current there is being diverted to ground, and the lower the output. When the load is higher, less current is diverted to ground.

[ashcat_lt]

Ouch...OK...I'm trying....the analogy makes sense, but can you explain in a little more detail?

Maybe V=IR? ...increasing the resistance yields a higher voltage? 

Almost, but let's rearrange it.

I = V / R

If the voltage remains the same, and the resistance gets smaller, more current is demanded.  This demand for current is what we call the Load.  More current = greater load.

[edvard]

I should have added in my first reply that by impedance I mean the implied resistance between your guitar signal and ground, not the resistance into the circuit (although that may be taken into consideration as well...).
The input to your circuit is a high-impedance type, as jfets have a rather large internal impedance and the bias resistor is a nice comfy 1.5 Megs.
The 10k volume pot in question may be just fine, as it is preceded by the jfet acting as a low-output-impedance buffer, which would be much less susceptible to loading effects.
The 'slight volume drop' mentioned would then be a function of the parallel resistance formed by the pot and the 3.9k source resistor, which would be ~2.8k at full volume.
With a 100k pot, you'd get ~3.8k.
The best thing to do would be to wire it up and see how it sounds.

- 100n from pin 7 to ground

According to National Semiconductor's datasheet, pin 7 has the function of "bypass" and almost always has a capacitor to ground, sometimes an unspecified value.
I suppose it may help suppress oscillations at higher gains, but I could be way off...

[earthtonesaudio]

- 100n from pin 7 to ground

According to National Semiconductor's datasheet, pin 7 has the function of "bypass" and almost always has a capacitor to ground, sometimes an unspecified value.
I suppose it may help suppress oscillations at higher gains, but I could be way off...

This cap partially bypasses the emitter resistance of the PNP-based long-tailed pair at the input.  Adding the cap in this location shunts much of the power supply noise to ground which reduces noise for the input stage, and also increases AC gain for the input stage.