Red Llama Mk.II gut shots.
- The Rotagilla
- Diode Debunker
I wouldn't have bothered but this thing is noticeably quieter at idle than other Llama layouts I've built. The IC is a MC14049UBCP and all the resistors appear to be standard Llama value except for the circled one which if I'm reading the color bands correctly is a 15k or 16k but measures out at 74.8k (it's also late here and I've had a shitty day so the margin for error is pretty large). Anyway, I'm curious if anyone has thoughts as to why the new one is so much quieter. Enjoy.
The television will not be revolutionized.
- Steven_M
- Resistor Ronker
I wonder why they chose to use half watt resistors and those huge axial caps.
Maybe the metal film resistors really reduce the noise. I've only ever used carbon comp in my Llama build, and didn't notice it being particularly noisy. I used the tonepad layout.
Maybe the metal film resistors really reduce the noise. I've only ever used carbon comp in my Llama build, and didn't notice it being particularly noisy. I used the tonepad layout.
- The Rotagilla
- Diode Debunker
Mojo of course! I've used various layouts and there's always been a fair amount of hiss/white noise (even with metal films) at idle with the Llama and like I said, this one is noticeably quieter. The 4049 seems to be wired up a little differently than what I'm used to seeing but I know they can be wired different ways to achieve the same end result. I suck at tracing so if anyone wants to take a crack at it, be my guest.Steven_M wrote:I wonder why they chose to use half watt resistors and those huge axial caps.
Maybe the metal film resistors really reduce the noise. I've only ever used carbon comp in my Llama build, and didn't notice it being particularly noisy. I used the tonepad layout.
Last edited by The Rotagilla on 15 Mar 2012, 17:17, edited 2 times in total.
The television will not be revolutionized.
- The Rotagilla
- Diode Debunker
OP edited.Duckman wrote:Wich one is the circled one?
The television will not be revolutionized.
- The Rotagilla
- Diode Debunker
I appears they are running to a ground plane, yes. Same plane as the negtive side of the 100uf is connected to.Güero 2.0 wrote:Are the unused 4049 pins grounded?
I can't see from the pics.
The television will not be revolutionized.
- The Rotagilla
- Diode Debunker
The mystery resistor is a 75k, just needed some good light to see it.
The television will not be revolutionized.
Information
Duckman wrote:Seems like this to me (for a start) Pins 7, 9, 11 and 14 are not grounded
I agree. It looks like they are tied to 9v.
- Duckman
- Opamp Operator
Is that a better way to get rid of the unused OA's?madbean wrote:Duckman wrote:Seems like this to me (for a start) Pins 7, 9, 11 and 14 are not grounded
I agree. It looks like they are tied to 9v.
- The Rotagilla
- Diode Debunker
I am assuming C3 = 51pf and C4 = 100pf as in the original. C5 = 100uf and C6 = 10uf tantalum.Duckman wrote:Hey Rotagilla, can you post the values for C3, 4, 5 & 6? Just to complete the picture.
The television will not be revolutionized.
- Duckman
- Opamp Operator
Ok, that's nice. But, since the unit has some custom tracing, can you confirm it?... please?The Rotagilla wrote:I am assuming C3 = 51pf and C4 = 100pf as in the original. C5 = 100uf and C6 = 10uf tantalum.Duckman wrote:Hey Rotagilla, can you post the values for C3, 4, 5 & 6? Just to complete the picture.
- The Rotagilla
- Diode Debunker
All values have been confirmed.Duckman wrote:Ok, that's nice. But, since the unit has some custom tracing, can you confirm it?... please?The Rotagilla wrote:I am assuming C3 = 51pf and C4 = 100pf as in the original. C5 = 100uf and C6 = 10uf tantalum.Duckman wrote:Hey Rotagilla, can you post the values for C3, 4, 5 & 6? Just to complete the picture.
The television will not be revolutionized.
- Groovenut
- Resistor Ronker
Depending on the circuit (ie the number of power connections) the filtered 9V rail can be considerable quieter than ground. Since it is a stable dc voltage, it has the same function as ground to signals.Duckman wrote:Is that a better way to get rid of the unused OA's?madbean wrote:Duckman wrote:Seems like this to me (for a start) Pins 7, 9, 11 and 14 are not grounded
I agree. It looks like they are tied to 9v.
- jonasx26
- Breadboard Brother
Information
Doesn't matter if you tie unused gates to ground or VCC.Is that a better way to get rid of the unused OA's?
The whole point of terminating the unused gates is to keep them at a fixed high/low-state.
Without a solid input signal the high impedance gates can pick up random noise and oscillate.
There MIGHT be a difference in whether you terminate them at VCC or ground within a certain pcb layout.
Look at the internal schematic of a 4049 inverter, see where the currents flow, and keep this in mind while designing the pcb..
@jonasjberg http://futileresistancefx.blogspot.com
- Liquids
- Breadboard Brother
You connect unused CMOS stage's inputs to one of the rails to reduce overall current draw, in addition to the above. Leave the output of an unused CMOS stage's output connected to nothing if unused.jonasx26 wrote:Doesn't matter if you tie unused gates to ground or VCC.Is that a better way to get rid of the unused OA's?
The whole point of terminating the unused gates is to keep them at a fixed high/low-state.
Without a solid input signal the high impedance gates can pick up random noise and oscillate.
There MIGHT be a difference in whether you terminate them at VCC or ground within a certain pcb layout.
Look at the internal schematic of a 4049 inverter, see where the currents flow, and keep this in mind while designing the pcb..
V+ is rarely as stable a voltage as we'd like it to be or try to make it given signal transients - which in some cases are clipping rail to rail. Ground is usually (if everything is properly wired) a bit closer to stability and 'less prone to noise' (!)....ground has less current potential than any other voltage.
In the end, connecting the unused CMOS stages input to either rail or the other is far superior to leaving unused CMOS gates 'floating.'
Some familiar designs utilize bipolar (+/-) voltages, and the CMOS devices see ground as their V+ and a negative voltage on their 'most negative rail voltage' pin. I'd still go with utilizing ground as the best place to connect unused inputs in this case.
As a side note, I hope no one reading this assumes that best-practice is the same for an unused op amp stage: an unused op amp stage should be wired like a buffer (Connect output to negative input), and the non-inverting input is best tied to the CENTER point between the two rails - in a typical 9v pedal, that's VREF (4.5v) etc, in a bipolar configuration that is ground, etc. Or even better, use that op amp stage. Because while CMOS gates are often more plentiful, they are not as 'swiss army knife' as an op amp stage, which can do and improve many things in a circuit.