What’s going on with the inductor in the Bobtavia?

[soggybag]

The Bobtavia was the first or second pedal I built. Sounds great easy to get parts for it.

What's going on with the inductor/transformer here? In particular what is happening at the center tap?

[Vivek]

At first look

It appears to be a full wave rectifier and that creates 2f components

The connection at the centre tap seems to bias the diodes a little bit, to get them to conduct faster upon receipt of Audio signal

And there might be a very slight effect of saturating transformer core


Plus, some kind of a ground reference for the transformer output, but I wonder why it's high impedance. Wonder what would happen if there was a cap across the 100k

[Vivek]

Level pot might have DC going through it, leading to a bit of crackle / faster wearing away of pot

[ElectricDruid]

It might need some diode bias to overcome the double forward voltage drop of the series diodes. Which is an odd choice that I don't understand. Usually with Octavia style circuits, you want diodes with as *low* a forward voltage as you can get - hence germaniums are often used. Using silicon and then using a series pair of them seems to be going in entirely the wrong direction! But maybe after that 386 amp there's enough level it doesn't care?

[Vivek]

Voltage Level on transformer's secondary depends upon turn ratio of transformer

That bias on 2 diodes could lead to a different final output waveshape than without bias. Maybe some kind of signal cancellation at low outputs, leading maybe to something like a crossover distortion.

I will try to SPICE it when I next have time.

[duck_arse]

four diodes coming off a transformer winding - doesn't it suggest a full wave bridge rectifier drawn wrong?

[soggybag]

I'm curious about the basic concepts involved.

If I understand correctly, the AC signal coming in on the left side is coming out inverted on the right side.

I think this transformer is 1k:8ohms. Looks like the 386 is feeding the 8ohm side. So the signal should be stepped up on the 1K side. What happens here? Is the AC signal boosted by 1000/8 ?

I haven't been considering the center tap so far. With the center tap is output at the top and bottom of te left side just offset by the voltage at the center tap?

I'm guessing we get the octave since an inverted copy of the signal is sent through each of the two sets do diodes. The diodes block the negative half of each wave form. The result is passively mixed at the volume pot.

[PRR]

Output jack "tip" surely drawn wrong?

The diodes are DC-biased to conduct all the time, only more/less left/right with signal. No DC in pot wiper so no crackle. Gain from input to 386 out is 20, to one side of transformer about 100, so a 10mV guitar signal will be switching double-diodes off?

[Vivek]

Normally each side of a transformer has wires of different thickness

Hence a 1:X resistance ratio does not imply 1:X voltage ratio



If the diodes are already biased on, the transformer does not need to surpass the total Vf

It is operating in the range of "slightly on" to "slightly more on"


It is not operating in the region of "off" and "on" where it needs to exceed total Vf of the diodes.

[soggybag]

It's the center tap creating the DC bias for the diode?

If that's the case would it help to have a trimmer on the center tap to find a sweet spot?

[antonis]

It's the center tap creating the DC bias for the diode?

It's the power supply (Battery (+)) through 100k resistor and 50k Level pot..
So, a DC current of about 50μΑ is permanently flowing through diodes, for them to be slightly "on"..
(same case as in almost all push-pull outputs for crossover distortion prevention..)

[soggybag]

Give me the math, how did you figure that out?

[antonis]

9V - 2 X Diode forward voltage drop = I X (100k +50K) => I =  0.000052A..
Equally splitted into both windings (of negligible resistance) and summed on diodes cathodes (Level pot lug 3)

[Vivek]

But Antonis, I feel that in this config, the DC bias will create more crossover distortion. I feel more bias current will lead to more crossover distortion



In any case, I feel that a cap across the 100k would help put more AC voltage across the pot

[Rob Strand]

But Antonis, I feel that in this config, the DC bias will create more crossover distortion. I feel more bias current will lead to more crossover distortion

Some circuits just use germanium diode to reduce the diode drops and dead zones.

This design goes in the other direction and uses two silicon diodes in series, much higher drop, but then tries to undo the dead-band issue by biasing the diode.   I think this was done very much intentionally.  Perhaps it sounded better to the designer.

You could go back to using germaniums but I suspect that's something the designer already tried.  At the end of the day you could do your own experiment.   You could also try increasing the 100k to 1M or 10M.  Try single silicon diodes or schottkys.

The DC through the pot could be handled differently.   For example replace it with a fixed 47k to ground then capacitively couple that to the pot; technically a larger value but it probably doesn't matter.

There's no resistor to ground on the 0.047uF cap that goes to the footswitch.

[PRR]

Normally each side of a transformer has wires of different thickness
Hence a 1:X resistance ratio does not imply 1:X voltage ratio

It is not about wire thickness. At all.

Transformer transforms Voltage AND Current.

2X voltage and 1/2 current is 4X resistance.

Even that is meaningless here, except as a cheap way to get >>1.2V voltage swing with equal-but-opposite voltage sources.

Yes, it distorts!! Isn't that the point?

Why do we want "more AC voltage across the pot"? Is it not loud enough?

[Vivek]

Does a 1:X resistance ratio of a transformer guarantee a 1:X voltage ratio ?

[Rob Strand]

Does a 1:X resistance ratio of a transformer guarantee a 1:X voltage ratio ?

1:X voltage ratio  => X:1 current ratio =>  1:X^2 resistance ratio

*Both* voltage ratio and resistance ratio are idealized concepts.   If the voltage ratio approaches ideal then the resistance ratio will approach ideal.

A real transformer has winding resistances (R1, R2), leakage inductance (Xl1, Xl2),  magnetization inductance (Bm), and core-losses from eddy currents and hysteresis (Gi).  The ideal transformer is embedded inside these elements.
(BTW, the labels on this diagram are impedances/admittances.  The part values would be Ll1, Ll2, Lm, Ri or Rm).


With no load the voltage ratio of the transformer is differs from the turns ratio due to the divider formed by the magnetization inductor and core losses.   When you pull current you get additional voltage drops through R1, X1 before the ideal transformer and R2 and X2 after the transformer.

This should be no surprise as you see the output voltage of a power transformer drop under load.   The turns ratio is fixed but the external voltage drops gives the appearance of different input to output voltage ratios under load.

The whole idea of the transformer in this circuit is to boost the voltage *so that* the voltage at the output of the transformer is high compared to the diode drops.   This is a very normal thing to do on old-school circuits which use diodes for rectification (not just pedals but things like capacitance meters where the voltage needs to be accurate).   The full-wave rectifier only has one diode drop instead of two diode drops for a bridge rectifier.   That's the motivation behind choosing this type of circuit.

[PRR]

Does a 1:X resistance ratio of a transformer guarantee a 1:X voltage ratio ?

No.

That may sometimes be optimum. It sure is not inevitable.

[Vivek]

The original poster wrote:

I think this transformer is 1k:8ohms. Looks like the 386 is feeding the 8ohm side. So the signal should be stepped up on the 1K side. What happens here? Is the AC signal boosted by 1000/8 ?


To which I replied :

X:1 resistance ratio of transformer does not imply X:1 voltage ratio