Hi,
I built a germanium fuzz face according to this schematic, and I am trying to understand its behavior in the soft clipping region.
In particular, at around 415Hz 10uA p-p input, gain at minimum setting, this kind of thing happens in my circuit:
Yellow is Vc1/Vb2, blue is Ve2.
I'm very surprised that this can happen in a circuit with lots of negative feedback!
If I ignore the feedback resistor, I can maybe convince myself that it makes sense:
As Vc1 approaches ground, Ie2 must decrease to keep Ve2 at a similar voltage due to the 1K emitter resistor. However, this means Ib2 must decrease proportionally. When Ib2 is small enough, Q2's logarithmic BE diode drop starts to shrink appreciably, so positive peaks of Ve2 will be soft-clipped at Vc2 (meaning negative peaks in Ib1 will be soft-clipped so long as Ve2 is still less negative than Q1's BE voltage).
Maybe another type of soft clipping also happens as Q1 is entering saturation.
So... I can believe that Ib1 is getting soft clipped once or twice by the time it becomes Ie2. However, the negative feedback seems to present a problem. It seems the amount of negative feedback is fairly high (beta1*33K/100K ~= beta1/3). If Vc2 is an inverted, soft-clipped Ib1, then I would think Ib1 must have exaggerated negative peaks compared to the input current, driving any sources of soft clipping an order of magnitude harder than it would without NFB. Surely this must be enough to reach hard clipping! How does this happen?
The fuzz face site says the following:
I'm not sure how well I understand this. Clipped positive peaks of Vc2 would generally make the bias Ib1 (DC) more negative than the quiescent bias. However, I guess exaggerated (via NFB) positive peaks that were then clipped may be a different story. They may have a larger total area despite a smaller maximum value than the negative peaks, and then result in average Ib1 being less negative than the quiescent bias. This would then move the negative peaks of Ib1 away from the clipping region, softening the output.
(Which is to say, B may have more area than A.)
I also found the same behavior down at 20Hz, so I'm hesitant to start thinking about parasitic capacitances.
This is my best guess as to what could be producing the output I see, but I have no idea if I'm even close to the right track! I came up with several alternative explanations before this one, but then decided they didn't make sense. Any input would be helpful here.
I built a germanium fuzz face according to this schematic, and I am trying to understand its behavior in the soft clipping region.
In particular, at around 415Hz 10uA p-p input, gain at minimum setting, this kind of thing happens in my circuit:
Yellow is Vc1/Vb2, blue is Ve2.
I'm very surprised that this can happen in a circuit with lots of negative feedback!
If I ignore the feedback resistor, I can maybe convince myself that it makes sense:
As Vc1 approaches ground, Ie2 must decrease to keep Ve2 at a similar voltage due to the 1K emitter resistor. However, this means Ib2 must decrease proportionally. When Ib2 is small enough, Q2's logarithmic BE diode drop starts to shrink appreciably, so positive peaks of Ve2 will be soft-clipped at Vc2 (meaning negative peaks in Ib1 will be soft-clipped so long as Ve2 is still less negative than Q1's BE voltage).
Maybe another type of soft clipping also happens as Q1 is entering saturation.
So... I can believe that Ib1 is getting soft clipped once or twice by the time it becomes Ie2. However, the negative feedback seems to present a problem. It seems the amount of negative feedback is fairly high (beta1*33K/100K ~= beta1/3). If Vc2 is an inverted, soft-clipped Ib1, then I would think Ib1 must have exaggerated negative peaks compared to the input current, driving any sources of soft clipping an order of magnitude harder than it would without NFB. Surely this must be enough to reach hard clipping! How does this happen?
The fuzz face site says the following:
I'm not sure how well I understand this. Clipped positive peaks of Vc2 would generally make the bias Ib1 (DC) more negative than the quiescent bias. However, I guess exaggerated (via NFB) positive peaks that were then clipped may be a different story. They may have a larger total area despite a smaller maximum value than the negative peaks, and then result in average Ib1 being less negative than the quiescent bias. This would then move the negative peaks of Ib1 away from the clipping region, softening the output.
(Which is to say, B may have more area than A.)
I also found the same behavior down at 20Hz, so I'm hesitant to start thinking about parasitic capacitances.
This is my best guess as to what could be producing the output I see, but I have no idea if I'm even close to the right track! I came up with several alternative explanations before this one, but then decided they didn't make sense. Any input would be helpful here.
The soft clipping is only working on the positive half of the cycle.
I found a soft limiter in an old Wireless World mag.
It uses two transistors in the feedback loop of an op-amp.
The limiting is controlled by a 2m2 pot.
I found a soft limiter in an old Wireless World mag.
It uses two transistors in the feedback loop of an op-amp.
The limiting is controlled by a 2m2 pot.
Germanium transistors have a very nonlinear, curved knee region, and will soft limit quite nicely when asymmetrically biased toward saturation. OP mentioned NFB, but we need to keep in mind that NFB only works correctly under linear operation. As soon as you get into much nonlinearity, all bets are off.
Sent from my phone with Tapatalk. Please excuse any typpos.
Sent from my phone with Tapatalk. Please excuse any typpos.
Hmmm... I guess there is an important difference between a gain stage that computes, say, A*(x+2x^3) (which is definitely nonlinear, but I would completely trust NFB to linearize very accurately), and a germanium with current gain of less than 100, in a feedback setup with current gain of (-beta/3). The difference being the first example has high positive gain, however variable, throughout the operating region, but the second example does not. As instantaneous current gain decreases due to saturation, it will quickly realistically become small enough that there is no more high gain limiting case. So then while NFB will try very hard to sharpen the knee by boosting (or I guess more accurately refusing to attenuate) the size of the input peak, it can't overpower the natural output in the same way as before, and may only get halfway there, which could still explain the curves I see. Even if the NFB waveform was hard-clipped, the output of Q1 would still be soft-clipped.
So I can see why transistors with relatively low beta are selected for Q1!
And I guess the soft clipping I am seeing is largely the result of Q1's saturation and not so much the Q2 diode thing.
I'll post some more conclusions I have reached...
If anyone can confirm, deny, or fill in some of my question marks, that would be super helpful. With multiple sources of nonlinearity, it's hard for me to know if I am right or not.
If my investigations can be useful to someone, that's great, but please don't take these as fact! I actually have no idea what I'm doing.
1.
Q1 is often operating with very small small-signal current-gain. In one test, gain pot at minimum, I sent low voltage signals through a 33K input resistor and found Vc1(AC) to be around twice the magnitude of the input (rather than the expected near-thrice) regardless of input frequency. This would seem to indicate that my Q1's differential current gain, under these conditions, is around 6. (hfe*(vin - vf/3) = vf and vf=2vin). For comparison, the measured beta (with leakage current measured and subtracted) under other conditions was over 50. Maybe this could be explained by differences in bias current, or by the measuring method being flawed in assuming leakage and legitimate current are just added linearly (which is clearly not true if beta is not constant).
On heating Q1, hfe1 apparently further decreases to around 3, despite no obvious clipping effects. The direction and magnitude of this change are both contrary to expectations. The next point will attempt to explain this.
2.
The biasing arrangement mostly seems to be trying to fix Vc1 to around 2 diode drops, and Vb1 to around 1 diode drop. However, quiescent base current is not enough to saturate the BE diodes, so there is only around -80mV per BE junction. Increasing hFE1 and/or Q1's leakage current should make VBE1 less negative in general.
I measured around 10mV across the 100K resistor (0.1uA), yet in order for there to be nearly 9V across the 33K collector resistor, Ic1 must be around 250mA (much more than 1uA*hFE). This indicates to me that the vast majority of Ib1 is collector-base leakage, and Q1's current gain plays an insignificant part in determining the DC bias VBE1.
Heating Q1 should exponentially increase the leakage Ibc and thus decrease -VBE1. In my experiment, touching Q1 briefly indeed sent VBE1 from -95mV to -80mV.
This change in bias current could be modifying hfe, but it could also possibly modify input impedance by a significant amount since the bias voltage is so small. To eliminate this possibility, I fed a ~1V signal through a 1M input resistor, turned the gain up (but not to the point of much distortion), and still observed that touching Q1 quickly leads to around 50% attenuation of vc2.
So, I concluded that the extreme decrease in hfe is probably the result of increased leakage leading to colder biasing via NFB.
3.
Q1 will fully saturate around ic1 = 250uA. When hfe is relatively small, NFB has a diminished linearizing effect and soft clipping can happen. If it isn't small to start out, it will become small as ic1 increases. Also, when hfe is relatively small, changes in hfe will change the current gain of this stage to something lower than 3 rather than it being asymptotically fixed at 3 by NFB. So a leaky, low-gain transistor should provide the softest clipping here.
Before this, Q2 may also show signs of soft clipping. This should be in the form of vbe shrinking as vb2 approaches vbe2 (and ve2 approaches 0). Vbe should in general become smaller as Q2's gain increases, although it's logarithmic, so the change might be small. I think Q2's leakage does not matter much, because however the proper collector current is created, through external base current or through leakage from base to collector, both ways involve the base-emitter junction in their path and therefore should lead to the same base-emitter diode drop. This is somewhat confirmed by experiment as heating Q2, which should greatly modify the leakage, does not seem to change the circuit's behavior in any way I can notice.
4.
When gain is turned up, there are some effects due to the finite time constant of the gain pot/cap. The effects are subtle, so for this scope plot I replaced the 20uF cap with 1uF to exaggerate them.
Blue is ve2, yellow is vc2/10.
Let g be the setting of the gain pot. As Q2 is able to keep ve2 near vb2, the capacitor voltage will tend toward g*ve2 with time constant (g*R || (g-1)R)C. For small signals, the visible effects on phase and magnitude are relatively small.
The interesting part occurs when Q2 enters cutoff. Since Q2 can't sink negative current, ve2 will be approximately the capacitor voltage, rather than vb2 or 0, and tends toward 0 with time constant g*R*C. Q2 will be in cutoff so long as vb2 approaches 0 faster than this capacitor discharge curve. (So, having a large positive slope matters but being close to 0 also matters). As Q2 exits cutoff (vb2 ~= ve2 ~= vcap), the capacitor, charged to vb2, will be fairly negative with respect to g*vb2, so next it begins drifting toward there like before. This leads to current starting to flow through Q2, and vbe becoming nonzero. The visible weirdness that results from this are that while ve2 has a positive slope during cutoff, ic2/vc2 have 0 slope, and then ve2 also has a positive slope while the capacitor re-approaches equilibrium after cutoff, but so does ic2/vc2 (which is generally supposed to be inverted).
The lack of inversion may not make a difference in the shape of the output for a sine wave input (which is symmetrical under time-reversal), but on something like a sawtooth wave input, it could. Whether there is an audible difference produced on instrument input is unknown, but it's worth noting that transients and bowed strings have a lot in common with sawtooths.
Something similar happens when Q2 enters saturation.
5.
On the positive peaks of the input signal, at higher levels, Q2 will saturate and clip hard. Because the clipping is asymmetrical, and the negative peaks are clipped to a smaller magnitude, the DC component of the feedback will move more negative to try to correct it, causing many regions (including originally positive regions) of the input to be clipped to a negative voltage, and relatively few regions to be clipped to a positive voltage, even at square wave levels.
This effect is not necessarily unique to DC feedback, and a similar effect probably occurs in many other circuits that have an asymmetrical clipping gain stage AC-coupled to a gain stage that clips the other side. It is however pretty interesting to me that the fuzz face accomplishes it with a different topology. If I wanted to design a distortion circuit that messes with the width of peaks, I would almost certainly think of 2 gain stages and a coupling capacitor, not the fuzz face circuit.
If anyone can confirm, deny, or fill in some of my question marks, that would be super helpful. With multiple sources of nonlinearity, it's hard for me to know if I am right or not.
If my investigations can be useful to someone, that's great, but please don't take these as fact! I actually have no idea what I'm doing.
1.
Q1 is often operating with very small small-signal current-gain. In one test, gain pot at minimum, I sent low voltage signals through a 33K input resistor and found Vc1(AC) to be around twice the magnitude of the input (rather than the expected near-thrice) regardless of input frequency. This would seem to indicate that my Q1's differential current gain, under these conditions, is around 6. (hfe*(vin - vf/3) = vf and vf=2vin). For comparison, the measured beta (with leakage current measured and subtracted) under other conditions was over 50. Maybe this could be explained by differences in bias current, or by the measuring method being flawed in assuming leakage and legitimate current are just added linearly (which is clearly not true if beta is not constant).
On heating Q1, hfe1 apparently further decreases to around 3, despite no obvious clipping effects. The direction and magnitude of this change are both contrary to expectations. The next point will attempt to explain this.
2.
The biasing arrangement mostly seems to be trying to fix Vc1 to around 2 diode drops, and Vb1 to around 1 diode drop. However, quiescent base current is not enough to saturate the BE diodes, so there is only around -80mV per BE junction. Increasing hFE1 and/or Q1's leakage current should make VBE1 less negative in general.
I measured around 10mV across the 100K resistor (0.1uA), yet in order for there to be nearly 9V across the 33K collector resistor, Ic1 must be around 250mA (much more than 1uA*hFE). This indicates to me that the vast majority of Ib1 is collector-base leakage, and Q1's current gain plays an insignificant part in determining the DC bias VBE1.
Heating Q1 should exponentially increase the leakage Ibc and thus decrease -VBE1. In my experiment, touching Q1 briefly indeed sent VBE1 from -95mV to -80mV.
This change in bias current could be modifying hfe, but it could also possibly modify input impedance by a significant amount since the bias voltage is so small. To eliminate this possibility, I fed a ~1V signal through a 1M input resistor, turned the gain up (but not to the point of much distortion), and still observed that touching Q1 quickly leads to around 50% attenuation of vc2.
So, I concluded that the extreme decrease in hfe is probably the result of increased leakage leading to colder biasing via NFB.
3.
Q1 will fully saturate around ic1 = 250uA. When hfe is relatively small, NFB has a diminished linearizing effect and soft clipping can happen. If it isn't small to start out, it will become small as ic1 increases. Also, when hfe is relatively small, changes in hfe will change the current gain of this stage to something lower than 3 rather than it being asymptotically fixed at 3 by NFB. So a leaky, low-gain transistor should provide the softest clipping here.
Before this, Q2 may also show signs of soft clipping. This should be in the form of vbe shrinking as vb2 approaches vbe2 (and ve2 approaches 0). Vbe should in general become smaller as Q2's gain increases, although it's logarithmic, so the change might be small. I think Q2's leakage does not matter much, because however the proper collector current is created, through external base current or through leakage from base to collector, both ways involve the base-emitter junction in their path and therefore should lead to the same base-emitter diode drop. This is somewhat confirmed by experiment as heating Q2, which should greatly modify the leakage, does not seem to change the circuit's behavior in any way I can notice.
4.
When gain is turned up, there are some effects due to the finite time constant of the gain pot/cap. The effects are subtle, so for this scope plot I replaced the 20uF cap with 1uF to exaggerate them.
Blue is ve2, yellow is vc2/10.
Let g be the setting of the gain pot. As Q2 is able to keep ve2 near vb2, the capacitor voltage will tend toward g*ve2 with time constant (g*R || (g-1)R)C. For small signals, the visible effects on phase and magnitude are relatively small.
The interesting part occurs when Q2 enters cutoff. Since Q2 can't sink negative current, ve2 will be approximately the capacitor voltage, rather than vb2 or 0, and tends toward 0 with time constant g*R*C. Q2 will be in cutoff so long as vb2 approaches 0 faster than this capacitor discharge curve. (So, having a large positive slope matters but being close to 0 also matters). As Q2 exits cutoff (vb2 ~= ve2 ~= vcap), the capacitor, charged to vb2, will be fairly negative with respect to g*vb2, so next it begins drifting toward there like before. This leads to current starting to flow through Q2, and vbe becoming nonzero. The visible weirdness that results from this are that while ve2 has a positive slope during cutoff, ic2/vc2 have 0 slope, and then ve2 also has a positive slope while the capacitor re-approaches equilibrium after cutoff, but so does ic2/vc2 (which is generally supposed to be inverted).
The lack of inversion may not make a difference in the shape of the output for a sine wave input (which is symmetrical under time-reversal), but on something like a sawtooth wave input, it could. Whether there is an audible difference produced on instrument input is unknown, but it's worth noting that transients and bowed strings have a lot in common with sawtooths.
Something similar happens when Q2 enters saturation.
5.
On the positive peaks of the input signal, at higher levels, Q2 will saturate and clip hard. Because the clipping is asymmetrical, and the negative peaks are clipped to a smaller magnitude, the DC component of the feedback will move more negative to try to correct it, causing many regions (including originally positive regions) of the input to be clipped to a negative voltage, and relatively few regions to be clipped to a positive voltage, even at square wave levels.
This effect is not necessarily unique to DC feedback, and a similar effect probably occurs in many other circuits that have an asymmetrical clipping gain stage AC-coupled to a gain stage that clips the other side. It is however pretty interesting to me that the fuzz face accomplishes it with a different topology. If I wanted to design a distortion circuit that messes with the width of peaks, I would almost certainly think of 2 gain stages and a coupling capacitor, not the fuzz face circuit.
> only around -80mV per BE junction.
This is reasonable for many Ge devices at these small currents.
Especially in the area where Ico leakage dominates design bias.
If built with LOW-leakage (Si) devices, voltage across the 100K can be accurately estimated. However with leaky Ge this voltage may go either way. With device change, and with temperature. This is a "good" circuit when the "100K" can be scaled down to 10K, as in line or tape-head inputs, and is often done (without a divider in Q2 Collector) as a general amplifier. When scaled to part-mA currents it does get device dependant.
> think of 2 gain stages and a coupling capacitor, not the fuzz face circuit.
It "is" two gain stages and a cap. The Q2 Emitter cap will charge up/down with signal asymmetry, just like a coupling cap will. Ignoring the need for DC bias, a cap of 1/hFE the value could be put in series with Q2 Base and give similar time-constant.
This is reasonable for many Ge devices at these small currents.
Especially in the area where Ico leakage dominates design bias.
If built with LOW-leakage (Si) devices, voltage across the 100K can be accurately estimated. However with leaky Ge this voltage may go either way. With device change, and with temperature. This is a "good" circuit when the "100K" can be scaled down to 10K, as in line or tape-head inputs, and is often done (without a divider in Q2 Collector) as a general amplifier. When scaled to part-mA currents it does get device dependant.
> think of 2 gain stages and a coupling capacitor, not the fuzz face circuit.
It "is" two gain stages and a cap. The Q2 Emitter cap will charge up/down with signal asymmetry, just like a coupling cap will. Ignoring the need for DC bias, a cap of 1/hFE the value could be put in series with Q2 Base and give similar time-constant.
But in this case, DC being present in the feedback seems to cause #5 even when the gain pot is at 0! This suggests the emitter cap is not involved at all, and that's why it was pretty exciting to me.
I found that turning the gain pot up does not noticeably affect the peak widths either, despite creating a lot of clipping at Q2's collector. On the other hand increasing input voltage amplitude through the same input resistor by an equivalent amount does change the peak widths considerably.
I think this is because Q2's emitter is a (sort of) faithful voltage buffer in the soft clipping region, regardless of gain pot setting. The #4 effect doesn't change the DC voltage by that much. Then at higher settings, when Q2 saturates on the opposite peaks, the emitter follower still isn't doing that much of DC interest (now both peaks have some weird diagonal RC thing that approaches a relatively low level triangle-ish wave). There is definitely distortion, and probably a DC component, but turning the gain pot to produce 10 times the output still leaves ve2 at the same order of magnitude. So the feedback effects are pretty negligible compared to amplifying the input signal (which is directly mirrored by ve2).
Anyway, I think the change in peak widths is caused by a current path I originally neglected - the input coupling cap beginning to charge toward V = VE2 with tau = 2.2uF*(Rf+Rpickup) >~= 220ms, a condition which seems to end up being met much faster than O(220ms) in reality because as the capacitor voltage moves toward VE2, the exponential-law voltage amplification of the Q1 stage also works to move VE2 toward the capacitor voltage. In my experimentation, the output waveform already seemed to reach its approximate final shape, with asymmetric peak widths, after 1 cycle of a 415Hz input being turned on.
I wasn't thinking about what was charging and discharging until your post. I just said "the input is AC coupled to B1 and the feedback passes both AC and DC", and then imagined what happens to an AC input when the DC level changes. But I guess this isn't really correct, because inputting an AC signal at DC 0 through only a resistor would not produce the same result - clearly there is no way peaks could affect valleys if there are no reactive components anywhere.
Something I neglected in #2 is the temperature dependence of the V-I relation for the BE1 diode.
Contradictions in my previous thinking include: Increased leakage causing IBE to become closer to 0 doesn't seem to make much sense (NFB should reduce the effect of increased leakage, rather than reverse it). Also, IBE becoming less negative (meaning Vc1 becoming more negative, at a much bigger rate than Vb1 changes) should increase hfe by bringing Q1 farther from saturation. But a decrease in hfe is observed.
Temperature dependence means IBE is not necessarily shrinking even as VBE is observed to do so. If, as I concluded before, the majority of IBE1 is BC leakage, then IBE is probably increasing with temperature (diminished to some extent by NFB). If this increase in IBE1 is enough to bring Vc1 closer to Vb1 (despite Vb1 being closer to 0), then it would explain the decrease in hfe.
Experiment seems to confirm:
Quiescent, room temp: VB1=-.095V | VC1=-.18V
Quiescent, heated by touch: VB1=-.080V | VC1=-.11V
Contradictions in my previous thinking include: Increased leakage causing IBE to become closer to 0 doesn't seem to make much sense (NFB should reduce the effect of increased leakage, rather than reverse it). Also, IBE becoming less negative (meaning Vc1 becoming more negative, at a much bigger rate than Vb1 changes) should increase hfe by bringing Q1 farther from saturation. But a decrease in hfe is observed.
Temperature dependence means IBE is not necessarily shrinking even as VBE is observed to do so. If, as I concluded before, the majority of IBE1 is BC leakage, then IBE is probably increasing with temperature (diminished to some extent by NFB). If this increase in IBE1 is enough to bring Vc1 closer to Vb1 (despite Vb1 being closer to 0), then it would explain the decrease in hfe.
Experiment seems to confirm:
Quiescent, room temp: VB1=-.095V | VC1=-.18V
Quiescent, heated by touch: VB1=-.080V | VC1=-.11V
Welcome to the real world where feedback ain't all it's cracked up to be.
FB requires gain, and lots of it. The FF cct has plenty of regions where that ain't necessarily so. Demonstrating where is left as an exercise for the reader.
well you said you have ALOT Feedback could be too...Less generates all kind of distortion's and your increasing a lot bass responses.
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