Input sensitivity, gain, and confusion.

[thorpej]

I've been thinking a lot about gain recently, partially because of the NFB issue on my 8802 clone and partially because I have a couple of projects in the brainstorming / design phase (one guitar, one Hi-Fi).

As I understand it, the garden variety line level output for consumer audio gear (e.g. a CD player) is -10dBV, or roughly 315mVrms, which would represent the maximum signal.

Now, let's assume an EL84 running at the design center operating point of 250V plate voltage, a 5K load, and a -7.5V bias. Looking at that load line, driving with roughly 4Vrms gets you right to the edges of the linear operating area. So it seems that to drive to full power, one only needs a drive stage with a gain of 14. And certainly there would be no value in having a signal > 7.5Vpp because you become grid-limited.

Have I got that right so far?

So, I'm looking at a schematic for a SE mono block that has a 6V6 preceded by both halves of a 6SN7. The description of this amplifier claims it has an input sensitivity of 500mV (which I'm assuming is rms). The first half of the 6SN7 has a gain of 17 (assuming my math is correct). That would be 8.5Vrms at full signal, which is already quite sufficient to drive a 6V6. But this is then amplified once more by the second half of the 6SN7, a stage with a gain of 12 (again, assuming my math is correct). On the face of it, it seems that is enough to quite nearly overdrive the second 6SN7 stage (it wouldn't get it into cut-off, but it would certainly get very near driving the grid to 0V). But of course 8.5Vrms * 12 = 102Vrms, and that just seems absurd (it would certainly overdrive the 6V6).

The schematic does have NFB in play -- on an 8 ohm tap, 1.5K feeding the cathode of the 2nd 6SN7 stage, which is in turn biased by a 1K resistor. Again, assuming my math is right, that would feed 40% of the output signal voltage back to the cathode, which is like -12dB, right? No matter how you slice it, it still seems like the 6V6 is being overdriven.

I almost feel like I am fundamentally misunderstanding something.

 

[jazbo8]

As I understand it, the garden variety line level output for consumer audio gear (e.g. a CD player) is -10dBV, or roughly 315mVrms, which would represent the maximum signal.

Most modern gear such as CDP's have much higher output, ca. 1.5Vrms~2Vrms. Please post the schematic of the SE amp, so we can see what you are seeing...

 

[gadget73]

yeah, you'd need to confirm the output levels of your device. You can often skip an active preamp stage these days if running more modern sources.

 

[kward]

From a design perspective, I find it easiest to think about the problem sort of in reverse--starting from the output stage and moving backwards through to the previous stages of the amp. In other words, in your example, starting with a 6V6 output stage, assuming it's biased at -7.5V grid with respect to cathode, that means more-or-less you need 7.5V peak drive voltage to drive the output stage to full power. 7.5V peak is about 5.3V RMS. So that's the drive amplitude that the previous stage needs to produce in order to cleanly drive the output stage to full power.

A 6SN7 with a mu of 20, when wired up a s a common cathode gain stage typically has a stage gain of about 12 if the cathode is not bypassed. If there is feedback applied, that reduces the gain of the stage by roughly the amount of feedback added. Let's say your design calls for 10 dB of feedback (quite a small amount). 10 dB is equivalent to a factor of 3.16. So after feedback is applied, the forward voltage gain of the stage is 12/3.16 = 3.8.

Input sensitivity is defined as the amount of input voltage necessary to drive the output stage to full power, thus in our example it's 5.3V RMS/3.8 = 1.39V RMS if 10 dB of feedback is applied. In other words an input signal with amplitude of 1.39V RMS will be amplified by the first stage by a factor of 3.8 to give an output signal with amplitude of 1.39 * 3.8 = 5.3V RMS, which is exactly what's needed to drive the output stage to full power.

If no feedback is applied, then the input sensitivity is 5.3V RMS/12 = 0.44V RMS.

Granted this is a "back of the napkin" way of estimating, but I find it's close enough to get a good read on the design is meeting drive requirements or not.

The same process can be applied to a 3 stage or even 4 stage amp, and also to push-pull amps with minor tweaks. For example, in a push-pull design, the phase inverter either adds some additional gain if it's a differential pair, or it subtracts a just a little gain if it's a split load inverter.

Concerning the amount of feedback applied, there are three variables at play: the feedback resistor size, the cathode resistor size in the lower portion of the feedback voltage divider circuit, and the cathode current of the tube to which the feedback line is connected. It is the voltage created at that junction that sets the amount of feedback at play. It can be calculated by modeling the amp and feedback circuit mathematically and solving a set of simultaneous equations, but it's a bit complicated, and I find it just as easy to get the amp on the bench and measure the amount of feedback applied.

 

[thorpej]

yeah, you'd need to confirm the output levels of your device. You can often skip an active preamp stage these days if running more modern sources.

Yah, I realize that active preamps are essentially not necessary these days, and this is mostly a mental exercise for me trying to make sure I have a solid understanding of what's going on.

 

[thorpej]

From a design perspective, I find it easiest to think about the problem sort of in reverse--starting from the output stage and moving backwards through to the previous stages of the amp. In other words, in your example, starting with a 6V6 output stage, assuming it's biased at -7.5V grid with respect to cathode, that means more-or-less you need 7.5V peak drive voltage to drive the output stage to full power. 7.5V peak is about 5.3V RMS. So that's the drive amplitude that the previous stage needs to produce in order to cleanly drive the output stage to full power.

Yes, that's how I like to think about it, and I looked at this particular 6V6 SE amp and worked forward as a way to validate my thinking.

A 6SN7 with a mu of 20, when wired up a s a common cathode gain stage typically has a stage gain of about 12 if the cathode is not bypassed. If there is feedback applied, that reduces the gain of the stage by roughly the amount of feedback added. Let's say your design calls for 10 dB of feedback (quite a small amount). 10 dB is equivalent to a factor of 3.16. So after feedback is applied, the forward voltage gain of the stage is 12/3.16 = 3.8.

AHA! And that's the bit I was missing. I was misunderstanding the effect of the NFB (I was subtracting rather than dividing).

So, with that corrected in my mental model, my 6V6 SE example looks more like:

.5 * 17 [input stage, cathode fully bypassed] -> 8.5 * 3.8 [drive stage, cathode not bypassed] -> 32.3 [output stage]​

...which still seems high and would overdrive the 6V6, but isn't absurd on it's face. And, I am likely miscalculating the amount of NFB. But it makes me wonder why the drive stage is even needed in this instance, as the input stage seems to be sensitive enough to directly drive the 6V6.

I'll follow-up with the schematic later; it's in a book, and I'll need to scan it.

 

[kward]

Ah, but hold the phone, though.

If you convert gain factors to a logarithmic scale (e.g. decibels) then you DO add or subtract. In other words., back to our original example of the stage gain of the 6SN7 (non bypassed) with a gain of 12 but with 10 dB feedback applied:

gain of 12 is 21.58 dB. Now that I have both numbers in the same scale, I can subtract the amount of feedback applied (10 dB):

21.58 dB stage gain - 10 dB feedback = 11.58 dB forward voltage gain. 11.58 dB converted back into a gain factor is 3.8.

 

[thorpej]

Ah, but hold the phone, though.

If you convert gain factors to a logarithmic scale (e.g. decibels) then you DO add or subtract. In other words., back to our original example of the stage gain of the 6SN7 (non bypassed) with a gain of 12 but with 10 dB feedback applied:

gain of 12 is 21.58 dB. Now that I have both numbers in the same scale, I can subtract the amount of feedback applied (10 dB):

21.58 dB stage gain - 10 dB feedback = 11.58 dB forward voltage gain. 11.58 dB converted back into a gain factor is 3.8.

Yah, but I attempted a conversion away from dB before doing the math first (so in addition to subtracting a non-dB value, I probably also got the conversion out of dB incorrect as well because I keep forgetting it's a ratio).

Thanks for setting me straight

 

[kward]

.5 * 17 [input stage, cathode fully bypassed] -> 8.5 * 3.8 [drive stage, cathode not bypassed] -> 32.3 [output stage]​

...which still seems high and would overdrive the 6V6, but isn't absurd on it's face. And, I am likely miscalculating the amount of NFB. But it makes me wonder why the drive stage is even needed in this instance, as the input stage seems to be sensitive enough to directly drive the 6V6.

I'll follow-up with the schematic later; it's in a book, and I'll need to scan it.

It does seem high, unless the amp was setup for a low level input like mic, tape, or phono or something.

 

[thorpej]

It does seem high, unless the amp was setup for a low level input like mic, tape, or phono or something.

Yah, and even if my math were wrong and it was -20dB feedback, it would just be close to unity gain and, well, what would be the point? In guitar amps, more stages = more distortion = more better! But in Hi-Fi, seems like you always want to use a few stages as you can get away with.

 

[thorpej]

The same process can be applied to a 3 stage or even 4 stage amp, and also to push-pull amps with minor tweaks. For example, in a push-pull design, the phase inverter either adds some additional gain if it's a differential pair, or it subtracts a just a little gain if it's a split load inverter.

I'm glad you mentioned phase inverters! This thread has given me some better insight as for how to choose a given PI topology for a given output stage. For example, if you have output tubes like 6V6 or EL84, you don't need a ton of driving voltage, so a PI with an overall gain of 0.90 or 0.95 like a cathodyne would work quite well (and has the other advantage of being easy to balance and frugal on tube count). Furthermore, this explains why one sees e.g. 6FQ7 as a common PI tube, because as a low mu tube, it's often biased in the -8v range and thus can handle the signal coming from a preceding drive stage (while still providing a ~7v signal on the other side, perfectly suitable for 6V6 or EL84).

And this is why you'd choose something like paraphase or LTP for bigger output tubes like 6L6 or EL34 (at least in the case of paraphase, whatever gain you have from the drive stage, you simply attenuate by the same factor to feed the PI).

Am I on the right track?

 

[kward]

Yup you're on the right track. Thus (among many other things) it's an analysis of how much gain and voltage swing you need to drive the output stage cleanly to full power such that the output stage clips first, way before the driving stage or the input stage(s).