Calculating Negative Feedback

[James Freeman]

As my first (serious) post I want to gift an easy way to calculate negative feedback with a simple equation.
That's V=sqrt(P*R) to get voltage after the output transformer, then a simple voltage divider using the NFB and Tail resistors R2/(R1+R2), this is fed back into the phase inverter.

sqrt (Watt * Tap) * (Tail / (Tail + NFB))


Examples.

100W from 4Ω tap with 100k NFB resistor ('73 1959 Super Lead, 2203):
sqrt(100*4)*(4.7/(4.7+100)) = 0.89

50W from 8Ω tap with 100k NFB resistor (1987x)
sqrt(50*8)*(4.7/(4.7+100)) = 0.89

50W from 4Ω tap with 100k NFB resistor (2204):
sqrt(50*4)*(4.7/(4.7+100)) = 0.63

100W from 8Ω tap with 47k NFB resistor (1959 Plexi):
sqrt(100*8)*(4.7/(4.7+47)) = 2.57

100W from 16Ω tap with 82k NFB resistor (JVM410H):
sqrt(100*16)*(4.7/(4.7+82)) = 2.168
The correct NFB resistor for 2203 spec (0.89 as above) without changing the output tap is 205k.
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You can copy paste that equation into google and it will solve it for you.
Hope that saves you some time and puts decades of confusing forum posts to rest.

PS.
Sticky please.

 

[Pete Farrington]

That's V=sqrt(P*R) to get voltage after the output transformer

I don’t see the relevance of the amp’s nominal power output here?
See https://en.m.wikipedia.org/wiki/Negative-feedback_amplifier

Open loop gain, feedback ratio and closed loop gain are all ratios.

If putting absolute numbers helps to get your head around things, I find that a reference nominal 1W output level at the relevant load impedance can be useful, it allows eg the 5F6A bassman’s feedback arrangement to be compared to that of the apparently very similar JTM45.

Unfortunately that approach ignores the OT primary impedance, voltage ratio and their affect on open loop gain, so it’s not the full picture.

The old school technique was to build the amp, measure the small signal open loop gain, the closed loop gain, and express the difference as xdB of negative feedback.

Maybe @arthur.lowery could explain thongs better ?

 

[Ronquest]

So, what happens if I only play my 1959 Plexi at 50W ? Do I move the feedback to the 16 Ohm tap?

 

[James Freeman]

Unfortunately that approach ignores the OT primary impedance, voltage ratio and their affect on open loop gain

Hi.

Yep, we assume that 100W actually delivers 100W over a certain load, so we assume that the output transformer was designed correctly and for simplicity it's ideal, same with 50W, so sqrt(P*R) is our ideal open-loop voltage gain.
The rest is trying to maintain the "feedback summing point" voltage at the input of the phase inverter, these are the bold numbers in the first post.

From Aiken's "Designing for Global Negative Feedback" (I can't post links yet).

Another way to look at this is that the required voltage at the feedback summing point must remain the same, but only half the original voltage is present on the input of the divider. The new value of Rf can be calculated by determining the original voltage

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This calculation is slightly more accurate than using the R*sqrt(2) at half the tap, and also you can find the correct NFB Resistor to maintain the same NFB amount between different amp Wattage.
Again, assuming the output transformers are ideal.

For example, to maintain the same NFB amount from 4Ω to 16Ω taps you need to go form 100k to 205k NFB resistor, so it's not exactly x2 like most would suggest using sqrt(2).

 

[2L man]

NFB effect is very easy to measure using oscilloscope or good quality voltmeter (reads multimeter costing less than 100). Amp drive is sine wave using for example cell phone app. Output set to about 20% power the output AC voltage measure is taken. Then NFB loop is disconnected and also this higher output voltage is taken. Then both voltages are set to "voltage to db converter" which are many in inet.

Math of the NFB loop resistors is simple but obviously there are other circuit variables which are not as simple? Precence which effect depends of the frequency and obviously stage gains?

 

[Pete Farrington]

so sqrt(P*R) is our ideal open-loop voltage gain

That’s the nominal max unclipped rms output voltage, not the voltage gain?

Assuming the same design, valves and operating point, I think 50 and 100W amps will have the same open loop voltage gain.

 

[66 Kicks]

Maybe @arthur.lowery could explain thongs better ?

Good, because women wearing thongs is easily explainable, but men wearing thongs is a total mystery to me.

 

[66 Kicks]

As my first (serious) post I want to gift an easy way to calculate negative feedback with a simple equation.
That's V=sqrt(P*R) to get voltage after the output transformer, then a simple voltage divider using the NFB and Tail resistors R2/(R1+R2), this is fed back into the phase inverter.

sqrt (Watt * Tap) * (Tail / (Tail + NFB))


Examples.

100W from 4Ω tap with 100k NFB resistor ('73 1959 Super Lead, 2203):
sqrt(100*4)*(4.7/(4.7+100)) = 0.89

50W from 8Ω tap with 100k NFB resistor (1987x)
sqrt(50*8)*(4.7/(4.7+100)) = 0.89

50W from 4Ω tap with 100k NFB resistor (2204):
sqrt(50*4)*(4.7/(4.7+100)) = 0.63

100W from 8Ω tap with 47k NFB resistor (1959 Plexi):
sqrt(100*8)*(4.7/(4.7+47)) = 2.57

100W from 16Ω tap with 82k NFB resistor (JVM410H):
sqrt(100*16)*(4.7/(4.7+82)) = 2.168
The correct NFB resistor for 2203 spec (0.89 as above) without changing the output tap is 205k.
---

You can copy paste that equation into google and it will solve it for you.
Hope that saves you some time and puts decades of confusing forum posts to rest.

PS.
Sticky please.

So.....that will give you the rms voltage from the voltage divider when the amp is operating at full rated power. Then what?

Try this:

 

[Pete Farrington]

Try this:

I’m a bit stuck on this

Would you mind walking me through that section?
I don’t understand why the LTP gain is being divided by the Vbias?
And it’s going to be specific to large signal conditions, whereas my feeling is that the small signal conditions might be more appropriate

Edit; here’s a link to Aiken’s page mentioned previously https://www.aikenamps.com/index.php/designing-for-global-negative-feedback

 

[66 Kicks]

I’m a bit stuck on this

Would you mind walking me through that section?
I don’t understand why the LTP gain is being divided by the Vbias?

The amp is at full rated power when the voltage from the LTPI equals the absolute value of the DC bias voltage. This is a peak voltage and we want rms because the rated power is rms, so the bias voltage is divided by 1.41. The negative sign is used because the LTPI inverts and we need a positive open loop gain for the other equations. We know the gain of the LTPI and the voltage that it puts out, so we can find the input voltage for the open loop. Since we know the input voltage and the output voltage, we know the open loop gain.

 

[66 Kicks]

Here is some hen scratch showing how Vbias ended up in the denominator:

 

[66 Kicks]

Assuming the same design, valves and operating point, I think 50 and 100W amps will have the same open loop voltage gain.

Does the 50W amp have half the number of power valves and twice the OT primary impedance as the 100W amp?

 

[Ronquest]

Seriously asking here. Why should I care about calculating my negative feedback? Is this just an exercise "to maintain the same NFB"? Am I missing something, should I be calculating when designing or building a guitar amp? Don't we all pretty much go with a known NF circuit and then perhaps tweak until pleased? Add resonance, change the presence pot and cap values, maybe a 3 way/bypass switch? And now calculate everything again? Does it matter?, If I like the sound, I'm going with the sound.

 

[Pete Farrington]

Does the 50W amp have half the number of power valves and twice the OT primary impedance as the 100W amp?

My thinking is -

As the anode resistance tends to be a fair bit higher than the typical load used, pentode stage gain is roughly valve transconductance gm x the load impedance RL.

And connecting a pair of valves in parallel doubles the gm (and halves the combined anode resistance).

So all else being equal, and a 50W OT with twice the impedance ratio of a 100W OT, the 50W output stage has half the gm and twice the RL compared to the 100 watter.
So stage gain should even out

 

[Pete Farrington]

Seriously asking here. Why should I care about calculating my negative feedback?

For the same reason we care about anything technical?
Rather than trusting to luck / experience (that may be based on an incorrect understanding) / known good prior design, having a grasp of what factors are actually in play may be beneficial in getting the desired results.
With specific regard to NFB, the theory allows us to understand why the presence control of a JTM45 has so much more range of control than that of eg a SL.

 

[66 Kicks]

My thinking is -

As the anode resistance tends to be a fair bit higher than the typical load used, pentode stage gain is roughly valve transconductance gm x the load impedance RL.

And connecting a pair of valves in parallel doubles the gm (and halves the combined anode resistance).

So all else being equal, and a 50W OT with twice the impedance ratio of a 100W OT, the 50W output stage has half the gm and twice the RL compared to the 100 watter.
So stage gain should even out

I'm with you on that, but what about the different OT turns ratios giving the 100W OT a gain that is 1.414 times the gain of the 50W OT?

 

[Pete Farrington]

I'm with you on that, but what about the different OT turns ratios giving the 100W OT a gain that is 1.414 times the gain of the 50W OT?

Oh yeah, thanks, I never considered that

 

[myersbw]

Good, because women wearing thongs is easily explainable, but men wearing thongs is a total mystery to me.

Actually, my FIRST thought was that Pete was on his third ale! Or, staring at a woman at the bar wearing something just a bit revealing! LOL

 

[myersbw]

My thinking is -

As the anode resistance tends to be a fair bit higher than the typical load used, pentode stage gain is roughly valve transconductance gm x the load impedance RL.

And connecting a pair of valves in parallel doubles the gm (and halves the combined anode resistance).

So all else being equal, and a 50W OT with twice the impedance ratio of a 100W OT, the 50W output stage has half the gm and twice the RL compared to the 100 watter.
So stage gain should even out
Click to expand...

I'm with you on that, but what about the different OT turns ratios giving the 100W OT a gain that is 1.414 times the gain of the 50W OT?


Ok...then here's where my engineering field tech mind kicks in and thinks,"Who ever pushes their amp to 100 watts...even to 50?" So, I choose a medium value...insert a pot and vary it to find the degree of actual NFB that suits my playing levels, feedback impact and practical amp power dissipation.

If you go back and reflect on all the "Master Volume" choices Marshall made and your max volume really starting to peak at 2-3 on the dial...I'm thinking,"Really???" and then change the voltage divider ratio to best reflect what I expect it would be to at least get 80% rotation on the MV dial and a LOT more control over effective audio range and discernment. (My typical thought on that was...are they trying to be impressive with kids thinking...look how loud MY amp is just on 2!

Cheers!