Low Frequency Response

[Super_Reverb]

So I have an express build in progress and am a bit puzzled with the high pass design through the preamp gain stages.

The first two preamp stages have bypassed cathode resistors with -3dB points (corners) < 10 Hz. Then there is the third stage which has unbypassed Rcathode and output coupling capacitor and a smallish shunt resistor that, together with 2nd stage Ro creates a high pass filter with -3dB point at about 850 Hz. Since these are all single pole filters with rolloff @ about -20dB/decade, the slope is fairly gentle.

So, why this design approach? Why carry low frequency content through first two stages only to filter it out at later stage? Wrecks aren't the only amps that do this. It just seems unusual. Why not limit bandwidth at the first gain stage and then slightly widen it in later stages to avoid stacked corners?

Am I missing something obvious, or is it just something "we have always done this way" and it works?

cheers,

rob

[dave g]

Did you forget a factor of 2 when converting from rad/s to hertz? The -3dB freq. should be around ~400 Hz.

That said, emphasizing the midrange and treble is how the late 60's superleads got their "crunch". Designs with "better" low end response (like Blackface Fenders) don't do the crunch thing as well because the excess bass tends to make the amp flabby and farty.

Usually, you want to use a big bypass cap on the cathode of the input stage of an amplifier to reduce noise. As to why the 2nd stage also has a big bypass cap? My guess has always been that KF took a Blackface Fender preamp with a few tweaked values, added an extra gain stage, then fed that into a 50 watt plexi-ish output section...and there's the TW express.

[Super_Reverb]

I figured the voltage being fed to the grid of stage 3 is: Vgr3=Vplate2*150K/[Ro||100K + R150K + 1/(2*pi*f*C)], neglecting input impedance of stage3.

If you assume Ro from stage2 = 70K, then at 10KHz, Vout/Vin=150K/(41K+150K+8K)=150/199=0.754

So -3dB is the same as a voltage gain of ~0.707, so calculate frequency at which Vout/Vin from above = 0.754*0.707=0.533. When capacitor impedance rises to 90.4K Ohms, gain drops by factor of 0.707 (-3dB). My calcs shows that happens at 880Hz.

Usually, you want to use a big bypass cap on the cathode of the input stage of an amplifier to reduce noise.

Don't understand this. By limiting bandwidth, noise should be reduced. Larger bypass cap increases bass response which theoretically increases noise. Considerthe practice of limiting telephone bandwidth as an example.

cheers,

rob

[dave g]

With no bypass cap, there is degenerative feedback through the cathode resistor, which increases bandwidth at the cost of gain. Think of it this way: if the gain bandwidth product is constant and adding a bypass cap increases gain, then bandwidth goes down! You are thinking about bandwidth the wrong way...

However, that's not even the point. The whole point is to reject heater hum! Remember that the cathode is an input too, so that 60Hz high current signal on the heaters is injecting quite a bit of noise. In this case, the bypass cap actually acts as a shunt to ground for this hum. This is much less of an issue in later stages.

You're calculating the cutoff frequency the long way and losing a factor of 2 somewhere. Assume 40k output impedance from the previous stage. Add this to the 150k to get 190k. So we've got R = 190k and C = .002uF which we can just plug into the standard formula for a first order filter:

Fc = 1 / (2*pi*R*C) = 1 / (2*pi*190e3*2e-9) = 419 Hz

[Super_Reverb]

However, that's not even the point. The whole point is to reject heater hum! Remember that the cathode is an input too, so that 60Hz high current signal on the heaters is injecting quite a bit of noise. In this case, the bypass cap actually acts as a shunt to ground for this hum, improving the PSRR of the stage. This is much less of an issue in later stages.

You're calculating the cutoff frequency the long way and losing a factor of 2 somewhere. Assume 40k output impedance from the previous stage. Add this to the 150k to get 190k. So we've got R = 190k and C = .002uF which we can just plug into the standard formula for a first order filter:

Fc = 1 / (2*pi*R*C) = 1 / (2*pi*190e3*2e-9) = 419 Hz

You're right about the long way. Apparently, I missed the class where we derived the formula for -3dB points. I assumed 20log(Vo/Vi) for -3dB point, when I should have used 10log(Vo/Vi) for the "half power" frequency, so ~400 Hz it is.

a few comments after pondering: I never stopped to think of the proximity of cathode to heater and the large bypass cap as a shunt element for the 60 Hz noise. This makes sense.

Think of it this way: if the gain bandwidth product is constant and adding a bypass cap increases gain, then bandwidth goes down! You are thinking about bandwidth the wrong way...

With regard to negative feedback increasing frequency response, it does so by reducing open loop gain: as a result, the dominant high frequency and low frequency poles are moved wider due to reduction in passband gain. I don't think the analysis that adding a capacitor which increases gain causes a decrease in bandwidth is kosher

Gain bandwidth product being fixed is applicable when you wrap feedback around a system and decrease its open loop gain. It is not applicable when there are reactive elements in the feedback network, e.g., cathode bypass capacitor, which moves the low frequency pole AND adds series feedback.

cheers,

rob

[FunkyE9th]

Using 20log(Vo/Vi) should have been the correct thing to do. You are dealing with voltages. 10log(P1/P2) is used if you are dealing with power. P already has the voltage squared that's why it's only multiplied by 10 instead of 20.

Edited/corrected due to my wrong math..

Oops!!! Nevermind my voltage divider math was wrong. Sorry it should have been...

Vo/Vin = -mu(150K/(Rp||100K +1/j2*pi*f*C + 150K))

Rp||100K = ~38K

Which is what you had.

Now I don't understand why you have to use 10log(Vo/Vin) to get the right numbers...

The equivalent circuit I had looks like this (simple voltage divider)...

[vibratoking]

Now I don't understand why you have to use 10log(Vo/Vin) to get the right numbers...

20*log10(Vo/Vin) is the correct formula to determine the gain in dB.

The -3dB point is 1/tau where tau is the time constant and is equal to the sum of the total resistance times the capacitance, in this case.

Both formulas give the same answer.

At high frequencies, the cap looks like a short and the resistive voltage divider dominates. So 20*log12(150k/188k) = -1.96 dB. The -3dB point is where the magnitude of the response is
-4.96. Here is a plot.

[FunkyE9th]

Thanks! Now I get it. I got the same plot as you did. My problem was when I did the math via 20log(vo/vin) it was relative to 0dB so I was getting ~810Hz. When really it should have been 3dB down from -1.96dB as you pointed out. It maxes out at -1.96dB because of the resistive divider so you'll never hit 0dB.

[Ian444]

So, why this design approach? Why carry low frequency content through first two stages only to filter it out at later stage? Wrecks aren't the only amps that do this. It just seems unusual. Why not limit bandwidth at the first gain stage and then slightly widen it in later stages to avoid stacked corners?

Maybe they want to keep the harmonics generated from the lower frequencies, but not the lower frequencies themselves.

[Super_Reverb]

A few things that I have learned from this thread is the formula f=1/(2*pi*R*C) is not correct in determining the corner frequency for the model of the 2nd-3rd stage coupling in an express preamp.

This formula is based on the concepts that 1) the resistor value and capacitor impedance are equal at the corner frequency and 2) the value of the *voltage* response is one half of the flatband gain. The output impedance of the 2nd stage term complicates things somewhat.

If you write a voltage loop for the circuit modeled in earlier post, you get:

Vout/Vin=R2/(R1+R2+1/(2*pi*f*C)), set Vout/Vin=0.5 and crunch through the algebra: the result is f=1/(2*pi(R2-R1)C). The idea of the equation is that the frequency at which voltage across R2 = one half applied voltage, is the corner frequency. You can see that the relationship only works if R2 is larger than R1.

So if you plug accepted values into this equation, you get f=1/(6.28*(150K-40K)*2nF) = 723 Hz.

Further, since this equation is based on the idea that the voltage amplitude falls to half, then you have to use the half power concept of 10log(Vo/Vi) to calculate -3dB point, because log(0.5) = -0.301, so 10log(-0.301) = -3dB.

This is considerably different that what is indicated by other sources, so check my math.

Cheers,

rob

[vibratoking]

A few things that I have learned from this thread is the formula f=1/(2*pi*R*C) is not correct in determining the corner frequency for the model of the 2nd-3rd stage coupling in an express preamp.

That equation is correct for the approximation that has been made so far in this thread. The main error in the approximation is due to the fact that several small internal tube capacitances have been ignored. Those capacitances don't change the -3dB point in any significant way. The equation is always correct for a ciruit with only one capacitive element.

This formula is based on the concepts that 1) the resistor value and capacitor impedance are equal at the corner frequency and 2) the value of the *voltage* response is one half of the flatband gain. The output impedance of the 2nd stage term complicates things somewhat.

The cutoff point is the -3dB pont by definition. There is nothing magical about it. It used by convention.

If you write a voltage loop for the circuit modeled in earlier post, you get:

Vout/Vin=R2/(R1+R2+1/(2*pi*f*C)), set Vout/Vin=0.5 and crunch through the algebra: the result is f=1/(2*pi(R2-R1)C). The idea of the equation is that the frequency at which voltage across R2 = one half applied voltage, is the corner frequency. You can see that the relationship only works if R2 is larger than R1.
So if you plug accepted values into this equation, you get f=1/(6.28*(150K-40K)*2nF) = 723 Hz.

You have an error. It should be R2+R1, so f=1/(2*pi(R2+R1)C). It is the total resistance and not the difference.

Further, since this equation is based on the idea that the voltage amplitude falls to half, then you have to use the half power concept of 10log(Vo/Vi) to calculate -3dB point, because log(0.5) = -0.301, so 10log(-0.301) = -3dB.

I think you are a little confused about that equation. We are interested in voltage relationships when examining the transfer function between preamp stages. What follows is 'common' engineering knowledge.

10*log10(Px/Py) is the definition of a power ratio expressed in dB. Px relative to Py. Px and Py are both power and result is in dB. One way to understand this in terms of voltage is as follows:
P = V*I and I = V/R. Substituting V/R for I gives P = V^2/R. Now let Px = Vx^2/Rx and Py = Vy^2/Ry. In our case, Rx and Ry are one and the same. Substituting both of those into 10*log10(Px/Py) gives the following:
10*log10(Vx^2/Vy^2) which can be rearranged to 20*log10(Vx/Vy) because the 2 moves out in front due to the principles of logs and Rx and Ry cancel in the division because they are equal. This explains exactly how the the ratio of power equation is transformed to the ratio of voltage equation.

I hope this helps you to come away with the proper conclusion.

[guitarmike2107]

The first two stages of the express never really distort (Audibly and visually), so there is little harmonic distortion content coming from these two stages compared to the latter stages also the tone stack after the first stage is lossy, hence another good reason why you would want to fully bypass stage 2.

In this case, if you cut the bass with a cathode cap/resistor on the second stage i.e. 2.7k + 0.68uf I suspect you will get more low mids and it may therefore sound slightly muddier compared to using the designed coupling cap method of cutting the bass to similar levels. Just a thought, need to try it myself.

[Super_Reverb]

You have an error. It should be R2+R1, so f=1/(2*pi(R2+R1)C). It is the total resistance and not the difference.

How is it the total resistance? The corner frequency is the frequency at which the voltage across the 150K Ohm resistor is half of the applied voltage. There are three passive elements in the equation.

When the voltage across (2nF cap + Ro) is equal to the voltage across the 150K resistor, that is the corner frequency, as the voltage has fallen by half.

f=1/(2*pi*R*C) works is you have one cap and one resistor or if you are measuring the voltage across multiple resistors in series with a coupling cap (high pass filter) or if you are measuring the voltage across the cap (low pass filter) with several resistors in series on the source side.

Let's consider an example:

The first stage Ro is 40K, R2 is 50K with 2nF coupling capacitor. With your proposed solution, the corner frequency calculates to f=1/(2*pi*(90K*2nF) = 884Hz. At 884Hz, the capacitor impedance calculates to 90K So, with your approach, voltage divider gives voltage across R2 at 884Hz calculated corner frequency as 50K/(50K+40K+90K)= 0.278 this is not 50% of applied voltage , which is what the -3dB point is based on (see my derivation in earlier post). So we have proven that f=1/(2*pi*Rtotal*C) doesn't work in this instance when calculating corner frequency between 2nd and 3rd stages in an express.

Now, using the equation for corner frequency, f=1/(2*pi*(R2-Ro)*C)=1/(6.28*(50K-40K)*2nF)=7.96kHz. Cap impedance calculates to 10K at 7.96kHz. So the voltage divider tells us the R2 voltage = R2/(R1+R2+1/(2*pi*f*C)=50K/(40K+50K+10K)=50K/100K=0.5

So, we have proven that the voltage across R2 is 1/2 of the applied voltage at the corner frequency using this equation: f=1/(2*pi*(R2-Ro)*C), which is valid for calculating the corner frequency between 2nd and third stages in an express amp.

cheers,

rob

[dave g]

Rob,

You are fundamentally misunderstanding the notion of a cutoff frequency here.

It is "wrong" in this context to define Vi as the AC signal on the plate of V2a, and then try to calculate the frequency for which Vo/Vi = .707. Why? Because the 40k output impedance of the 2nd stage effectively shifts the entire bode plot down by a factor of 15/(15+4) ~= -2dB. So even for frequencies which are high enough to make the capacitor look like a short circuit, there is still a loss of -2dB just from the divider formed between the 150k and the output impedance of the 2nd stage.

To find the cutoff frequency, you should actually be calculating the -5dB point of the circuit (given the way you've defined your variables).

I have attached a schematic and bode plot I made with a computer simulation to illustrate what I am talking about. Cursor 'A' marks the passband response (at 10 kHz [XA on the plot], we are down -2.06dB [YA on the plot]), while cursor 'B' marks the cutoff frequency of the circuit (-3 dB down from the passband gain, or -5 dB....right at 419 Hz, like I originally calculated [XB and YB on the plot]).

[FunkyE9th]

Attached is a file that shows you the math on how to calculate the 3dB point using the transfer function. It gives the same result as using Tau (i.e using the time constant formula)