Designing Common-Cathode Triode Amplifiers

General


The most commonly used amplifier stage is the common-cathode amplifier.  It features high input impedance, medium-to-low output impedance, relatively high gain, and good frequency response.  The frequency response can be tailored by either the input and output coupling capacitors or the cathode bypass capacitor.  The output signal of the common cathode amplifier is inverted with respect to the input signal.

The common-cathode amplifier configuration – fully bypassed cathode


Following is a schematic of a typical common-cathode amplifier stage with a bypassed cathode resistor:




    Capacitor Ci is the input coupling capacitor.  It is used to isolate the grid circuit from the DC voltage at the output of the previous circuit.  This capacitor, in conjunction with the grid resistor, controls the frequency response of the stage.


    Rg is the grid resistor, which is used to provide a reference voltage for the grid circuit (ground in this case).  It is usually a high value, such as 1Megohm.  This resistor controls the input impedance of the stage.


    Rk is the cathode resistor, which is used to develop the cathode bias voltage.  The flow of cathode current through this resistor creates a voltage drop across it, resulting in a positive voltage at the cathode of the tube.  Since the grid resistor references the control grid to ground potential, this positive cathode voltage creates an effective negative grid voltage with respect to the cathode, providing the bias operating point for the tube.  This resistor controls the headroom of the stage (output before clipping) and linearity, or distortion level, of the stage.  As the bias point is shifted, the amplifier will clip more on the top or bottom portion of the waveform.  If the cathode resistor is unbypassed, this resistor also controls the stage gain to a certain extent.


    Capacitor Ck is used to bypass the cathode resistance to ground for AC signals, which results in a higher gain.  Without Ck, there is negative feedback, or degeneration, which reduces the gain of the stage and increases the output impedance.  If Ck is not large in comparison to Rk, it will affect the frequency response of the stage, by introducing a “shelving” response, where the stage gain is boosted at higher frequencies compared to lower frequencies.  The “breakpoint” of the frequency response is controlled by the value of Ck, in conjunction with the cathode impedance.


    Resistor Rp is the plate load resistor.  The output signal voltage is developed across this resistor, by the action of the plate current flowing through it.  Because of this, the value of the plate resistance directly controls the gain of the circuit.  It also controls the output impedance of the stage.


    Capacitor Co is the output coupling capacitor.  It is used to isolate the plate DC voltage from the next stage it is driving.  This capacitor, in conjunction with the input resistance of the following stage, also controls the frequency response of the stage.


    Resistor Rl is the load resistor.  It usually is also the grid resistor of the next stage.  It controls the midband gain of the amplifier, because, for AC signals, the effective plate resistance is the parallel combination of Rp and Rl.  If  Rl is made at least ten times larger than Rp, it can usually be ignored for gain calculations.
     



  • Biasing



    The bias point of the common cathode stage is set by the value of the cathode resistor, Rk.  The best way to calculate the value of the cathode resistor is by the use of a load line.  The plate resistor is chosen first, typically around twice the value of the internal plate resistance of the tube for most linear operation, although higher values can be used for more gain and output voltage swing.  A load line representing the plate load resistance is then drawn on the characteristic curves, and a bias point is chosen that gives the desired operating point.  The bias point is the intersection of the grid bias voltage curve and the plate current axis at the chosen plate voltage point that gives the desired range of plate voltage swing as the grid voltage varies from cutoff to saturation.   The value of the cathode resistor is then calculated using the following equation:
     



      Rk = Vg/Ip


    where:



       
      Vg = grid bias voltage at the bias point
      Ip  = plate current at the bias point
       


    For example, suppose a 12AX7 were chosen, with a plate supply voltage of 300V, and a 100K load line was drawn on the curves, and the desired operating point was the intersection of the -1.5V grid line and the 1.1mA plate current line, at a quiescent plate voltage of 190V.  The required cathode resistor would be:
     



      Rk = 1.5V/1.1mA = 1.36K (use 1.5K as the nearest standard value)




  • Gain (fully bypassed cathode)


    Some typical 12AX7 numbers:


      transconductance:               gm = 1600 micromhos
      plate resistance:                  ra = 62.5K
      amplification factor:             mu = 100

      The equation for the midband voltage gain for a fully-bypassed cathode (ignoring the effects of the following load resistance, Rl)  is:



    Av = (mu*Ra)/(Ra+ra)
         = (100*100K)/(100K+62.5K)
         = 61.5
         = 35.8dB

    Where:

    Ra = the total load resistance, which is Rp in parallel with the input resistance of the next stage, Rl. If there is no Rl, as in this case, Ra = Rp.
    ra = the internal plate resistance (62.5K for a typical 12AX7)
    mu = the mu of the tube (100 for a typical 12AX7)
    dB = voltage gain in decibels = 20*log(V1/V2) = 20*log(61.5/1) = 35.8dB

    Note that the above calculations are for a single stage alone.  They do not take into account the loading effect of the following stage.  For instance, if you have the above stage driving a 1Meg volume pot, the effective midband AC load resistance is the parallel combination of the plate resistor and the input resistance of the following stage, in this case, 1Meg.  The effective load resistance, Rl , is then 100K in parallel with 1Meg, or 90.9K. Therefore the midband gain would be:

    Av = (mu*Ra)/(Ra+ra)
         = (100*90.9K)/(90.9K+62.5K)
         = 59.3
         = 35.5dB

    As you can see, the stage gain has dropped from 61.5 to 59.3 due to the loading effect of the 1Meg pot.  Smaller input resistances on the following stage will load the first stage even more, further reducing gain.  If the load is at least ten times the value of the plate load resistor, it can usually be ignored for the purposes of gain calculations.


  • Input Impedance


    Since the grid circuit is a very high impedance, the effective input impedance of the common cathode stage is approximately equal to the value of the input grid resistor, Rg.


  • Frequency response due to input circuit


    Low frequency response:

    The low frequency response due to the input circuit is controlled by Ci and Rg.  These components act as a high-pass filter with a -6dB/octave (-20dB/decade) slope and a lower -3dB point that can be calculated as follows:

    f = 1/(2*pi*Rg*Ci)

    For example, if a 0.022uF input coupling capacitor is used, and a 1 Megohm grid resistor is used, the contribution to the overall stage frequency response would be:

    f = 1/(2*pi*1Meg*0.022uF) = 7.23Hz

    This means that the response would be down -3dB at 7.23Hz relative to the midband gain.  It would drop with a -20dB/decade slope as the frequency is lowered.

    High frequency response:

    The high frequency response due to the input circuit is controlled by the output resistance of the stage driving the common-cathode stage and the input capacitance of the stage.  The input capacitance is governed primarily by the Miller capacitance of the stage, and can be calculated as follows:

    Cin =Cgk + Cgp*(Av + 1)

    where:

    Cgk = the grid-to-cathode capacitance
    Cgp = the grid-to-plate capacitance
    Av = the stage voltage gain

    For example, a typical 12AX7 stage has the following capacitances and gain:

    Cgk = 1.6pF + 0.7pF stray = 2.3pF
    Cgp = 1.7pF + 0.7pF stray = 2.4pF
    A = 61

    Therefore, the total input capacitance would be:

    Cin = 2.3pF + (61+1)* 2.4pF = 151.1pF

    The input capacitance and output resistance of the previous stage form a first-order low-pass filter with a -6dB/octave (-20dB/decade) slope and an upper -3dB frequency point that can be calculated as follows:

    f = 1/(2*pi*Rout*Cin)

    For example, if the previous stage has an output resistance of 38.5K, and is driven through a series resistor of 470K with a 1Meg grid resistor, the upper -3dB frequency response point would be:

    f = 1/(2*pi*((38.5K + 470K)||1Meg)*151.1pF) = 3.13kHz

    Note that the symbol “||” means “in parallel with”.  The effective output resistance of the previous stage is the Thevenin equivalent of the driving circuit, which looks like the parallel combination of the grid resistance and the output resistance of the previous stage plus the series resistance.

    This is why large series resistive dividers are typically bypassed with small capacitors in order to boost the frequency response to compensate for the high frequency rolloff due to the input capacitance and the source resistance of the previous stage.


  • Output Impedance (output taken from the plate, bypassed cathode)


    Since the internal plate resistance is effectively in parallel with the plate load resistor, the output impedance (if the signal is taken off the plate) will be:

    R = ra || Rp
       = 62.5K || 100K
       = 38.5K

    Note: the symbol “||” means “in parallel with”.  Resistors in parallel add in reciprocal, i.e. 1/Rt = 1/R1 + 1/R2.
             Another method to calculate parallel resistances is: Rt = (R1*R2)/(R1+R2)


  • Frequency response due to output circuit


It can be shown1 that the low frequency response due to the output circuit is controlled by Co, , and the output impedance of the stage. These components act as a high-pass filter with a -6dB/octave (-20dB/decade) slope and a lower -3dB point that can be calculated as follows: 



      f = 1/(2*pi*(Rl+Ro)*Co)


    For example, if a 0.0022uF output coupling capacitor is used, and a 1Megohm grid resistor is used, and the output impedance of the stage is 38.5K as calculated above, the contribution to the overall stage frequency response would be:




      f = 1/(2*pi*(1Meg + 38.5K)*0.0022uF) = 69.7Hz


    This means that the response would be down -3dB at 69.7Hz relative to the midband gain.  It would drop with a -20dB/decade slope as the frequency is lowered.  When the frequency reaches the point where the input response is down-3dB, as controlled by the input coupling capacitor and the grid resistor, the slope of the response would steepen to -40dB/decade.


The common-cathode amplifier configuration – unbypassed cathode


Following is a schematic of a typical common-cathode amplifier stage with an unbypassed cathode resistor:




This circuit is similar to the bypassed cathode configuration with the exception of the removal of the cathode bypass capacitor, Ck.  Removing this capacitor adds degenerative negative feedback at the cathode, which lowers the gain of the stage and increases the output impedance at the plate.  The frequency response due to the input and output coupling capacitors remains the same as for the bypassed configuration.



  • Output Impedance (output taken from the plate, unbypassed cathode)



    The internal plate resistance will increase if there is negative feedback due to an unbypassed cathode resistor, and the voltage gain of the stage will decrease as well. Assuming a cathode resistor of 1.5K, the resistance seen looking into the plate when the cathode resistor is unbypassed is:

    ra'(unbypassed Rk) = ra + (mu + 1)*Rk
                                  = 62.5K + (101)*1.5K
                                  = 214K

    Therefore, the output impedance will be:

    R = ra’ || Rp
       = 214K || 100K
       = 68.2K


  • Gain (unbypassed cathode)


Now, the unbypassed gain is:


    Av(unbypassed Rk) = (mu*Rp)/(Rp+ra’)
                                   = (100*100K)/(100K+214K)
                                   = 31.85
                                   = 30dB


  • Cathode impedance (unbypassed cathode)



       
      It can be shown2 that the resistance seen looking into the cathode (Rk unbypassed) is:



    Rk’ = (Rp+ra)/(mu+1)
          = (100K + 62.5K)/(101)
          = 1.61K

    Therefore, the total cathode resistance, or the output impedance if you take the signal off the cathode, is the parallel combination of the cathode resistance, Rk’, and the cathode resistor, Rk, as below:

    R  = Rk’ || Rk
        = 1.61K || 1.5K
        = 777 ohms


The common-cathode amplifier configuration – partially bypassed cathode


If the cathode bypass capacitor is not large in comparison to the cathode resistance, the circuit is said to have a partially bypassed cathode.  The result of a partially bypassed cathode is a shelving frequency response.  This can be used to provide a small amount of treble boosting (or bass cut) for an amplifier stage.
 



  • Gain and Frequency Response (partially bypassed cathode)

 
For example, suppose you wanted to partially bypass a cathode to get a bit of treble boost, and you chose a 0.1uF cathode bypass capacitor and a 1.5K cathode resistor.


    The minimum gain that you can get from this stage (at the lowest frequency) is equal to the value of the unbypassed gain, which was calculated above to be 31.85, or 30dB. The highest gain you can get is the value of the fully bypassed gain, which was 61.5, or 35.7dB.   The gain transitions smoothly between these two values at a frequency set by the cathode bypass cap and the equivalent cathode resistance (it will be -3dB down at that transition frequency breakpoint).

    What this means is that the partially bypassed cathode results not in a first order highpass filter with a -6dB per octave slope extending down to DC, but rather a shelving equalizer, in this case with a response that starts at 30dB, is 33dB at 1.2KHz, and then rolls off to a maximum of 36dB above that. It looks like this:
     



In this case, there is only a 5.8dB boosting of the high frequencies (or cutting of the lows, depending on how you want to look at it). On the other hand, a plate coupling capacitor, in conjunction with the next stage input impedance, acts as a true first order highpass filter, and has a -6dB per octave (-20dB per decade) slope extending down to DC. When shaping the frequency response of an amplifier, it is important to know which one to use, depending upon the desired result. 

If the plate resistor, cathode resistor, and bypass capacitor values were different than the ones used in the example, the “center” point of the shelf would shift left or right, and the ratio between the top and bottom of the “shelf” would change.   Basically, the lower gain part of the “shelf” is set by the unbypassed gain of the stage, and the upper gain part of the shelf is set by the fully-bypassed gain of the stage, and the transition frequency is then set by the value of the bypass cap in relation to the equivalent cathode resistance.



  Appendix A:  The math behind the output frequency response highpass filter:


1Transfer function determination:


The output of the tube can be modeled as a voltage source in series with a resistance equal to the internal plate resistance in parallel with the plate load resistor as mentioned previously.  The entire output stage circuit can then be modeled as a voltage source with this output resistance, R1, in series with the coupling capacitor, C,  followed by the load resistance, R2, to ground.

Using the voltage divider rule, the transfer function can be derived as follows:


Vout =       Vin*R2    
              R1+R2+1/sC

Substituting H(s) for Vout/Vin, we get:

H(s) =         R2        
            R1+R2+1/sC

Multiplying top and bottom by sC, we get:

H(s) =         sCR2          
            sCR1+sCR2+1

       =          sCR2          
             sC(R1+R2)+1


Dividing top and bottom by C(R1+R2), we get:

H(s) =          sCR2  
                C(R1+R2)      
             s +          1      
                    C(R1+R2)

Canceling out theC terms in the numerator, we get:

H(s) =           sR2  
                 (R1+R2)      
             s +          1      
                    C(R1+R2)

This is now in the form of a standard highpass filter transfer function as shown below:

H(s) =    Ks
            s + a
           
where K = a constant indicating the magnitude the transfer function approaches at very high frequencies, and
           a  = the cutoff frequency of the network (in radians)

So, for the example above,

            K =      R2    
                   (R1+R2)
and

             a =        1        
                   C(R1+R2)

Thus, the magnitude approaches  R2/(R1+R2) at high frequencies, and the cutoff frequency (in radians) is 1/C(R1+R2).

Example:  In the case of the fully bypassed cathode, the output impedance was 38.5K ohms.  If we had a circuit with a 0.022uF coupling capacitor and a 100K load resistance following it, R1 would be 38.5K, R2 would be 100K, and C would be 0.022uF, and the maximum gain would approach 100K/(38.5K+100K) = 0.722 times the maximum gain if there were no coupling capacitor or load resistance, which is equal to -2.83dB loss, and the frequency response -3dB point would occur at 1/(0.022uF*(38.5K+100K)) = 328.19 radians, or 52 Hz (to convert radians to Hz, divide by 2*pi).

The schematic and frequency response plots for this example are shown below.  As you can see, the response is 2.8dB down from the 0dB line, and the -3dB cutoff point is at 52Hz.





Appendix B:  The math behind the equivalent cathode impedance:


2Equivalent cathode resistance determination:

If you inject a voltage of 1V at the grid, with the cathode grounded, you get an equivalent “internal” voltage source of mu*Vgk, which then produces a current of mu*Vgk/ra (this is effectively a transconductance – the voltage at the grid produces a corresponding current change in the plate circuit).  There is no current flowing in the grid circuit, so the 1V at the grid contributes no direct current of its own except through the transconductance of the tube.

However, if you inject a voltage of Vgk at the cathode, with the grid grounded, you get a current of mu*Vgk/ra plus the current due to Vgk injected at the cathode, which now appears directly across (ra+Rp), since they are in series.  This means the equivalent voltage source is now mu*Vgk + Vgk, which is equal to (mu+1)*Vgk, so the current produced by this source is (mu+1)*Vgk/(ra+Rp).

This gives an impedance, seen “looking into” the cathode, of:

        Vgk/((mu+1)*Vgk/(ra+Rp))

which is equal to:

        (ra+Rp)/(mu+1)

since the Vgk terms cancel.

Example:

With a plate resistor of 100K, and internal plate resistance of 62.5K and a mu of 100, the 12AX7 will have a cathode impedance of  (62.5K+100K)/(100+1) = 1609 ohms.  This impedance is then in parallel with the actual cathode resistor, so if you used an 820 ohm cathode resistor, the actual cathode resistance would be 1609||820 = 543 ohms. 


Appendix C:  The math behind the lower cutoff frequency due to a partially-bypassed cathode

3This section under construction – thanks to Jean-Pierre Trolet for pointing out an error in the original document.  I will update it when I get some time to review and rewrite the document.

Read More