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Drawing Tube Load Lines

& Using Them to Design Tube Circuits

By Rob Robinette, last edited 10/13/2017

Have comments or corrections? Email rob at: 


To really understand how tubes work you need to learn to use the tool used by circuit designers to build amp circuits--load line charts. You can't help but learn about tube circuit operation by playing around with these charts. You'll get to know how all the basic parts interact like, "What happens when I increase the value of the plate load resistor? How about decreasing plate voltage? What supply voltage will give me the cleanest signal with the most headroom?" These charts will answer those and many other questions about tube operation.

Looks complicated but it's not. Plate voltage on the bottom, plate current along the left and grid voltage curves inside. Saturation (valve wide open) and grid clipping occurs near the 0V grid voltage curve. Cutoff occurs at the bottom of the chart at 0 amps. Notice the curved grid voltage lines? They are curved so they are nonlinear and always induce nonlinear distortion which also causes harmonic and intermodulation distortion. This is why tube amps tend to sound "warm". The distortion fattens up the tone of anything being amplified by a tube circuit.

Notice how the grid voltage lines are curved? Since the lines aren't straight we say they are "nonlinear" and those curves cause nonlinear distortion even when the amp is played clean--but that's not a bad thing. Nonlinear distortion generates harmonic and intermodulation distortion and is what makes a guitar amp sound warm and thick compared to when you plug a guitar into a Hi-Fi audio amplifier (if you haven't tried it it sounds terrible--thin and anemic). If a tube operates in the lower, "curvier" part of the chart we'll get more nonlinear, harmonic and intermodulation distortion.


5E3 Deluxe Second Gain Stage

Let's draw the load lines for the 5E3 Deluxe second gain stage (V2A). It's a very common gain stage that's used in many amps. The V2A gain stage uses a 12AX7 triode with a 250V plate supply voltage, 100k plate load resistor and 1.5k cathode resistor. The following stage's impedance load is made up of a 1M grid leak resistor + a 56k tail load resistor.

We'll be drawing the load lines for the second preamp stage V2A (center left).

To get your blank plate characteristics chart just Google "12AX7 datasheet" and find the "Average Plate Characteristics" chart then copy and paste the chart into your drawing program. I really like the free irfanview drawing program.

Here's a blank 12AX7 Plate Characteristics chart you can use with your drawing program to plot lines.

12AY7 chart   5751 chart   12AU7 chart   12AT7 chart   12BH7A chart

To draw other tube load lines just Google their datasheet and follow the directions below. Of course you can always use paper, pencil and a ruler too.

The Completed Load Line Chart

That's a lot of lines! This is what we'll end up with when we finish drawing. Note the color coded legend at upper right. Load line in red, cathode load line in magenta, AC load line in yellow and operating point in green.


Draw the Load Line

We start our exercise with the DC plate load line. When you see just "load line" that's what is meant, the "DC plate load line". The bottom point of the line is the plate supply voltage (B+3) of 250 volts. We use the supply voltage because when current flow is cutoff by the tube there well be no voltage drop across the plate load resistor so plate voltage will equal supply voltage. The bottom of the chart is the 0 current (cutoff) line and at cutoff plate voltage = supply voltage.

We calculate the left end of our line by using Ohm's law and dividing the supply voltage by the plate and cathode resistors (100k plate load resistor + 1.5k cathode resistor)

        250V / 101,500 = .00246A or 2.46ma

We add the resistance of the cathode and plate resistors because all the current that flows through the tube has to flow through those two resistors.

If we we're drawing a phase inverter load line we would include the "tail resistor" and presence pot resistance in the equation.

Here's a simple Excel spreadsheet that will calculate the load line, cathode load line and AC load line points: Calculate Tube Load Lines.xls

We then draw our load line from 250 volts on the bottom of the chart to 2.46ma on the left side of the chart.

Load Line

The plate load line is drawn from our power supply voltage of 250V to our calculated plate current of 2.46 milliamps.

Increasing the value of the plate load or cathode resistor will reduce plate current and make the load line more horizontal.

Reducing the value of either resistor will increase plate current, make the load line more vertical and reduce gain.

Increasing the supply voltage will shift the line to the right but it will stay pretty much parallel to the original load line. If we change the supply voltage to 350 volts the load line would cross more grid voltage curves. That means we'd have more grid voltage swing available with more headroom.

Reducing supply voltage will shift the load line left and offer less headroom.

The slope of the load line is the primary indicator of gain. The steeper (more vertical) the load line the lower the gain. Note how the cathode resistor plays an insignificant role in gain. It's size barely changes the slope of the line due to its small size compared to the plate load resistor.

If we were designing a gain stage from scratch we could draw a line from our desired supply voltage to our desired plate current then calculate the external plate resistance (plate load resistor + cathode resistor) by dividing our chosen plate supply voltage by the plate current. Using the numbers from the chart above we divide 250V supply by .00246A = 101.6k ohm plate load which is very close to our actual plate load made up of the 100k plate load resistor and 1.5k cathode resistor.


Draw the Cathode Load Line

Cathode Load Line

I plotted two points, one on the -1V grid voltage line and one on the -2V line, then connected them to draw the cathode load line.

Our 5E3 amp's second gain stage uses a 1500 cathode resistor. We'll plot two points, one on the -1V grid line and another on the -2V grid line.

        -1V grid line = 1V / 1500 = .67ma and plot this point on the -1V grid line.

        -2V grid line = 2V / 1500 = 1.33ma  and plot this point on the -2V grid line.

We then draw the cathode load line between the two plotted points.

The size of the cathode resistor sets the operating point (intersection of the plate and cathode load lines detailed in the next section) along the red plate load line which affects headroom. For the most clean headroom we want the operating point halfway between the 0 volt grid line (saturation and grid clipping) and the bottom of the chart (cutoff). If we want less clean headroom we can shift the operating point away from the mid point. The Cold Clipper further down the page does just that by placing the operating point very low in the chart.

When we increase the value of the cathode resistor the cathode load line will become slightly more horizontal and offer a little more gain.

When we decrease the cathode resistor the cathode load line will become more vertical and offer a little less gain. A smaller cathode resistor will warm the bias but it will not add gain.

When plotting these points you may have to use other grid voltage lines. For example when I plotted the Marshall cold clipper stage further down the page I had to plot using the -2 and -3V grid voltage lines to get the cathode load line to cross the plate load line. For an 820 ohm cathode resistor I had to use the -.5 and -1.5V grid lines.


Draw the Operating Point

The operating point is the intersection of the load line and cathode load line. The operating point is also known as the bias point, idle point, no signal point, quiescent point and Q point.

Operating Point

The operating point is the intersection of the plate load line and cathode load line. Notice how the bias point is just about in the middle of the load line from the 0 grid volt line to the bottom of the chart. That means this triode is close to being "center biased" which minimizes distortion and allows the maximum plate voltage swing before symmetric clipping begins. This chart says our triode should idle at the green lines, 159 plate volts and .88 milliamps of plate current. Beam resistance at idle is: 159V / .00088A = 181k.

A lower value cathode resistor will "warm the bias" by shifting the operating point to the left and up along the red load line. A higher value cathode resistor will "cool the bias" and shift the operating point down and to the right. Moving the operating point away from the mid point between the 0v grid line and bottom of the chart will reduce clean headroom which is often desired in guitar amplifiers.

If we boosted the supply voltage to 300 volts but left the plate load at 100k and cathode resistor at 1.5k our load line would run between 300V and 2.96ma (300v / 101.5k = 2.96ma). The chart's grid voltage curves are straighter along this 300V load line so we'd get a little less distortion. We'd also get a little more headroom from the extra length along the load line between the 0V grid line (saturation) and the bottom of the chart (cutoff) for more available grid voltage swing and headroom.


Internal Plate Resistance

To calculate internal plate or anode resistance, called  ra  or  rp or  rip we'll move +/-25 plate volts from the operating point paralleling the nearest grid voltage curve and read the plate current at +25 and -25 plate volts.

We then divide the plate voltage change of 50 volts by the plate current change of .85ma to get the internal plate resistance:

         50V / .00085A = 58.8kΩ rp

Internal Plate Resistance

The cyan lines translate +/- 25 plate volts from the operating point to plate current. 1.69ma - .84ma = .85ma of plate current change for a 50 plate volt change. Note this chart is from the "Warm Biased 12AX7" section below.


Draw the AC Signal Load Line

Before we can determine our triode's AC signal voltage gain we must first plot the AC load line. Why is the AC plate load line different than the DC plate load line? Because the coupling cap connected to the plate stops the flow of DC. The DC load line isn't affected by anything beyond the coupling cap but the AC guitar signal passes through the coupling cap and gets loaded down by the impedance in the following circuit so we must consider the impedance beyond the coupling cap to draw our AC load line.

AC Load Line

The AC (signal) load line in yellow is slightly steeper (meaning less voltage gain) than the DC load line.

The AC load on our triode equals: plate load resistor + cathode resistor in parallel with the impedance of the following circuit. In the 5E3 the following circuit is the phase inverter with a 1M grid leak resistor and a 56k phase inverter tail resistor.

        plate load 100k + cathode resistor 1.5k || 1M grid leak + 56k phase inverter tail resistor. The || means "in parallel with".

        The equation for parallel resistance is: 1 / (1/101.5k + 1/1056k) = 92k  (hint: use your calculator's 1/x inverse key)

We then calculate the AC current: 250V / 92,000 = 2.72ma

We now draw the AC load line from 250V on the bottom of the chart to 2.72ma on the left.

Now this is the tricky part, we must slide the AC load line to the left until it intersects with our operating point (shifted line shown above in yellow). This is easy to do in some programs like Adobe Photoshop but others like irfanview don't allow it. Keep the AC load line's slope the same, just slide it to the left. You can also draw a second AC load line paralleling the first, but run it through the operating point, then erase the first line if possible.

The heavier the load of the following circuit, the steeper the AC load line becomes which reduces voltage gain.

Note: Since the 5E3's V2A gain stage's following circuit is a "bootstrapped" cathodyne phase inverter its input impedance is in reality much higher than our calculated 1056k so the true AC load line pretty much overlies the DC load line.


Draw the Voltage Gain

Voltage Gain

The operating point is at -1.3 grid volts. Go .5 grid volt left and right of the operating point along the yellow AC load line (gold lines) then drop down to get the plate voltages: -.8 grid volts = 130 plate volts. -1.8 grid volts = 188 plate volts. 188 volts - 130 volts = a voltage swing of 58, so for a 1 AC volt peak-to-peak change on the grid we'll get 58 volts of plate voltage change. In other words, 1 volt in is amplified into 58 volts out.

We can see from this chart that a heavy load from the following circuit, such as a TMB tone control, will make the AC load line more vertical and offer less voltage gain.


Max Voltage Swing

Max voltage swing (in light blue) runs along the yellow AC Load Line from the 0 Volt grid line on the left (representing saturation or grid clipping) to 0 milliamps of current at the bottom of the chart (representing cutoff). 372V - 64V = 308 volts of maximum voltage swing from this max effort preamp stage with 410V plate voltage, 270k plate load resistor and 4.3k cathode resistor. We have 188 plate volts from the 252v idle point to saturation at 64v and 120v from idle to cutoff  at 372v. A perfectly centered bias would have the operating point at -2.3 grid volts allowing a -2.3v swing to the 0v grid line and a +2.3v swing to cutoff.


Max Plate Dissipation Curve

Some Average Plate Characteristics charts include a maximum plate dissipation curve but many do not. If you need to plot the dissipation curve just divide the tube's max dissipation watt rating found in the datasheet by the plate voltage to get the current and then plot the points.

To plot the max dissipation at 500 volts you divide the 12AY7's max dissipation rating of 1.5 watts (each triode) by 500 volts:

        1.5W / 500V = 3 milliamps and plot the intersection of 500V and 3ma on the chart

        1.5W / 450V = 3.33ma and plot the intersection of 450V and 3.33ma on the chart

        1.5W / 400V = 3.75ma and plot it

        and so on, then connect the dots.

        To find where the curve hits the top of the chart at 7ma you can divide 1.5W by 7ma = 214.3 volts and plot 214.3V and 7ma.

12AY7 Max Plate Dissipation Curve

You can see the max dissipation line is very high on the chart and isn't normally a concern unless you are using the tube as a power tube.


Transconductance (gm)

Transconductance expresses the relationship between voltage change on the grid to the change in plate current. We see below that moving from -1 volt on the grid to -2 volts results in plate current going from 1.39ma to .16ma or 1.23ma per volt.

Transconductance

A 1 volt change on the grid (-1V to -2V) gives you a change of 1.39ma - .16ma = 1.23ma of plate current change per volt (shown in orange).

.00123A change * 1V change = .00123 mho of transconductance.


mu ()

mu is the tube's amplification factor and is a constant. The 12AX7 datasheet lists it as 100. mu is determined by the physical construction of the tube. A fine, closely spaced control grid placed very close to the cathode will generate the maximum mu. You don't need to graph the mu when you plot load lines, I just want to show how mu fits into the Average Plate Characteristics chart.

mu

These mu lines show how it fits into the Plate Characteristics chart. We keep plate current constant at .88ma and move from the -1V grid line to the -2V grid line and get 100 volts of change. For a 12AX7 triode mu is always 100. mu represents the maximum possible gain the tube can deliver. To get a voltage gain of 100 the AC load line would have to be horizontal which is practically impossible. This is why a tube circuit's gain is always less than the tube's mu.


The Finished Product

Load Lines and Operating Point

With no AC guitar audio signal on the grid the tube idles at the operating point shown in green.

When an AC guitar audio signal hits the grid the tube's plate voltage and plate current will move along the yellow AC load line.

When the positive voltage half of the guitar signal hits the grid the grid voltage increases which means electrons are pulled from the grid wire. With fewer electrons on the grid to repel the free electrons given off by the cathode more of them will flow through the grid to the plate--the electrons are attracted to the plate's high voltage. More electrons flowing onto the plate = more plate current. More plate current increases the voltage drop across the plate load resistor--this change in voltage drop is the amplified guitar audio signal.

When the negative voltage half of the AC guitar signal hits the grid the grid voltage drops meaning electrons are pushed onto the grid wire. These extra electrons on the grid repel more free electrons and the flow of electrons through the grid decreases which reduces plate current. Less plate current causes the voltage drop across the plate load resistor to decrease. Again, this change in voltage drop is the amplified guitar audio signal.

A small guitar audio signal on the grid is amplified by the tube and plate load resistor. The plate load resistor changes the tube circuit from a current amplifier into a voltage amplifier.

For more info on how tubes work see How Tubes Work.

 

Real world 5E3 voltages. V2A's B+3 plate supply voltage is 247v, the plate voltage is 164v and the cathode voltage is 1.2v.

Our load line chart predicts an idling (no signal) plate voltage of 159v compared to 164v measured. It also predicts a cathode voltage of 1.3v with 1.2v measured. This is pretty close considering the imprecise nature of carbon composition resistor values.


The Marshall Cold Clipper

Lets do a load line plot for the Marshall cold clipper gain stage. It's designed to generate early distortion and asymmetric clipping by clipping on the cold side of the operating point. It uses a 12AX7 with a 280V supply voltage, 10k cathode resistor and a 100k plate load resistor. The following circuit load is a 470k attenuator resistor and 470k grid leak resistor in series for a total impedance of 940k.

V1A Cold Clipper In the Marshall JCM800

The cold clipper gain stage is shown at center. Note the 10k unbypassed cathode resistor.

Draw the Load Line

We start with the DC plate load line. The bottom point of the line is the plate supply voltage of 280 volts. We calculate the left side point by dividing the supply voltage by the total resistance (100k plate load resistor + 10k cathode resistor)

        280V / 110,000Ω = .00255A or 2.55ma.

We then draw our load line from 280 volts on the bottom of the chart to 2.55ma on the left side of the chart (shown below in red).

Marshall Cold Clipper Load Lines

The operating point (intersection of green lines) is very low in the curvy end of the grid voltage lines so the negative half of the guitar audio signal is distorted even before clipping occurs. Guitar audio signal clipping will occur with the negative lobe of the signal voltage much earlier than the positive lobe. This gain stage's cold bias will lead to early sweet sounding asymmetric cutoff clipping.

Draw the Cathode Load Line

Our cold clipper stage uses a 10,000 cathode resistor. To draw the cathode load line we'll plot two points, one on the -2V grid line and another on the -3V grid line. When I first drew this chart I tried the -1 and -2 grid voltage lines but the cathode load line didn't cross the plate load line so I had to use the -2 and -3V grid voltage lines to find the triode's operating point.

        -2V grid voltage line = 2V / 10,000 = .20ma and plot this point on the -2V grid line.

        -3V grid voltage line = 3V / 10,000 = .30ma  and plot this point on the -3V grid line.

We then draw the cathode load line between the two plotted points (shown above in magenta). Notice how this very high cathode resistor value flattens the cathode load line--it's almost horizontal.

Draw the Operating Point

The operating point (no signal idle, quiescent point or Q point) is the intersection of the plate load line and cathode load line. It sits at 250V and .252ma (shown in green above). Notice how low the operating point is in the chart. There's very little headroom below and to the right of the operating point causing signal distortion and clipping which is the purpose of the cold clipper.

Draw the AC Load Line

Before we can determine cold clipper voltage gain we must first plot the AC load line. The AC load on the cold clipper equals: plate load resistor + cathode resistor in parallel with the impedance of the following circuit. In the JCM800 the following circuit's impedance is 940k as discussed above.

        plate load 100k + cathode resistor 10k || following stage 940k

        The equation is: 1 / (1/110k + 1/940k) = 98k

We calculate the AC current: 280V / 98k = 2.84ma

We now draw the AC load line from 280v on the bottom of the chart to 2.84ma on the left.

We then slide the AC load line to the left until it intersects with our operating point (shown above in yellow). This is easy to do in some programs like Adobe Photoshop but others like irfanview don't allow it. Keep the AC load line's slope the same, just slide it to the left.

Chart the Voltage Gain

Go .5 grid volt left and right along the AC load line and drop down to get the plate voltages. 267 volts - 225 volts = a voltage gain of 42, so for a 1 AC volt peak-to-peak change on the grid we'll get 42 volts of plate voltage change. 1 volt in is amplified into 42 volts out. This voltage gain is quite a bit lower than our 5E3 gain stage above of 58.


The Lead Channel Mod

The "Lead Channel Mod" for the 5E3 Deluxe, blackface, silverface and many other amps changes one channel's first gain stage to use a 12AX7 with a 2.7k cathode resistor, 220k plate load resistor, smaller .68uF cathode bypass cap and a smaller .0047uF coupling cap. The mod adds some gain and early harmonic distortion and makes asymmetric clipping distortion more likely.

The lower "Lead" channel has a 2.7k cathode resistor and 220k plate load. The following circuit impedance is made up of two 1M volume pots in parallel. The schematic shows a 12AY7 tube but a 12AX7 is a common substitute for higher gain players.

Draw the Load Line

We start with the DC plate load line. The bottom point of the line is the plate supply voltage of 250 volts. We calculate the left side point by dividing the supply voltage by the total resistance (220k plate load resistor + 2.7k cathode resistor)

        250V / 222,700Ω = .00112A or 1.12ma.

We then draw our load line from 250 volts on the bottom of the chart to 1.1ma on the left side of the chart (shown below in red). Notice how this very high plate load resistor flattens out the load line for more voltage gain.

Lead Channel Mod Chart

12AX7 with 2.7k cathode resistor and 220k plate load resistor.

Draw the Cathode Load Line

The Lead Channel stage uses a 2,700 cathode resistor. To draw the cathode load line we'll plot two points, one on the -1V grid line and another on the -2V grid line.

        -1V grid line = 1V / 2,700 = .37ma and plot this point on the -1V grid line.

        -2V grid line = 2V / 2,700 = .74ma  and plot this point on the -2V grid line.

We then draw the cathode load line between the two plotted points (shown above in magenta).

Draw the Operating Point

The operating point (no signal idle, quiescent point or Q point) is the intersection of the plate load line and cathode load line. It sits at 135V and .5ma (shown in green above).

Notice how there's 2.7 volts from the operating point to the 0V grid curve and 3.4 volts from the operating point to the bottom of the chart (cutoff). A larger cathode resistor of 3.3k would get us closer to center bias for a little more headroom but the purpose of the Lead Channel mod is to add gain AND distortion.

Draw the AC Load Line

Before we can determine Lead Channel stage voltage gain we must first plot the AC load line. The AC load equals: plate load resistor + cathode resistor in parallel with the impedance of the following circuit. In the 5E3 the following circuit's impedance changes with the volume setting. With the Lead Channel volume pot at max the impedance is 500k (impedance of both 1M volume pots in parallel) so we'll use this for the chart.

        plate load 220k + cathode resistor 2.7k || 500k

        The equation is: 1 / (1/222.7k + 1/500k) = 154k

We calculate the AC current: 250V / 154k = 1.62ma

We then draw the AC load line from 250v on the bottom of the chart to 1.62ma on the left.

We must slide the AC load line to the left until it intersects with our operating point (shifted AC load line shown above in yellow). This is easy to do in some programs like Adobe Photoshop but others like irfanview don't allow it. Keep the AC load line's slope the same, just slide it to the left.

Using a higher value plate load resistor will boost gain but it also adds impedance making our guitar audio signal "thinner" with less current backing up the signal voltage swing.

Chart the Voltage Gain

Go 1/2 grid volt left and right along the AC load line and drop down to get the plate voltages. 164 volts - 103 volts = a voltage gain of 61, so for a 1 AC volt peak-to-peak change on the grid we'll get 61 volts of plate voltage change. 1 volt in is amplified into 61 volts out. Using a 220k plate load resistor gives us slightly more voltage gain than the factory gain stage of 58.


Warm Biased 12AX7

Let's draw the load lines for another common gain stage, a 12AX7 with an 820 ohm cathode and 100k plate load resistor.

Draw the Load Line

We start with the DC plate load line. The bottom point of the line is the plate supply voltage of 280 volts. We calculate the left side point by dividing the supply voltage by the total resistance (100k plate load resistor + 820 cathode resistor)

        280V / 100,820Ω = .00278A or 2.78ma.

We then draw our load line from 280 volts on the bottom of the chart to 2.78ma on the left side of the chart (shown below in red).

Warm Biased 12AX7 Gain Stage

12AX7 with an 820Ω cathode resistor, 100k plate load resistor and 280V supply voltage. The bias point is further up the load line at -1V on the grid so the AC guitar signal can swing from 0 grid volts to -2V without clipping. The positive half of the AC guitar signal will be clipped earlier at saturation or grid clipping before the negative half of the signal is clipped at cutoff.

Draw the Cathode Load Line

Our 12AX7 is warm biased with an 820 cathode resistor. To draw the cathode load line we'll plot two points, one on the -.5V grid line and another on the -1.5V grid line.

        -.5V grid line = .5V / 820 = .61ma and plot this point on the -.5V grid line.

        -1.5V grid line = 1.5V / 820 = 1.83ma  and plot this point on the -1.5V grid line.

We then draw the cathode load line between the two plotted points (shown above in magenta).

Draw the Operating Point

The operating point (no signal idle, quiescent point or Q point) is the intersection of the plate load line and cathode load line. It sits at 154V and 1.25ma and -1V grid voltage (shown in green above).

Notice how there's 1 volt from the operating point to the 0V grid curve and 2.6 volts from the operating point to the bottom of the chart (cutoff). The operating point is higher and to the left in the warm bias region. An overdriven guitar signal will clip at saturation and grid clipping (0 volt grid) before it is clipped at cutoff (bottom of the chart).

Draw the AC Load Line

The AC plate load equals: plate load resistor + cathode resistor in parallel with the impedance of the following circuit. To keep things generic we'll say there is a 1M grid leak in the following circuit.

        plate load 100k + cathode resistor 820 || 1M

        The equation is: 1 / (1/100,820 + 1/1M) = 92k

We calculate the AC current: 280V / 92k = 3.06ma

We then draw the AC load line from 280v on the bottom of the chart to 3.06ma on the left.

We must slide the AC load line to the left until it intersects with our operating point (shifted AC load line shown above in yellow). This is easy to do in some programs like Adobe Photoshop but others like irfanview don't allow it. Keep the AC load line's slope the same, just slide it to the left.

Chart the Voltage Gain

We go .5 grid volt left and right along the AC load line and drop down to get the plate voltages. 184 volts - 122 volts = a voltage gain of 62, so for a 1 AC volt peak-to-peak change on the grid we'll get 62 volts of plate voltage change. 1 volt in is amplified into 62 volts out. Using a warm bias with an 820Ω cathode resistor gives us slightly more voltage gain than the typical gain stage with a 1.5k cathode resistor's voltage gain of 58.


12AX7 Max Gain Stage

Here's a max effort to get the most gain and clean headroom out of a preamp stage by using a 12AX7 with a high supply voltage of 410 volts, a large 270k plate load resistor and 4.3k cathode resistor. Although the 12AX7 datasheet lists the maximum plate voltage at 300 volts these tubes run fine with much higher plate voltages (see the Fender tremolo oscillator below with 415V on the plates). We'll say the following circuit has an impedance of a 1M grid leak resistor.

A very high supply voltage, large value plate load resistor and cathode resistor yield a voltage gain of 69 and we have a very wide grid voltage swing available of 0 to -4.7 volts for lots of clean headroom. This gain stage is center biased at -2.4 grid volts allowing a 4.7V peak-to-peak signal before symmetric clipping occurs.

DC plate current = 410V / (270k + 4.3k) = 1.50ma

Cathode load line points

        -2V / 4300Ω = .47ma

        -3V / 4300Ω = .70ma

AC plate current = 410V / (1/( 1/(270k + 4.3k) + 1/1M)) = 1.90ma

Voltage gain = 283V - 214V = 69


AB763 Tremolo Oscillator

Fender AB763 blackface and silverface amps use a 12AX7 as a tremolo oscillator with a very high supply voltage of 415V, a large 220k plate load resistor and 2.7k cathode resistor. The following circuit has an impedance of 429k. The 12AX7 datasheet lists the maximum plate voltage as 300 volts. This is the hottest 12A*7 gain stage I know of.

AB763 Tremolo Oscillator

An extremely high supply voltage, large value plate load resistor and cathode resistor yield a voltage gain of 64, not that much higher than the typical gain stage's 58. This gain stage is near center bias at -2.2 grid volts allowing a swing from 0 to -4.4 grid volts before saturation or grid clipping occurs near 0 grid volts.

DC plate current = 415V / (220k + 2.7k) = 1.86ma

Cathode load line points

        -1.5V / 2700Ω = 5.56ma

        -2.5V / 2700Ω = 9.26ma

AC plate current = 415V / (1/( 1/(220k + 2.7k) + 1/1M)) = 2.28ma

Voltage gain = 269V - 205V = 64

This stage is pushed well beyond the max plate voltage of 300 volts but the tubes hold up well in blackface and silverface amps.

 

By Rob Robinette

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