Designing an amplifier for tube rolling requires special power supply considerations. The power transformer must be able to adequately supply current to power the filaments of the various tubes plugged into the circuit. The plate and screen B+ supplies must be a voltage below the maximum plate and screen ratings of the tubes plugged into the circuit. The plate and screen B+ supplies must provide adequate current without overloading the power supply. It will be necessary to calculate circuit values that allow various output tubes to function properly. In general, most circuit values can be approximated.

There are two approaches to designing an amplifier capable of using different output tubes. Use switches that would select operation modes for different types of output tubes, or keep the design simple by only accepting tubes that have close operating parameters. Switchable circuits tend to develop contact corrosion over time. Switching modes with corroded or worn contacts could cause some serious problems. Also, you might forget to switch modes and destroy a tube or damage circuit components. For the following design example, tubes that have identical pin connections and close operating parameters will be used.

Pentode and beam power tubes have the widest selection of tube types suitable for tube rolling. Although several power tube types may have the same pin connections and similar operating characteristics, how they perform depends on how well they match circuit component values. A sample of output tubes with the same pin connections includes 6V6GT, 6L6GC, 5881, 6CA7/EL34, 6550, KT66, KT77, and KT88. The following lists the tubes in order from the lowest filament current to the highest filament current. All the listed tubes have the same filament voltage, 6.3 volts.

Select a group of tubes with filament currents that are within a .6 amp range. For instance, you could use a 6V6GT, 6L6GC, and a 5881 as alternate output tubes. The rest of the tubes have a much larger filament current value compared to the 6V6GT. Another set of tubes with close filament current ratings are 6L6GC, 5881, KT66, KT77, and 6CA7/EL34. Tubes with higher filament currents within a .6 amp range include KT66, KT77, EL34, 6CA7/EL34, KT88, and 6550.

Maximum Power and Plate/Screen Voltage
Make a list of Class A maximum outputs to use as a guide for each tube's power output capability. Include the datasheet's stated maximum plate and screen voltages for each tube.

The list gives you an idea of each tube's power output capability. If you are designing a single-ended output amplifier, the power output listed above is close to what you will have. In the case of a push-pull output, power output will be approximately three or more times as high as listed above. The actual power output will vary depending on the power tube type used and the operating voltage values.

In a group of tubes, it is important that the power supply B+ voltages do not exceed any of the tube's maximum plate and screen voltages. Use the tube with the lowest maximum voltage rating as a limit. For example, if you intend to use either a 6V6GT, 6L6GC, or 5881 output tube, the plate and screen B+ supply voltages cannot be higher than the 6V6GT's maximum plate or screen voltages.

In most cases, the limiting factor for maximum output is not to exceed a tube's maximum plate or screen dissipation. Plate dissipation is the power dissipated as heat from the plate. Screen dissipation is heat dissipated by the screen grid.

Consider plate dissipation as the difference between the power supplied to the plate and the power delivered by the tube to a load. For example, a single-ended Class A 5881 output tube with a plate voltage of +385 volts and a plate current of 0.075 amps (75 mA) delivers a full output power of 12 watts to an eight-ohm load. The power dissipated by the plate is equal to the full 29 plate watts.

For instance, an output tube with a plate voltage of 385 volts and 75 mA of current. (385 X 0.075 = 29 watts), minus the 12 watts of power delivered to the load, or 17 watts. Simply multiplying the plate voltage by the plate current implies the tube delivers no power to the load. The true plate dissipation at full output is the full output plate watts minus the power delivered to the load 29 − 12 = 17 watts.

In the selected group of tubes, if the output tube with the lowest maximum B+ voltage rating is used as a reference, all the other tubes in the group should remain below maximum plate and screen dissipation. The tube with the lowest maximum plate and screen voltage rating is used to set the B+ voltage for all the output tubes in the group.

The secondary winding of a transformer will produce its rated voltage when loaded at the winding's ampere rating. For instance, a 6.3-volt @ 2-amp winding will produce 6.3 volts with a 2-amp current load. If the winding is loaded at 1 amp, the voltage will be higher than 6.3 volts. If the winding is loaded at 3 amps, the voltage will be less than 6.3 volts. It is not good practice to exceed a transformer's secondary current rating. Excessive loading of a secondary winding can cause a transformer to run hot and possibly fail.

Although most tubes will operate with a filament voltage that is 10% higher or lower than their rated voltage (for a 6.3-volt filament, 5.7 volts to 6.9 volts), the RCA tube manual states the following.

This would seem to indicate that for the longest tube life the filament voltage should be as close as possible to its rated voltage.

Similar to the filament winding, loading also affects the high-voltage secondary of a transformer. A 600-volt center-tapped (300-0-300) secondary rated at 200 mA will deliver 600 volts when loaded at 200 mA. If loaded at 150 mA, the voltage will be higher than 600 volts, and if loaded at more than 200 mA, the voltage will be less than 600 volts.

The group of tubes chosen that have filament currents within a 0.6-amp range are 6L6GC, 5881, KT66, KT77, and 6CA7/EL34. Two output tubes are required for a push-pull output amplifier. The phase inverter and voltage amplifier will be a 12BH7 or 12AU7. A high-gain preamplifier will be a 12AT7, 12AX7, or 12AY7. The filament voltage of the 12BH7 is 6.3 volts, and the current load is 0.6 amps. A 12AU7 has a filament voltage of 6.3 volts with a current load of 0.3 amps. The filament voltage of the 12AT7 or 12AX7 or 12AY7 is 6.3 volts, and the current load is 0.3 amps. The amplifier will have a total of four tubes: two output tubes, a combination phase inverter and voltage amplifier, and a pre-amplifier tube.

The current load that a transformer's filament winding must be rated for is calculated by using the output tubes that will draw the highest filament current. In this example, the 6CA7/EL34 is the highest at 1.5A. For a push-pull output, two tubes are required. For two output tubes, the current load will be 3.0 amps (1.5A + 1.5A). Add to that the 12BH7 and pre-amplifier filament loads, and the total filament load current is 3.9 amps. A power transformer with a 6.3-volt filament winding rated at 4 amps would have sufficient current to power the selected tubes.

Just as important as the maximum filament current is the minimum filament current. This would be the total filament current load using the output tubes with the lowest current load, in this example, the 6L6GC tubes. Two 6L6GC tubes have a total filament load of 1.8 amps. Add to that the 12BH7 and pre-amplifier filament loads for a total filament load of 2.7 amps. With the lower filament current load, the transformer's 6.3-volt filament voltage will be higher than 6.3 volts. It is important that the filament voltage stay below 6.9 volts, regardless of what output tubes are used. Use the following formula to approximate how much higher the 6.3 volts will be with a 2.7 amp load.
EuL = {[( 1 − (IaL/IfL)) X .1] X EtR} + EtR

The transformer's 6.3 volt filament winding is rated at 4 amps, but with an actual load of 2.7 amps.

Write out the formula, then plug in the numbers.
(Where appropriate, round out numbers to two digits after the decimal place.)
EuL = {[(1 − (IaL/IfL)) X 0.1] X EtR} + EtR
EuL = {[(1 − (2.7/4)) X 0.1] X 6.3} + 6.3

(Solve the inner most brackets first, then work your way out doing the next inner most brackets)

EuL = {[(1 − (2.7/4)) X 0.1] X 6.3} + 6.3
2.7 divided by 4 = 0.68

EuL = {[(1 − 0.68) X 0.1⁠] X 6.3} + 6.3
1 minus 0.68 = 0.32

EuL = {[0.32 X 0.1] X 6.3} + 6.3
0.32 times 0.1 = 0.032

EuL = {0.032 X 6.3} + 6.3
0.032 times 6.3 = 0.20

EuL = 0.20 + 6.3
0.20 plus 6.3 = 6.5
EuL = 6.5 volts

6.5 volts is below the 6.9-volt upper voltage limit. The selected output tubes have filament loads that can be used with a 4 amp transformer filament supply. The operating voltage range is about 6.3 volts to 6.5 volts.

The design for this discussion is a cathode-biased push-pull ultra-linear output. Cathode resistors are used in a cathode bias arrangement. Cathode bias is a simple, less critical method than fixed bias and better suited for tube rolling. The cathode resistor should be a value that will produce a voltage drop across the cathode resistor equal to the required grid bias voltage. The value can be calculated by dividing the required grid bias voltage by the output tube's zero-signal current (idle current). If KT66, KT77, or KT88 tubes are used as output tubes, each output tube must have its own cathode resistor and bypass capacitor. The cathode bypass capacitor can have a value of 100 uF up to 220 uF and be rated at 100 volts.

By tradition, guitar amplifiers do not use ultra-linear outputs. However, ultra-linear matches up better between various tube types for tube rolling. An ultra-linear output transformer with a primary screen tap is required.

For the 6CA7/EL34 output tubes, the socket pin terminals must have pin 1 connected to pin 8. Except for the 6CA7/EL34, the other output tubes listed with identical pin connections do not use pin 1.

Since this is an ultra-linear output, the only maximum voltage consideration is the plate voltage. When operated ultra-linear, the screen voltage of an output tube may be operated at the plate's maximum voltage rating. This should allow plugging in any of the selected tubes without exceeding maximum ratings. The output tube with the lowest maximum plate voltage is used to determine the maximum B+ voltage. In this example, of the selected output tubes, the 5881 tube has the lowest maximum plate voltage at 400 volts. The power supply B+ for the output tubes cannot be any higher than 400 volts with 5881 output tubes plugged in. The B+ voltage should be slightly below the maximum rating, in this case at about 380 volts. This should allow plugging in any of the selected tubes without exceeding maximum ratings.

The B+ high-voltage supply must be able to supply the required current to power the output tubes that draw the most current. Add the maximum plate and screen dissipation of each tube in the selected group. For instance, using the plate and screen dissipation values listed earlier, the total maximum plate and screen dissipation of a KT77 is 31 watts (25 watts on the plate plus 6 watts on the screen). Of the selected tubes, the 6L6GC actually has the highest maximum plate and screen dissipation at 35 watts. Using 380 volts as the target high voltage value, use the following math to approximate the maximum current of two 6L6GC tubes operating in a push-pull output circuit:

The maximum current of the output tubes at full volume is 184 mA. Allow an additional 4 mA for each section of the 12BH7 (4 mA X 2 = 8 mA) and 2 mA for each section of the pre-amplifier tube (2 mA X 2 = 4 mA). The total current load on the transformer's high-voltage winding is 184 mA + 8 mA + 4 mA, or about 196 mA. The closest standard transformer current rating is 200 mA.

Capacitor Loading
Factoring in 19% capacitor loading applies to power supplies using solid-state rectifiers. For power supplies using a vacuum tube rectifier, datasheet load charts for the rectifier should be used.

Factoring a 19% capacitor loading effect for a 380VDC 200 mA solid-state supply: (380 x 0.707 = 269)
Transformer no-load value = 269 VAC
Factor 19%
Multiply the no-load value by 0.19.
(269 X 0.19 = 51)
Add to the no-load transformer value.
(269 + 51 = 320)
A transformer's high-voltage secondary rated at 640 VCT (320-0-320) @ 200 mA would be a close match.

Numbers in the above calculations were rounded off.

The power transformer's high-voltage secondary's actual voltage will vary depending on the load. Tubes that draw less current will result in an increase in the secondary voltage. A higher secondary voltage means a higher B+ voltage. For example, the high-voltage winding required for the amplifier is 640 VCT (320-0-320) at 200 mA. However, when 5881 output tubes are used, the current load is lower than the other output tubes in the selected group.

The maximum current of the 5881 output tubes at full volume is 136 mA. Allow an additional 4 mA for each section of the 12BH7 (4 mA X 2 = 8 mA) and 2 mA for each section of the preamplifier (2 mA X 2 = 4 mA). The total current load on the transformer's high-voltage winding with 5881 output tubes is 136 mA + 8 mA + 4 mA, or about 148 mA.

Refer to Figure one and follow the steps to calculate the percentage difference between 200 mA and 148 mA.

Figure One

Figure one is a graph that can be used to estimate how much higher the DC high voltage will be if a transformer with a higher current rating than required is used. To use the graph, you must first find the percentage difference between the full load current rating and the actual load current. In this case, 200 mA and 148 mA. The following formula is used to calculate the percent difference between two numbers:
% = [{(Lv − Sv) / Lv} X 100]
Lv = Larger value
Sv = Smaller value
− = Subtract
/ = Divide by
X = Multiply by
(solve the innermost brackets first, then work your way out.)
% = [{(Lv − Sv) / Lv} X 100]
% = [{(200 − 148) / 200} X 100]
% = [{52 / 200} X 100]
% = [0.26 X 100]
% = 26

The results are a 26% load current difference. To use the graph in Figure one, find 26%. On the left side of the graph. Follow the 26% point to the right until it meets the red line, then go straight up. A 26% load difference indicates about a 3.7% increase in B+ voltage. This will increase 380 VDC to approximately 394 VDC (380 X 0.0037 = 14, then 380 + 14 = 394). This would be about 394 VDC out of a solid-state rectifier. There should be a choke transformer coil after the rectifier. Depending on the choke DC resistance, there will be a voltage drop across the choke coil. The voltage drop across the choke coil will reduce the B+ voltage to the output tubes. If the voltage drop is significant, it may be necessary to increase the power transformer's high-voltage secondary voltage to compensate.

The output transformer's primary load resistance can be selected to match a preferred tube. If there is no preferred tube, then use an average of the selected tubes. Keep in mind that the output transformer must be power rated for the tubes that produce the highest power.

Figure two is a Class AB push-pull circuit. Power output is around 25 to 30 watts, depending on which pairs of output tubes are used. The limiting factor is the 5881 output tubes. Because of the 5881 tubes, the B+ supply voltage is limited to 380 VDC.

Figure Two

Figure three is a high-gain preamplifier. Input jacks are provided for two guitar pickups. The 100K resistors in series with each input are for isolation. The preamplifier has a voltage gain of 116 to 122, depending on the type of tube used. Some gain is lost through the isolation resistors. If there is a need to reduce gain, increase the value of R1. Reducing the value of R1 will increase the gain a bit. Any tone control circuits would be inserted between the preamplifier output and the Figure two master level control input.

Figure Three

Creating an amplifier where you are able to safely swap out various output tubes is an involved process. Swapping output tube types in an amplifier not specifically designed for a group of output tubes requires careful attention to tube datasheet specifications. Maximum plate (and screen) voltage plus plate dissipation maximums should not be exceeded.