When the first installment of the article, “Brief Comparisons of 7591 Vacuum Tubes,” was written, one thing became crystal clear: Commercial testers are not doing well at assessing the quality of output tubes. While some modern testers might be able to measure Gm accurately, it turns out that Gm is not a good indicator of the power capabilities of such tubes. [Gm is a measure of tube gain for small signals.] Reduced power capability is directly related to deterioration of the cathode, which is the primary wear out mechanism for power tubes. Moreover, we found that even high quality tube testers (with a couple rare exceptions) from the old days could not even test Gm accurately!
Fortunately, my friend Dave Gillespie had constructed a custom-designed, tube power tester and we used that to evaluate various versions of 7591 tubes. This article shows you how to how to measure a tube’s power capability using common lab equipment. It’s based on Dave’s test methodology and I’m grateful to him for sharing that. The principle behind the tester is simple: The tube is operated in a basic amplifier stage, intended to represent half of a push-pull pair, as shown at right.
A value of load resistor, RL, is chosen to get roughly maximum power out of a new tube, while staying within its ratings. For the test, grid bias is set for a specified quiescent plate current and the AC drive to the grid is increased until the output waveform clips at top and bottom. The DC voltage (Vpp) read at the output of the peak-to-peak detector indicates the output power that the tube under test can deliver. A math expression converts Vpp to watts. With that expressed as a percentage of the power that a new tube can deliver, we have the “score” of the tube.
Equipment used in the 7591A test included:
- Octal relay socket
- 400V power supply w/ 80uF cap
- (2) Lower voltage power supplies
- (2 - 4) Digital multimeters
- Dual trace oscilloscope
- 1650 ohm load resistor 25W (for 7591)
- Protoboard with P-P detector circuit
- Audio generator
- Misc (clip leads, 10K res, 0.1uF cap)
The schematic of the test system shown below in the procedure section labels the specific test equipment I used, just for reference. It’s all fairly standard gear but here are some tips about specific pieces:
High Voltage Power Supply
While average plate current is under 100mA for the 7591A, peak plate current is in the 200mA ballpark. Earlier when I tried to use the solid state B&K 1602 (rated at 200mA), its latching overload shutdown was being triggered on peaks. Adding the 80uF capacitor on its output only helped a little because the supply has a very low output impedance and was still attempting to supply much of the transient current. [This could probably have been alleviated by adding a resistor between the supply and the external cap, perhaps 10-30 ohms, 1-2W.]
The Fluke, vacuum-tube 407D is good to 300mA and worked fine without the 80uF capacitor for the 7591A. However, the 6550 test has average current under 200mA and peaks in the 300mA area, causing the Fluke to limit. Fortunately, its higher output impedance allows the 80uF capacitor to handle the peaks and it works well with that. Looking at the popular Heathkit IP-17 (similar to IP-2717), its 125mA limit should suffice for the 7591, 6L6GC and lesser tubes but not for tubes in the 6550 class. This assumes that the external 80uF output capacitor works with it to handle current peaks. That seems likely, given its existing 20uF output capacitor.
The 6550 test requires plate voltage of 400V and screen voltage at 300V. If you use a second supply for the screen voltage, I recommend that you include a protection diode (as at left), in case you make the mistake of turning off the plate supply first or turning on the screen supply first. (See warnings below.) If the screen is powered when the plate is not, plate current will be diverted to the screen and could well damage it. To prevent that, connect a 1N4007 diode from the plate supply to the screen supply, with the cathode towards the plate supply. Normally, the diode will be reverse biased and won’t conduct current. If a fault occurs in which the screen supply is higher than the plate supply, the diode will pull up the plate supply. If the screen supply is unable to do that, it will be pulled down instead, still protecting the tube.
I used a second supply for the screen but for folks who don’t have that, the simplest solution might seem to be a 100V, 5W Zener diode between the plate supply and the screen, as shown at right. The problem with this is the typical 5% tolerance of the Zener is way too loose for our measurement. It’s desirable to keep the high voltages within 1V; definitely within 2V. Instead of the Zener, you could employ the simple circuit at left. A pot divides the 400V B+ down to the voltage we need and a MOSFET follower buffers it. The FDPF12N60 shown is available at Mouser. It must dissipate up to 2W, so it needs a small heatsink. A 1.5” square of aluminum should suffice. Since the MOSFET has an insulated mounting tab, no insulating hardware is needed but heatsink grease should be used. MOSFETs with metal mounting tabs are more available. Most 500V+ N-channel MOSFETs in a TO220 package would work but usually have those metal tabs. (Since mine is rated at 12A, using one at least 8A would be prudent — may affect output resistance.) For example, the NTE2385 is available at Fry’s. Since the mounting tab is connected to the drain, the heatsink will be electrified at 400V unless you use insulating mounting hardware. The kicker is that the thin insulators normally used will not work for 400V, even if the material itself is rated for it. There is breakdown around the edges of the mounting hole. Aluminum oxide insulators 0.07” thick worked for me. The pot can be a trimmer or larger type. I was able to set it accurately to 300V±0.2V and the output drops less than 0.5V with a 20mA load.
It takes a substantial amount of voltage swing to drive the grids of power tubes. The maximums I recorded were (Vrms): 7591–16V, 6550–20V, 6BQ5–8V, 6L6GB/C–18V. The HP function generator I have can only muster 7Vrms out and was not up to the task. Similarly, the low-distortion oscillator on my bench is limited to about 6Vrms. Then I remembered that the vacuum tube Eico 378 audio generator I restored has an unusually high output capability. Into a 10K load, it can drive up to 18Vrms! Thus, the lowly Eico proudly took its place on the bench, surrounded by high-class HP, Tektronix and Fluke equipment.
It worked well for the 7591A tests but when it came to the 6550, it was just short of sufficient output. So I put a restored Eico Cortina 3070 amplifier on top and had it drive a small, reverse-connected, filament transformer, as seen at right. That provided plenty of voltage swing. NOTE: I later found that my Eico 378 has an issue with slow, random variation of output level. Over a period of a minute it can vary 2.7% or so. With the amplifier added, generator output isn’t critical and the modified Heath IG-18SL worked well.
Ia - Plate (anode) current
Iamax - Plate current limit
Iq - Plate bias current
Pouti - Ideal power out
RL - Load resistor
Vb - "B+" plate supply voltage
Vg1 - Control grid voltage
Vg2 - Screen voltage
Vh - Heater voltage
Vpp - Peak-to-peak plate voltage
How Many Multimeters?
While having four meters is convenient, you can certainly get away with fewer. For example, once the heater settles down, the Vh meter could be used for Vb and Vpp. [By the way, the Vh meter leads should go directly to the relay socket terminals to minimize errors due to wire resistance.] You do need a dedicated meter for Ia. If your plate power supply is stable enough, you could get away with just two meters — one for Ia and one for Vpp, Vb and Vh. However, when you’re ready to do the test, the Vpp meter must already be in place. That’s needed to minimize the time that the tube is operated at high current. The multimeters don’t have strict accuracy requirements but both the Vb and Vpp meters need to handle at least 400V, which will exclude some models. For example, the HP3468A and HP3478A only allow 300V inputs. (HP, I always thought you walked tall! :)
The 1650 ohm value is appropriate for the 7591 and 6L6GB/C. Values for two more are given in the table. Additional ones must be determined on a case-by-case basis. The idea is to exercise the tube in the ballpark of its maximum output power, while staying within the maximum ratings for cathode current, screen dissipation and (less critically) plate dissipation. I try to use data book examples for two reasons: They guarantee that the tube isn’t over stressed and they tend to reflect the way the tube was typically used. Choosing the right value is a bit of an art and depends in part on being able to clearly establish the power that an exemplary, new tube would produce with that load. That’s needed to define the 100% reference. Hence, we need to have plate curves for the screen voltage used in the example.
As of this writing, only the ideal power value (Pouti) for the 7591A has been verified against trusted NOS. However, all Pouti values in the table have been tested on multiple, used tubes, with good results. Since it affects the power calculation, it’s important to set the load resistance accurately. Multiple resistors or variable resistors may be needed. The power rating should exceed the Pouti value in the table.
Peak-to-Peak Detector Circuit
Seen in the schematic below and the photo at right, the peak-to-peak detector gives us an accurate, DC reading of the voltage swing that the tube causes across the load resistor. It’s accurate, that is, after adding 1.5V to the reading to account for diode drops and peak clipping. Some tips and notes:
- An oscilloscope reading is not a very accurate way to get the peak-to-peak measurement.
- The meter used for Vpp must have an input resistance of 10Mohm or greater to minimize peak clipping.
- The P-P detector works well so long as the tube is strong enough to flatten the bottom peak of plate waveform. For “pointy” peaks, we have seen readings as much as 1% low, so weak tubes might be read slightly low.
- The smaller cap and the resistor shown in the photo have been eliminated. The white clip lead from the tube’s plate now goes to the junction of the diodes.
Procedure to Test Output Tube Power Performance
First, set up the test equipment as shown in the schematic below but without the tube in the socket. See the table below for the value of RL for tubes other than the 7591 and 6L6GB/C.
The first 5 steps are setting up the equipment; then we plug the tube in and let it warm up (2-steps). There are 6 steps of the actual measurement. In the last 3 steps, we turn it off and calculate the tube’s power and score. In the 6 steps of the actual measurement, we get the tube warmed up with the correct heater voltage. Then the plate supply is turned on and the grid bias is used to set the zero-signal bias current. The heart of the test is to bring up the drive to barely clip-flat the bottom peak of the plate signal. The DC reading from the peak-to-peak detector gives us the result.
- Set up the oscilloscope with two 10:1 probes, the plate channel AC coupled at 50V/div and the grid channel DC coupled at 10V/div. Accurately set the 0VDC point at one division down from the top.
- Set the heater supply (Vh) for 6.3V or the proper value. Will need to readjust after the tube is inserted.
- Roughly set the grid supply (Vg1) using the meter on the supply. (Temporarily use another meter if need be.) See the table below for values of Vg1.
- Set the plate supply (Vb) to zero.
- Set the audio generator frequency to 1kHz and the output to zero. We will call this the “grid drive.”
- Insert the tube under test in the relay socket and readjust Vh after 10-20 seconds or so.
- Allow a few minutes for warmup. (Five minutes is better.)
- Readjust Vh. Meter leads should be right on the relay socket.
- Increase Vb to the value in the table, while watching Ia. If it goes over (say) 50mA, make Vg1 more negative.
- Adjust Vg1 to set Ia to the quiescent current, Iq, given in the table.
- From this point on, try to minimize the time but don’t rush. Increase the grid drive while watching the plate signal on the scope. Keep an eye on Ia. Don’t let it exceed Iamax (from the table) by more than 20%. Set it only high enough so you can just see the bottom of the plate waveform clipped flat. However, don’t let the grid waveform exceed 0VDC — weak tubes might not clip. (The tube is now under stress.) The pic at right shows a little too much clipping at the bottom. If this is the first time you’ve tried this test, briefly check Vb with the scope to make sure ripple is below (say) 1V peak-to-peak.
- Readjust Vb. It may have dropped due to the load.
- Write down the value of Vpp. This indicates the power reading.
- Reduce AC grid drive, Vb and Vh to zero. You can breathe easy now, as the tube stress has ended.
- Correct Vpp = Vpp + 1.5.* Figure the output power as Pout = (Vpp)2/(4 x RL) For example, if Vpp=350 and RL=1650ohms, then Pout = 18.56W.
*The Vpp correction is for diode drops and clipping.
- From the ideal Pout (Pouti) listed in the table below, calculate the tube’s score, S, as S = 100 x Pout/Pouti. The example yields 100%.
Table of Test Parameters
The parameters given here are generally based on data book examples which operate the tube near its maximum output in class AB1 service*. This attempts to test the tube at typical, yet stressful conditions. For more information about adding other tubes to the table, please see the “How is the ideal output power (Pouti) determined?” section below and the “Load Resistor” section above.
This method can be used by advanced hobbyists to accurately test the true condition of their rare and expensive power tubes. I don’t know of any commercial testers, new or old, which have the capability to make these power measurements. The lab equipment required is fairly common and is affordable for most serious tube aficionados. However, it does require good laboratory skills and the high voltages demand caution and experience.
Questions About the Procedure
How do I interpret the tube’s power score?
The tube’s “score,” calculated in the last step of the procedure, is its measured power, expressed as a percentage of ideal. But how do we interpret the result? Tube expert, Dave Gillespie, uses these criteria: A tube is GOOD if it produces more than 80% of established power output. It’s BAD if it produces less than 64% and it’s WEAK if it’s in-between.
One way to look at this is in terms of decibels (dB), which are used to measure loudness. The table at right shows the dB loss for various percentages of power. The smallest noticeable level difference is often taken to be about 1dB but to hear for yourself, check out the excellent examples at this site. I just use the “Files being tested” buttons, which play a tone that increases or decreases by 1dB after one second. Can you hear the difference? Maybe. Now, near the top of the page, select the 0.5dB examples. Can you still hear it change? I could hear the up-change but not the down-change. In any case, this demonstrates how small a 10 or 20% loss in power is. Moreover, the instant change of a continuous tone is best case in terms of audibility. If the same tones were heard a couple seconds apart, it would be far more difficult to hear the differences.
To judge the score, we have to clarify what we are using it for. Possible purposes include: (1) deciding whether a brand new tube is okay and (2) deciding whether to replace an existing tube. We would want to hold a brand new tube to a stricter standard, as it is usually expected to be near perfect. For that, I would want to see performance near 90% or better of the tube’s ideal value. However, I would also take into account known limitations of modern manufacture. For evaluation of an existing tube, before shelling out hundreds of dollars for replacements, it’s nice to know that it will make a significant difference. Bear in mind that slightly weak power tests, unlike Gm tests, do not necessarily warn that circuit bias operation may be impaired. Bias should be adjusted individually for output tubes. A loss of a couple dB would barely be noticeable in the dynamic range of music, so 60% power could be considered the threshold for replacement. If I wanted to be picky, I might use 70% as a threshold.
There is another consideration though: Suppose just one side of a push-pull pair is weak. This becomes a more serious matter because it can result in more distortion. If the output stage has dynamic balance adjustment, as was added to the Eico ST70A, this can be compensated, to a certain extent. However, most amps don’t have that. Distortion tests could be used to investigate and determine whether replacement is needed. Generally, I wouldn’t want to see more than (say) 10% difference in power performance between the tubes in a push-pull pair.
Where does the output power calculation come from?
In a push-pull output stage (at right), each tube pulls down its plate voltage from the B+ value applied to the transformer center tap. It’s working against the load resistance presented by its half of the primary winding. We simulate that in the test with the plate supply and the load resistor, RL. Nominally, each tube drives the output for a half cycle. The output power of the push-pull stage is determined by how far the tube can pull down its side of the transformer. In operation with a sine wave signal it’s drive to the load (ideally) looks like the dark portion of the sine shown below right. If it were a complete sine wave, we know its RMS voltage would be Vrms = Vp/1.414 and the power into the load would be Vrms2/RL = Vp2/(2xRL). But that’s for a complete sine wave. For the half cycle that our tube produces, the power would be half of that, which is: P = Vp2/(4xRL). Notice that the peak-to-peak measurement in the test procedure (Vpp) is the same as Vp in this derivation, so P here is the same as Pout in the test procedure. Thus we are calculating the maximum AC power that the one tube can deliver to the load, when it is part of a push-pull pair. By the way, the load that each tube sees is one quarter of the total primary impedance, not half, as some might expect. So for example, our choice of RL=1650ohms for the 7591 tube simulates a transformer with a 6600ohm primary.
Why doesn’t the distorted waveform of the test spoil the power measurement?
For the test, we only need to establish how far down the tube can pull its plate voltage, with the given RL and supply voltage. If it can pull it down for the peak, it can just as well pull it down for any part of a sine wave, given suitable drive. For purposes of power measurement, it doesn’t matter whether we do this with a pulse, sine or continuous DC, so long as we can accurately measure the voltage swing across the load resistor. (This neglects second-order effects of average cathode current.) However, to minimize stress on the tube, it’s best to keep the duty cycle short, so continuous DC would not be acceptable. In the lab test, the waveform becomes asymmetric and the conducting period does indeed have a relatively short duty cycle, so that’s a good thing. It won’t look anything like a half sine though.
How is the ideal output power (Pouti) determined?
This uses the time-honored method of drawing a load line on the plate characteristics curves in the original tube datasheet. (Seen at right.) That was done in PhotoShop at high resolution (much higher than the illustration). The left end is on the vertical axis at Io = Vb/RL. For the 7591, Io = 400/1650 = 242.4mA. The right end of the load line is on the horizontal axis at Vb. All we really need to know is how far down the tube can pull its plate, given the supply voltage and load resistance. Class AB1 operation limits grid voltage to zero and negative values, so maximum current will be the curve labeled Ec1=0 volts. The load resistance is given by the chosen data book example, as discussed in the Load Resistor section above. Then the lowest instantaneous plate voltage (Va) which the tube can achieve is at the intersection of the load line and the Ec1=0 curve. For the 7591, that’s at Va=50V. So Vpp = (400-50) = 350V. Then ideal Pouti = Vpp2/(4xRL) = (350)2/(4x1650) = 18.6W. This is the maximum power that a single, nominal, new, 7591 tube can produce in a class AB1, push-pull pair, driving a 6600ohm plate-to-plate load.
How does the power test procedure differ from an emission tester?
Some might think that an emission tester performs a similar function to the power test procedure (PTP). After all, emission testers purport to test cathode emission and the PTP does, to a large extent, test the condition of the cathode. The difference lies in the following benefits of the PTP:
- It exercises the tube near maximum power, revealing the true condition of the cathode.
- To do that safely, it stimulates the tube with a duty cycle less than or equal to operation in an amplifier.
- Performs the output power test in a manner which is calibrated against original, published tube data. This includes accurate, regulated, DC sources.
- Provides a table of ideal performance values derived from original, published tube data. This results in a percentage score for the tube, versus ideal.
- Accurately measures the effect of plate saturation voltage on output power.
I would like to thank Dave Gillespie for sharing the power tube testing methodology he uses, as embodied in his custom built power tester. The techniques described here are based on his efforts. However, I take responsibility for any deficiencies or errors in this article, as this lab-equipment-based procedure and the accompanying Table of Test Parameters necessarily differ somewhat from his approach.
Many thanks go to Art Grannell for technical proofreading, corrections, useful suggestions and for generously making tubes available for testing.
June 08, 2017 - 03:05 pm|
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