Sometimes I get excited about my latest circuit and after looking for just the right tubes, output transformers, coupling caps, and low noise resistors, the power supply design becomes almost an A question about ripple factor in a bridge rectifier thought. Sometimes things turn out ok and there are no problems.

The truth is, there is no reason to suffer power supply set backs like this. The design of basic tube power supplies is actually very straight forward. What I’d like to do here is to walk through a design process first recommended by O. Schade1 in 1943 and presented in a slightly modified version by Herbert Reich2 in 1944 in his book “Theory and Applications of Electron Tubes”. Now, the state of electronics has changed somewhat in the 67 years since Schade first published his landmark work.

As such, I will present again, a somewhat modified version of this process more suited to today’s amp builders and hobbyists. I am going to work through the procedure by designing a supply for a mythical low power stereo amplifier. It’s a stereo single ended class A design I’ll refer to as the Ghost amp. I will first present the process for each step and then fill in some numbers to show how the process works. I urge readers to read the entire design process first and then go back and look closely at the math. It’s not advanced, but it can be a little intimidating the first time through. 2 Reich, Herbert J, “Theory and Applications of Electron Tubes”, 2nd Ed.

If you are building for a stereo amplifier, be sure to include both channels. 250v at 59mA at idle and 65mA at peak output. These will be our starting points. Now we need to choose a topology. There are other topologies of course, but these three are the most suitable for high fidelity amplifiers. For the Ghost amp I will choose a full wave dual diode rectifier.

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I am also going to choose to use a capacitor input filter. I could also use a choke input filter, however the choke would require a much higher transformer secondary voltage and a larger choke than the capacitor input filter. This would drive up the size and cost of the power supply and small and cheaper is better than large and more expensive. The first of these are the diode plate currents.

Once calculated these will allow us to choose a suitable rectifier tube for our amplifier. The peak plate current is more complex. Because the input capacitor only partially discharges on each half cycle, the diode can only conduct on that portion of the following cycle when the plate voltage exceeds the voltage to which the capacitor discharged between cycles. This means that the capacitor charging current must be significantly greater than the average plate current.

Here we are going to make an assumption. Now that we know the peak plate current it is time to choose a rectifier tube. Making this decision requires checking the tube data sheets, looking at max ratings and voltage drops, and applying some judgement and experience. There are some basic rules to apply which make this task easier. First, and most importantly, the maximum tube ratings must not be violated. We already have an estimate of the maximum plate current. But what about the diode peak inverse voltage.

There is one other important parameter which is the diode voltage drop. While the amp is at idle, the voltage drop across each diode will be one voltage, and when the amp is approaching full power the current draw will be greater and the diode voltage drop will be greater. This will result in what is known as power supply sag. It should be noted that minimizing sag is not always the best design choice. Particularly in guitar amplifiers, power supply sag can create unique distortion and tonal qualities which are actually quite desirable. But for musical reproduction this is not generally the case so, for the Ghost amp, we want to limit sag as much as possible.

This is a collection of vacuum tube rectifier voltage drops as a function of plate current. It shows just how much drop there is for each tube and allows us to get a feel for how much sag to expect for a given tube. The lines in figure 1 only extend to the maximum allowed tube current. We know the peak current for the Ghost amp is 260mA, so this eliminates the 6X4 and the 25Z5.

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Since we already decided we would like to minimize sag we could go with the 5AR4. However this is a large and rather expensive tube. We’ll keep this in mind but something cheaper and smaller would be nice. We have chosen a capacitor input filter so we need to check the allowed operational ratings. The chart below is directly out of the GE data sheet. It shows under what conditions it is permissible to use a capacitor or choke input filter with this tube.

This is just an educated guess at this point. We’ll check back later to make sure we’re still ok. Also, according to the data sheet, the typical conditions all assume a 50µf capacitor. There is a temptation at this point to attempt to maximize this first capacitor, however this can be an unwise decision. The larger the first capacitor, the less it will discharge between cycles. However, as designers become more experienced and familiar with the various rectifier tubes, most tend to settle on one or two favorites, and only stray when they really need something different.

Now that we have chosen an appropriate tube and input capacitor value, it’s time to begin estimating the performance of our design. Since we already have the peak current estimate from equation 2 we can use this along with the voltage drop from figure 1 to calculate the peak diode resistance. This information is also available on the data sheet. Up to this point we have been making simple calculations based on simple circuit theory and data sheets. Now we need to fall back on some of the work already accomplished by Schade back in 1943.

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So by following Schade’s solution we get the following. Now we need to find the total peak and average resistances which include the effects of the transformer winding resistances. Obviously this presents a slight problem as we haven’t yet chosen a power supply transformer. So again we will have to make an estimate. The effective transformer secondary winding resistance is given by the following equation.

N is the voltage step up ratio per section. For a modern vacuum tube power transformer, the resistance of a typical secondary winding of 250v to 300v, rated at 150mA, is on the order of 50Ω and the primary may be on the order of 10Ω. At this point it’s a good idea to make a check on our rectifier data sheet for the minimum allowed source resistance. If the source resistance is too low then the initial inrush current to charge the capacitor will exceed the limits of the rectifier. So lets return to the data sheet. However, if you look at the note on the graph you’ll find that this minimum resistance is only required if “hot switching” is involved. Hot switching is where the high voltage supply is activated after the tube filaments have come to temperature.

In reality, there is almost never a reason to include a standby switch in a piece of audio equipment with a vacuum tube rectifier. As such, we’ll make the decision to leave out the additional resistance and forgo using a standby switch in our design. This is another design constraint of which we’ll need to keep track. Getting back to our 93Ω estimated transformer resistance, we’ll now use this estimate to calculate the total effective values for the peak and average supply resistances. Here we simply add the transformer resistance estimate to those values calculated in equations 4 and 5. There are three more pieces if information we need before we go back to Schade’s work.

The first is equivalent load resistance presented by the amplifier RL. This is found by simply dividing the load voltage by the load current. The second piece of information we need is a “frequency time constant” for the rectifier load. This is a measure of the effective discharge time in radians for each half cycle and is shown in the graphs as ωRC.

And the final piece of information is the source resistance to load resistance ratios. These will allow us to check some of our previous assumptions, determine the efficiency of the rectifier, and to chose an appropriate transformer. All the way back in equation 2 we made the assumption that the peak diode current was four times the average plate current. I also said that we would refine this estimate base upon our initial results. 2, nωRC is 64, and the peak resistance ratio from equation 11a, in conjunction with the lower graph we find that the peak current is better estimated as 5 times the average plate current. In light of this, we now need to recalculate equations 2, 4, 5, 8, and 11 to arrive at better estimates of these parameters.

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The values found in equations 11a’ and 11b’ do not differ significantly from those found initially with 11a and 11b, so a second iteration is not required. In some instances where the load currents are high or VL is relatively low, another iteration may be required, but in general, one iteration of these equations is sufficient. If we check the 325mA peak current against the 500mA tube limit we find that there is still significant margin. These graphs allow us to determine the efficiency of the rectifier and hence the required output voltage of the transformer. Here is Reich’s reformatted graph for a dual diode, full wave center tapped voltage rectifier circuit.

In equation 11b’ we found the refined resistance ratio to be 0. This is shown mathematically as follows. Reformatting equation 12 allows us to determine the required transformer peak secondary voltage per section. To find the required RMS secondary voltage per section we simply divide the peak voltage by the square root of 2.

Before we go forward, it’s time to make one more check on our rectifier data sheet. From our calculated data and figure 5 we determined that the rectifier efficiency was 0. We need to check if this is acceptable. The following is the efficiency rating chart from the 6CA4 data sheet. This means that the 6CA4 has passed the last technical hurdle and it totally suitable for our power supply.

With the one design caveat of no hot switching allowed. It was simply assumed that whatever transformer was required would be wound for the voltage needed. Today we need to approach this aspect of power supply design a little differently. It is virtually impossible that we’ll be able to find a transformer which exactly matches our 239v requirement from equation 14. But in reality that’s not a problem because there are some more things to consider.

First, tubes are not like integrated circuits with respect to power supply requirements. Now you may be thinking that we set VL as 250v right at the start. How can it be ok to simply increase the voltage 20v to 30v without rerunning all the equations? If you have any questions, then you should rerun the numbers to convince yourself. With practice and experience you’ll begin to get a feel for how different parameters affect the design point and rerunning the equations usually won’t be required. So now we want a transformer with a secondary voltage of somewhere between 259v and 269v RMS per section or between 518v and 538v RMS for the entire secondary. A quick check of a suppliers website quickly shows that this is not a very common voltage range.

However what is a common voltage is 550v or 275v RMS per section. So, we settle on this voltage and will if we need to slightly reduce the voltage, we’ll add a dropping resistor in the filter. Returning to equation 12, we can rearrange the equation to get the dc output voltage as a function of the peak secondary voltage per section. Also, by rearranging equation 14 we can express the peak voltage as a function of the RMS voltage. This voltage is 38v higher than our target voltage of 250v.

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However, this is all right as it gives us some voltage drop to apply to our smoothing filter. The other question is now how much ripple are we going to have to filter. This is again a place to fall back on our friend Schade. As part of his work, he published a very nice plot showing ripple factor as a function of ωRC and the peak resistance ratio as calculated in equation 11a. This value will allow us to determine that total ripple output by our power supply. This level of ripple may look high, however it is actually very typical of this type of power supply design. I consider this an excellent result for input to our smoothing filter.

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Load regulation is good and we have ample rectifier margin to ensure good operation and long tube life. Supply hot switching will not be utilized to minimize inrush currents at startup. Our output voltage is 288v going into the smoothing filter leaving 38v for filter voltage losses therein. In the next section we’ll be discussing the smoothing filter. This will take the output of our basic power supply and make it suitable for feeding our high fidelity Ghost amp.

The most obvious question at this point is “How much filter do I really need? Unfortunately there is no easy answer to this question. In chapter 14 of his book “Theory and Applications of Electron Tubes”, Herbert Reich gives some good rules of thumb for just how much ripple may be acceptable for various types of circuits. So this is where we’ll start with our filter design. A smoothing factor is simply a numeric factor for how much the ripple is reduced. For example, a smoothing factor of 100 will reduce 5 volts of ripple to 0.

As an example, the output of out ghost amp power supply was 288v with 5. They may be passive or active, single or multi-stage, and closed or open loop. RC filter stage and the LC filter stage. The only catch is that equation 18 only holds if in each stage the reactance of the series impedance element at the primary ripple frequency is about 20 times the reactance of the shunt element at the same frequency.

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Normally this isn’t a problem but it is one thing that we’ll have to check later. With these thoughts in mind, lets look at the two types of filter stages stages mentioned above. This stage simply acts as a frequency selective voltage divider where the shunt impedance is frequency dependent. 8 is going to act as an open circuit and the DC output voltage will be the equal to the input voltage minus the voltage drop in R1. At AC the capacitor has a finite reactance given by the following relation.