Reprint from Sound practices Design 
and Easy Calculation of a Regulated Power Supply 
for a 300B Stereo Amplifier

The job of the power supply in a push-pull amplifier is facilitated by a symmetrical arrangement of the output devices that eliminates residual hum by cancellation. Moreover, if the push-pull amplifier operates in true Class A the current flow is constant and the power supply voltage does not fluctuate with the rhythms of audio modulation. Class A push-pull amplifiers present a relatively simple challenge from the standpoint of power supply requirements,

It is a completely different story with the single-ended triode amplifiers currently in fashion. In these amps, the residual hum is applied to the output transformer through the internal resistance of the tube (see Fig. 1). If we presume that the residual hum is 100 mV, a hum voltage on the order of 4.5 mV will appear at the output of the amplifier. Of course, this is not tolerable, especially when high efficiency loudspeakers are employed.

The classical cure is to add more filter capacitance, which can work well, or to add more inductance in the choke, which could be bad if it raises the internal impedance of the power supply. The consideration of internal supply impedance leads us to the second point of the argument, by far the most important. Let us look at the operation of a single ended output stage when driven almost to the point of clipping by a large sinusoidal signal. The positive half of the signal applied to the grid increases the current in the tube and creates a condition of maximum power dissipation in the supply. Then the negative signal causes the current to decrease until it is nearly cut off and at this moment the power dissipation is again at maximum. Where does the power come from ? Obviously, the power is taken from the energy stored in the primary inductance of the output transformer. Q=l/2 LI² There is an analogy in the work of the coil ignition of an automobile. In order to do a good job, the output transformer must have a substantial inductance, a well-known fact, but the power supply must also be able to reload the primary inductance effectively. In your car, no problem. The battery is husky and the charging current gushes forth. In an amplifier, it is easy to be far from the mark. For instance, a reservoir capacitor of 100 microfarads will have an impedance of 80 ohms at 20 Hz. That is not negligible, (see Fig. 2, the measured impedance of a 100 microfarad capacitor).

Principle of the Regulator Supply
In our regulated supply, the rectification and filtering are achieved in the classical way but the high voltage current is supplied to the amplifier through the regulator tubes, called the ballast, (see Fig 3).
The ballast in the regulator under discussion consists of several low plate resistance triodes (6080, 6AS7, 6336) connected in parallel. The triode grids are controlled by a differential amplifier which compares the instantaneous value of the high voltage to a reference voltage provided by zener diodes or a voltage regulator tube (85A2 OA2, OB2, etc..
From moment to moment, the variations of the high voltage supply are corrected through the action of the regulator. Since nothing is perfect, there is always a residual hum and some voltage variation dependent on the load.

Let us examine how we might keep these inevitable failings as negligible as possible.
Design of a 400V—250mA Supply
Figure 4 gives the diagram of a simplified regulator minus the various artful devices which we will add to achieve the desired quality. The parts shown in the block diagram of Figure 3 are replaced by electronic components:
The ballast tube T1 is the 6080 dual triode, a tube type specifically designed for this application.
The T2 tube, EF 184, is a sharp cut-off pentode normally used as an RF amplifier, but also perfect for the job here.
Resistors R2 and R3 feed the screen grid of the pentode and provide the current for .the zener diode voltage reference source. R1 is the load for the pentode.
Potentiometer PI acts as a voltage divider to supply a part of the high voltage to be compared with the reference voltage.
Let us move on to a practical exercise, a regulated power supply to power a stereo 300B amplifier with a required voltage of 400V and a current delivery of 250 mA.

Calculation of the Ballast (T1) Figure 5 shows the basic system with one triode section of the dual triode 6080 illustrated. Since these tubes have a control grid with a tight pitch placed very near the cathode, the consequent grid current can be very inconvenient. This is especially a problem with old tubes.

Therefore, it is prudent to maintain the instantaneous voltage of the grid very negative and that prescribes the first limitation: The bias will have to be above -4OV. That means that the voltage between plate and cathode will have to be no less than 100V in the worst case, that is to say when the line voltage is decreased by 10%.

The necessary voltage after rectification and filtering, therefore, must be:
U_F=U_R+U_T1= 400V + 100V = 500V

Now we find that, when the line voltage is increased by 10%, the rectified voltage will reach :
(500V x 120%) /100% = 600V

In this case, the voltage between plate and cathode of the ballast tube will be:
600V - 400V = 200V

That gives a dissipated power of
200V x 250 mA = 50W
for a current consumption of 250 mA.

We can see immediately that this power requirement, although a rarely seen extreme case, forces us to use several 6080 tubes in parallel because each half of a 6080 can only dissipate 1 W. Two 6080s (4 triode sections) are good for 52W. This is just enough, without any margin of safety.

An improvement to the basic scheme of Fig. 4 consists of fitting a resistor in parallel with the 6080 to handle part of the current (see Fig 5).

The choice of value of this resistor must be made with the 250V plate-cathode voltage limit of the 6080 in mind. A higher voltage can arc over, causing damage to the cathode coating.

Note also that the 300B tubes the supply is powering are quick-heating filamentary types and the 6080 are slow-heating cathode types. Without the parallel resistor R4, the voltage after turn-on across the ballast tube would reach the totally unacceptable value of 600V in the worst case situation of 10% high line voltage.

Since the normal current flow of a pair of 300B tubes is about 150 mA, R4 will have to drop 250V maximum at this rate of current flow.
R4=250V/150mA=1kW 7

The current through the 6080 will be notably decreased and will go down as the line voltage goes up, thereby helping to regulate the power dissipated by the ballast tubes.

Fig. 6  Reprint from SOVTEK 6 A8 7 DATA SHEET 

 

Resistance in cathode circuit of each triode separate, in ohms	Number of triodes at parallel operation
	1		2		4		6		10		over 10
	IA	PA	IA	PA	IA	PA	IA	PA	IA	PA	IA	PA
	mA	W	mA	W	mA	W	mA	W	mA	W	mA	W
0	130	13.0	93	9.3	74	7.4	68	6.8	64	6.4	56	5.6
50	130	13.0	101	10.1	87	8.7	82	8.2	78	7.8	72	7.2
100	130	13.0	106	10.6	95	9.5	90	9.0.	87	8.7	82	8.2
150	130	13.0	109	10.9	100	10.0	96	9.6	.94-	9.4	89	8.9
200	130	13.0	112	11.2	104	10.4	101	10.1	98	9.8	94	9.4
250	130	13.0	114	11.4	107	10.7	104	10.4	101	10.1	99	9.9

I_A - anode curent of one triode
P_A - anode dissipation of one triode

So for a minimum line voltage (-10%)
IR4 = YT1 / R4 = 100V / 1k W 7 = 59 mA

and the dissipated power in the tubes will be
P_T1 =100V x 191 mA = 19W1
for a nominal line voltage
U_F= 550V, U_T1 + U_F - U_R = 600V- 400V = 150V
I_R4-U_T1/R4 =150V / lk W 7 = 88mA
and the dissipation of the 6080 will be
P_T1 = 150V x 162 mA = 21W3

For the case of maximum line voltage (+10%):

U_F= 600V, U_T1 = U_T - U_R
= 600V-100V = 200V
I_R4=U_T1/R4
= 200V / 1k W 7 = 117mA

The current in the 6080 will be
P_T1 = 200V x 133mA = 26W6

It appears that on the basis of power dissipation alone, one could be content with only one 6080. However, prudence commands to use two tubes for security's sake and for two other reasons.

1—Two tubes improves the regulation from the standpoint of residual hum and lower internal impedance of the power supply.

2—The variation between 6080 tubes is astonishing. It is an idle dream to find two tubes that are matched between sides and between tubes. Therefore, we must also wire in balance resistors in series with the cathode of each tube. The chart in Fig.6, published by SOVTEK gives useful information. For four triode sections in parallel in the absence of precautionary measures, the maximum power handling of each tube is not more than 7W4 and 29W6 for the quartet, thereby erasing our security limit entirely. Installing 100 ohm series resistors in series with each plate improves the situation such that the power handling becomes 9W5 x 4 = 38W

This arrangement is the second improvement we propose (see Fig.9).

Design of the Differential Amplifier
The transconductance of the EF184 is 15000 micro-mhos in normal use. Here, the transconductance is highly reduced owing to the very low current flowing through the pentode in our special connection. Since the differential amplifier escapes calculation, it will have to be estimated empirically, but we are mainly interested in the gain (and regulation factor) we obtain. Maximum gain will be determined empirically by varying R1 using a 1M W potentiometer. On the other hand. The values of R2, R3, and the Zener reference voltage can be determined by calculation, considering the current through the pentode as negligible and without any practical effect.

Zener Reference Voltage Choice It is a compromise. The higher the voltage, the more of the potentiometer will be in circuit. Since U_R = U_R1 + U_T2 + U₂ the Zener voltage is a compromise between the voltage drop across R1, the voltage between the plate and cathode of T2, and the voltage on the Zener diode. Given the realities of the components that we are using, we have to provide a reasonable operating voltage for the pentode. In view of this requirement, the error voltage applied to the grid of T2 through P1 must be limited to a certain value, even if the efficiency of the regulation is thereby decreased

What is the voltage on the plate when the power supply is at nominal line voltage? We previously determined that in this condition, the current through the 4 x 6080 would be 162 mA (40.5 mA in each triode) and the voltage between cathode and plate would be 150V.

Referring to the curves of Figure 8, it is easy to determine that the grid voltage should be -70V. therefore, the plate voltage of the EF 184 is

U_R - 70V = 400V - 70V = 330V

If we consider that it is wise to keep about 200V between plate and cathode on the EF184, we have

330V-200V = 130V

for the reference Zener voltage. Since 62V Zener diodes are readily available we can use two in series to obtain

Uz = 62V x2= 124V

The values of R2 and R3 will determine the current through the zeners. It is necessary to be very prudent in this, because zeners definitely don't like heat. The wattage rating given by the manufacturer must be severely derated.

We use B2X 85C diodes and they are specified for 1W3 or about 2.5 W for the two diodes. Taking a factor of 5 as a security coefficient, we would only dissipate 0.5W for the two diodes and 0.25W per diode. For a half-watt dissipation, the current should be

P/U₂ = 0.5W/126V = 3mA9 ---- R₂+R₃ = U_R-U_Z/3mA9 = 70KW

We can choose to use 2 resistors of 39KW each, yielding a power dissipation of 0.22W per diode at 3.5 mA.

I believe that modern Zener diodes are better than glow discharge VR tubes in terms of noise and internal impedance. However, it is absolutely necessary to bypass them with a high value capacitor. We use a 100 microfarad electrolytic capacitor (C4) paralleled with a 0.47 polystyrene. This is the third important improvement on the basic circuit of Fig. 4, as pictured in Figure 10. The bypass capacitors eliminate residual hum on the regulated output and should be considered absolutely essential. It is not possible to adequately bypass a glow discharge VR tube because a VR tube and capacitor in parallel has a tendency to provide an excellent generator of sawtooth waves!

Choke Filter Elimination
The above discussion does not address rectifier and filter problems coming upstream of the regulator circuit we describe. A good regulator allows us to forego a choke and use only one capacitor after the rectifier.

The rectified and filtered voltage Up is wired to Grid #2 of the EF184 pentode through C5 and adjustable resistor R5. This capacitor should present a negligible resistance at 100 Hz (120 Hz for the USA) We use a 0.47 for C5. There is no sensible progress in increasing the value. In any case, R5 allows a fine adjustment of the compensation. This is the fourth improvement toward making the best of electronic regulation, shown in Fig 11.

Controversy on Rl Dteposition

Some writers have indicated other solutions to power the load resistor of the pentode tube, Rl.
1st Variant: The resistor is connected before the regulation on U_F In this case, the voltage U_F is always more than U_R, Therefore the current is increased through the pentode and, consequently, its transconductance is increased. And, the efficiency factor is increased also.
This is an advantage, but unfortunately, there is the problem of hum which does not exist in the U_R Connection.
2nd Variant: The resistor is connected to an increased external voltage source.
If the auxiliary voltage is very well-filtered, the result is indeed improved. However we have to pay for this improvement with a substantial increase in complexity if the auxiliary supply is not already in place. Therefore, we chose not to adopt that strategy.

Lastly, there is another simple way to improve the regulation. We install a capacitor C6 between the upper terminal of potentiometer P1 and its moveable contact so that any variations in U_R Are conducted directly to the grid of T₂. This represents the fifth and last improvement we propose (Fig. 12).

Results
All those efforts to put this regulated supply into place truly pay rewards. The benefits are clearly evident in the measurements, taken in actual conditions of use powering a stereo 300B single-ended amplifier and 2x 6BQ5 in the input stage. The actual current flow is 250mA with the output voltage at 400 V. The curve shown in Figure 13a indicates the regulation area. The voltage on the primary of the power transformer is varied with an auto-transformer. The voltage is stable at 400V under line voltage variations between 180V and 250V (and probably more, but 250V is the limit of our auto-transformer). This 20% security zone is very comfortable, even though we may never use it in practice.

Experience shows that when the hum stays lower than 10mV, the signal to noise ratio of the amplifier is not seriously affected. This hum level becomes evident with a line voltage decrease of about 12%. Taking this as the limit of the regulator, the security zone of the regulator is still quite considerable.

The practical results demonstrate the reliability of the calculations and suggest that the method outlined above would be useful for the design of other power supplies.

Finally, the measurement of internal impedance as a function of frequency given in Fig. 13b is most pertinent. The amplifier is driven to maximum power by a sine wave generator between 20 Hz and 10 kHz. The millivoltmeter is connected to U_R through a 100 microfarad capacitor, large enough not to affect the frequency response of the test setup.

At 20 Hz, the internal impedance of the power supply is less than 0.35 ohms. To obtain the same result with a capacitor would require 22,000 microfarads. This demonstrates that a regulated power supply offers a large improvement and at a cost that is practically the same (considering that the filter choke is removed). The result is near perfection.