Calculation of Singe Ended Triode Amplifier

One of the easiest way to put up a schematic for simple triode power amplifier is to open a tube manual, or some other source, containing the data of the tube and read what is written in it for the triode you have or intend to use. There might be an example for the chosen tube with all precalculated /suggested by the manufacturer/ optimum elements such as Rp, Rk, Rg, Ck, Cg - given for class A - single tube. All you have to do is to choose a proper transformer with proper tube load and ratio to match impedance of the speakers and the final stage is ready.

Basic schematic

for common cathode SE power amplifier is shown here. First "autobias" schematic requires some more passive elements and Ug is fixed only for standing tube current, for signal conditions Eg (Ug) is "floating" with the input signal. Fixed bias schematic requires separate voltage for feeding the -Eg on the grid of the tube and with the increase of the signal tube may go to A2 class of operation. Some authors (including me) say that autobias is better /musically/ others like fixed bias better. No matter what you choose you have to stay within tube limiting value parameter for Eg - optimum for this class.

If you are not happy with what the books say, or perhaps you may like to achieve higher power, or reduced distortions you may try to calculate your SE triode amp from the scratch. You need a family of curves related to the tube, or if you don't have it - try to find it. Otherwise you need a testing equipment to measure and draw this curves yourself - it is a time consuming job and it is not for beginner to start with. But if impossible to find diagram with family of plate curves all you have to do is to measure plate current for any given voltage on the plate and the grid and make notes. With all this data you plot the curves on paper and make diagram (if you have the nerves to do that). See below how it looks the diagram I made for my "made up tube". My goal is to make it as simple as possible.

Determining slope of the load line

Suppose we have already power tube chosen and we have also family of curves for this particular tube, taken from the manual or the dealer we obtain this particular tube. Next step is to determine Rp for this tube and my goal here is to simplify calculations. The following diagram - family of curves is made up from me and it is not for specific type power triode - it's an example for tubes from the type 2A3, 6B4G, 300B, AD1 and others used in DHT SE amps.

For example my "made up tube" have plate voltage of Up 0 = 250 V, amplification factor u=4, maximum plate dissipation Pp=15 Watt. Rg1 max= 750 K, Rp=800 Ohm

To read the diagram: on vertical axis we put plate current - Ip, horizontal axis - plate voltage Up, black curves shows Ip and Up for some values of grid voltage - Ug. Blue curve shows plate dissipation of the tube - above this curve tube may become overloaded and burned. Yellow line/on some monitors orange/ between points M-N is for helping us to determine the slope of the load line. The red line between points Q-S paralleled /with same slope/ to M-N help line is the needed loadline. But any line paralleled with M-N help line may be chosen for loadline for this chosen load. How to draw this?

All colors here are just to make things easy to spot and understand. Usually diagrams in the books are given in black and white.

 

Let's say the tube need reflected impedance Rl=2.4 K plate load as stated. To make it more clear: for example we decide to run the tube with load of Rl=2.3 K plate impedance to achieve a bit more power /few percentage points/ we have to see how this will be reflected on the family of curves.

Fastest way to determine the slope of the load line is to draw M-N help line. I choose plate current of 100 mA (0.1A) and look for corresponding Up to calculate Rl - which here is the impedance of the transformer primary. More for the output transformers and reflected load - click here.

Rl = Ua/Ia

Rl must be equal to our chosen load of 2300 Ohm (2.3K)

that is true and only one line(or line paralleled to it) you may draw for

Ua=230 V at Ia=0.1 A (100 mA)

/Reason I choose Ia=0.1 A is for faster calculation - otherwise you may choose different current, slope or load/

Rl = 230 V / 0.1 A = 2.3K /Kilo Ohm/

Next is to see for Ua 0=250 V we make a vertical line (Rl = 0) to the crosspoint with the plate dissipation limit (the blue curve) and determine the necessary Ug /Eg/. After that we draw line Q-S (in red) paralleled with M-N help line.

For maximum power Q-S must have same crosspoint at blue curve with Ua=250V. Keep Q-S line slightly in touch with the blue curve to achieve maximum power from the given tube for this particular load - and here we go - the line Q-S - is the loadline for our triode at Rl=2.3K.

If Q-S goes over the blue curve - that means we have exceeded power dissipation of the tube and have to choose other parameters - Rl, Up, Ug. If you keep this line below the blue curve you are safe to continue with your design and the calculations. If too far below you will get lower power. So keep it in touch without crossing it - that's the optimum without blowing the tube.

You may use this fast method to calculate loadlines for preamp tubes as well!

Above crosspoint is very important for the regime we put tube in. For class A the standing plate current Ip0 should be around half the maximum plate current for the tube. On the vertical axis from the crosspoint draw horizontal line (Rl = ~) we note that the standing plate current is Ip0=60 mA (or 0.06 A). The maximum plate current reading is Ip max=120 mA (0.12 A) taken from Q point connected with right angle horizontal line to the vertical axis of the diagram. From same reading for Q make right angle vertical line with the horizontal axis and for Ip max (maximum signal plate current) we measure on the diagram that Up min = 110 V.

The achieved output power will be:

Pout max=(Ip max - Ip0). (Ua0 - Up min)/2

And for the data taken from this specific diagram (for my made up tube) that makes for Pout max:

Pout max = (0.12-0.6)0.06.140/2 = 4.2 Watt

 

Ug from the crosspoint at the diagram at Uao= 250 V we have reading for Ug= - 45 V. That means that we must have grid - cathode potential of - 45 V. For fixed bias - make additional power supply to the grid of - 45 V and you are ready.

For selfbias we cannot make the grid negative due to lack of additional power supply, we have to make the cathode more positive. So we have to make voltage drop on Rk exactly 45 Volts.

Next is to calculate Rk for selfbias schematic. There is no current drawn from the grid so cathode current Ik0=Ip0=0.06 A.

Then:

Rk=Uk/Ip0

Rk = 45V /0.06 A = 750 Ohm

(Or choose closest standard value if you don't have standard result. You can combine several resistors to make the exact value as well. Another way is to wound the resistor yourself (bifilar or non inductive) from resistive wire if you know how to do it.

Dissipation for this resistor is P= Ip0 x Ip0 x Rk= 2,7 Watt /choose 5-10 W resistor minimum/. If not - excessive heat may be build up around this resistor and that's really bad for the construction and neighboring elements.

Rg1 is given as minimum and maximum value by the tube manufacturer. Rg1 is usually between 100 K to 1 M and it is also important for forming the load for the preamp-driver tube. Rg1 should be made as higher as permissible by the grid requirement of the power tube without introducing extra load on the driver tube.

Rg2 - that's resistor you can omit in some cases. Rg2 is suppose to protect the grid of the power tube from high frequency parasitic and oscillations. You may choose it anywhere between 1K to 10 K max.

I've seen some amp designers choosing very big capacitors for Ck and Cg. They probably have designed transistor equipment before getting into tubes, where big capacitors are heavily used. Transistors handles large currents so the capacitors have to be large. For tubes this is wrong - the real output transformer cannot pass very low frequencies without some distortion. Tubes and large capacitors can. That easily adds up extra distortions at all frequencies once you overload the transformer. So be careful choosing capacitors - not too big, but sufficient. Choose the middle way. Click here to read more on tubes/transistors.

Ck - choose it between 20 uF and 200 uf the higher Rk the higher Ck you choose. Keep in mind that if you choose very high value, that may overload the output transformer. For our design Ck=50 - 100 uF / 100V should suffice.

Cg choose between 0.02 uF to 0.33 uF keeping in mind same thing we said about Ck - do not overload the transformer with too much lower frequency. That may give additional harmonic distortions if Cg is too big. If Cg value is too small the lower frequency of interest may suffer from that.

For our design Cg= 0.05 uF/ 400 V should be enough. But if you can find not very expensive (let say Japanese made) large output transformers (most probably you won't be reading this if you have it) with lower frequency starting at 10 Hz /-0.5 dB/ you may at least triple chosen capacitors value and that should be O.K.

Performance and distortion

After all elements are calculated or chosen we have to see how this amp will perform on paper and what kind of distortion we should expect from it. As you most probably noticed we have two blue triangles (with Pout in the middle) build on the red loadline of diagram. They represent the output power for each half period of the input signal. If they have visually equal surface - zero distortion is introduced. But that's not happening in real life. So he have to calculate what distortion we should have at maximum rated power. Let's compare each triangle by plate currents. To have D tot = 0 %, we must have Ip max - Ip0 = Ip0 - Ip min. In our case:

Ip max - Ip0 = 120 mA - 60 mA = 60 mA

that is the current for positive half period of the input and

Ip0 - Ip min = 60 mA - 5 mA = 55 mA

that the current for negative half period of the input. Here we spot that those currents are not equal. So some distortion might be present.

To see the % distortion use the following basic formula:

After calculation for our example is performed /do it for exercise/ we find that D tot = 4.54 % - that all distortions at maximum chosen and rated power for our readings from the diagram.

How will perform? Build it and listen to it. That's best way to see that. Too many factors and parts will be involved in this process and each is equally important. Evaluate it with your hearing playing the music you know and love.

 

 

Regards

P.G.

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