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Wednesday, April 24, 2002 10:12:48 AM Loadlines Made Simple vol. 2 by Matthias M. and Carl B. This text is a try to give a rough idea of how a standard preamp stage works, especially how tube curves and the plate resistor influence voltage gain. I've tried to keep it simple and step-by-step, but still providing some insight, rather than giving a guide to plugging in numbers. Some understanding of tube basics, especially of a standard preamp stage (what does the cathode resistor do, what is biasing etc.) should be enough to follow. Dave Sorlien's great P1 theory paper explains these things very well, so these things are not repeated here. The circuit in this example looks like this: Of course you know that in reality we would have to add a cathode resistor for biasing, a coupling cap etc., but these things do not matter in this text. Now lets have a look at some curves from a ECC83 (=12AX7) datasheet: 1 of 7 Wednesday, April 24, 2002 10:12:48 AM For the first calculation, we choose a plate resistor Rp=100k, the supply voltage is 300v. Imagine what happens if we put a very negative voltage on the grid: The current Ia through the tube is completely cut off, so there is no voltage drop across Rp and the plate voltage Va goes up to the supply voltage (300v in this example). So (Va=300V, Ia=0mA) is the first important point in the loadline graph. Now, what would happen if the tube would present a short circuit? The plate voltage would go down to Va=0V and the plate current would only be limited by the plate resistor Rp=100k, so it is Ia = Vr / Rp = 300V/100k = 3mA. (Va=0v, Ia=3mA) is the second important point in the loadline graph. With this information, we can already draw the loadline for V=300V and Rp=100k, connecting the two points: 2 of 7 Wednesday, April 24, 2002 10:12:48 AM Ignore the black Vg=... lines for a moment. All this graph shows now is the voltage drop across both tube (Va) and resistor (Vr) for a given current, simply following Ohm's law. A given plate current (for example 0.5mA) causes a voltage drop Vr across the plate resistor Rp: Vr = Rp * Ia = 100k * 0.5mA = 50v Subtract this value from the supply voltage and you get the plate voltage that results from this particular plate current (300v - 50v = 250v). Do this calculation for all current values from 0 to 3mA and you will always find that the (Va,Ia) point is lying on the loadline. Got this? OK, let's go on then and forget the loadline for the next three sentences. Remember what these black Vg=... lines mean? They show all possible (Va, Ia) combinations for a given grid voltage Vg. For example, with Vg=-1v there will be Ia=0.5mA when Ua=100v, Ia=1ma when Ua=140v and so on. But wait, in this example circuit the (Va, Ia) combinations are always lying on the loadline, as we found out above. So, with a given grid voltage Vg=X, the (Va, Ia) point must be a) on the loadline, b) on the Vg=X line, so it must obviously be at the point where these two lines meet. For example Vg=-1v results in (Va=160v, Ia=1.3mA), Vg=-2v results in (Va=223v, Ia=0.75mA) and so on. As we want to amplify voltages, we don't care about the Ia values for our further calculations. For now we choose a operation point somewhere in the middle of the useful area of the loadline (the part close to and left of the Vg=0v line causes grid current which we want to avoid in this example), so we use Vg=-1.5v, Va=190v, Ia=1.1mA. 3 of 7 Wednesday, April 24, 2002 10:12:48 AM For a grid voltage swing of -1v..-2v (1v peak to peak), the resulting plate voltage swing is 160v..223v (63v peak to peak): 4 of 7 Wednesday, April 24, 2002 10:12:48 AM So the voltage gain of this stage is 63v/1v = 63. In reality, the output voltage gets loaded down a bit by the grid resistor of the next stage or by whatever else comes after that stage, so the gain is a bit smaller. Now, what happens if we use a 200k plate resistor with a 300v supply? 5 of 7 Wednesday, April 24, 2002 10:12:48 AM Basically the same thing: Finding the two points for drawing the loadline: (300V, 0mA) and (0V, 1.5mA) (because 300V / 200k = 1.5mA), choosing a operation point (Vg=-1.5v for simplicity), and so we get a (unloaded) voltage gain of (200v-128v) / 1v = 72. How do we calculate gain if the datasheet for the tube does not contain any plate curves? We use this formula: Voltage Gain = (gm / 1000000) * Ra * Rp / (Ra + Rp) Where gm is the transconductance figure in micro-mhos (for example 1250 as given for the 12AX7 at Va =100v). In european datasheets you might find something like S=1.25 mA/V instead, we can just multiply this figure by 1000 to convert it to micro-mhos. Ra is the plate resistance (Ri in european datasheets) which for the 12ax7 varies from 60k to 80k and Rp is that plate resistor. gm and Ra vary with plate voltage, so we must be careful to pick the right figure from the datasheet. Let's calculate the voltage gain for the first example with Rp=100k this way: We know the operation point for the stage (Va=190V, Ia=1.1mA) and these figures from the datasheet: 6 of 7 Wednesday, April 24, 2002 10:12:48 AM Transconductance and plate resistance are only given for 100v and 250v plate voltage, so we have to estimate a bit for Va=190v: gm=1.5mA/V=1500 micro-mhos and Ra=70k. (1500/1000000) * 70000 * 100000 / (70000 + 100000) = 61.8 This result is pretty similar to the result we got via the loadline method (63), so our estimates were probably not that bad. The effective output impedance from the stage is simply the plate resistor in parallel with the plate resistance. In the first example this would be 100k || 70k = ca. 40k. I hope this text gives some ideas, the loadline method should work with all preamp tubes (even with pentodes, if you keep the plate voltage higher than the g2 voltage). Excuse any errors, let me know and i'll correct them, corrections of my mediocre english are welcome, too ;). If you have things to add on how to choose operation points and Rp values for things like huge headroom, warm distortion, buzzy distortion, symmetrical and asymmetrical clipping, what happens at the ends of the loadline when the tube is overdriven and how and why all this colors the tone, it would make this text much more useful. If you want to use parts of the drawings and the text for a better explanation (or a better looking .pdf file), go ahead and publish it. Many thanks go to Carl B. for writing the first loadline tutorial, checking this text for errors and contributing important parts. -Matthias Please post questions, suggestions, corrections etc. on the AX84 Discussion Forum. 7 of 7