Learning to interpret the performance curves for KT88

[pingfloid]

Hi,

I am trying to walk my first steps into the performance curves. I will show what I have done so far, so if someone can correct my mistakes I hope that at the end it will also be useful to other people.

I am trying to figure out the working conditions of the KT88 in this single ended schematic:

http://www.bonavolta.ch/hobby/en/audio/kt88_1.htm

I will reproduce a part of that schematic here, just for clarity

[pingfloid]

1 - Plot the LOAD LINE

This line intercepts at the Y axis at Ip = Vb / (Rl + Rk) = 370 / (2.500 + 360) = 0,130 = 130 mA

Ip = Plate Current
Vb = Plate voltage = 370 V
Rl = Loar Resistance = 2.500 Ohm
Rk = cathode resistance = 360 Ohm

And it intercepts the X axis at Vb = 370 V

2 - The OPERATING POINT is the plate current/voltage point corresponding to the dc grid bias Vg. In this case Vg = 37,5 V

3 - When input signal Vi is appied, the plate current will move along the load line, so it will produce the corresponding amplified output signal across the load resistor Vo.

If the grid signal does not exceeds the cutoff voltage, the resulting output signal will have the same shape as the input signal (full cycle) this is Class A amplification.

One doubt: ¿how can I calculate the amplitude of the input signal in that example, just to make sure that I will not exceed the cutoff voltage?
(-40V) If I exceed this voltage, I guess that I will have to reduce the grid bias Vg, ¿maybe to 30V?, to run in pure Class A.

¿Did I said many nonsense?

[stigla]

Well, here are some comments on your work;

first off, you can see the screen of the KT 88 is tapped from the OutPut Transformer (OPT) at 40% (from the 375V).

This is called "Ultralinear" and is kind of a mix between Pentode and Triode (100% tapping from the 375V end would give you a triode connected pentode and 0% gives you a straight pentode. 40-43% is common with UL and gives least distortion (and 65% of the pentode power). 20% gives max power)

This UL connected KT88 have dedicated curves in the datasheet, so here is the first correction. You have used the pentode curves.

Next, you have a so called Auto bias, or Cathode bias. This form of bias elevates the cathode with respect to the grid at 37.5V (-37.5V seen from the grid.) This cathodevoltage must be SUBSTRACTED for the anodevoltage seen.

Anodevoltage = Anode to cathode voltage = 370V-37.5V = ~332V "plate voltage"

And this gives you a operating point at 332V at 100mA (on the UL curves).

The OPT is an inductive component and makes the anode voltage swing able to swing a LOT over the 332 platvoltage.

When calculating loadlines, I use this equation:

Load Impedance = (x-platevoltage)/plate current

and resolve for 'x' which gives you the second point (in volts at 0 anode current) on the straight loadline (the first one is the operating point).

Now you can draw your loadline.

As of the Class A aspect, it will only be valid if at twice the grid voltage (-37.5*2= -75V), the anode is still conducting current (on the loadline).

how can I calculate the amplitude of the input signal in that example, just to make sure that I will not exceed the cutoff voltage?

The needed swing from the driver will be 75V peak to peak. Or 37,5V positive and -37,5 negative or ca. 26,5V RMS.

[pingfloid]

Stigla, thank you very much for your reply.

So, if I'm not wrong again, the Plate Voltage when anode current is null will be:

x = (Load Impedance x Plate Current) + Plate Voltage
x = (2.500 x 0,1) +332 = 582 V

This gives the X interception point, being the second point (332V, 100 mA).

¿Maybe something like this?

[stigla]

you got it!

Now you can continue doing as you did in nr 3) in your 2nd post. This will give you the output voltage swing, on the anode (you must then calculate the step-down ratio to the OPT and then you can get the voltageswing on the 8ohm tap).

step down ratio = n =sqrt(Primary Z / secondary Z)



stigla

[Positron]

Hi Pingfloid,
A couple of things. The loadline is approx a straight line at "midfrequencies" and above. Below this the load line becomes more elliptical.

Second, the graph is only a fairly close approximation, esp when calculating distortion. The graphs are old, so always directly measure the distortion if possible.
Trying to calculate distortion is rather fruitless as interpolation is only an approximation. Being off by only 1 mil along the load line will change the distortion calculations rather "substantially". Also the newer tube manufacturers have improved the design of many tubes, including the KT88 family, so the distortion figures should be somewhat lower, depending on the manufacturer (Thus the newer curves would be very slightly different, doesn't take much change.) . Ranges of more than 2 to 1 in distortion figures with different brand KT88s are common.

Calculating distortions are a waste of time except in the most general sense; actual measurements are necessary for descent accuracy.

Hope this helps.