What's wrong with SE Cad?

[jane]

If I make the primary load equal to the plate resistance (Rload=rp), DF should be 1. At the secondary side DF should be 1 or less than 1 if we add up Rdc in the transformer. Why is SE Cad telling me that DF is greater than 1 and Zout is less than 8 ohms?
As far as I can see, the transformer has an 8 ohms secondary (10.9 ratio = 950/8).
Could someone please explain why SE Cad calculates such generous output impedance?



Jan E Veiset

[smoking-amp]

I haven't used SE Cad, but I see a Plate Zo figure of 475 Ohms listed. Maybe the operating point is at a different current than the tube's spec point for the 949 Ohm?

Don

[Svein_B]

I have not used John Broskie's SE Amp CAD either.

Id seems that it calculates the Z-out from the tube as Rp in parallel with the load Rl. This is as as I understand as per textbook learning - if were to measure at this point.

When we measure at the secondary load, the level with no load is twice the level with 8 ohm load. This should indicate that the internal output impedance is around 8 ohm.

I have simulated this in LtSpice to confirm my estimate.

When we apply a signal to the output load from another generator (with no input signal to the amp), 1 V with an internal impdance of 1k, we measure 4mv across the load, idicating that the output impedance is in fact about 4 ohm as predicted by John Broskies program.



SveinB.

[jane]

You are right. The value Plate Zo is rp in parallel with Rl (dcR included) and these value (Plate Zo) is used to calculate the output impedance and DF.
If I reduce dcR in the transformer to a minimum (still with rp=Rl), output impedance is calculated to 4 ohm / DF=2. Nice result, but I wouldn't call it output impedance [1].
With the load included in the calculations, DF will never be less than 1.

Jan E Veiset

[1] Output impedance - Wikipedia, the free encyclopedia

[smoking-amp]

"indicating that the output impedance is in fact about 4 ohm"

I assume this was measured with the load connected. In that case it would look like an 8 Ohm output Z in parallel with the 8 Ohm load giving the 4 Ohm result. But the definition of damping factor is Zload/Zsource (or Zload /Zout) so the damping factor should be around 1 (8/8). ?