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
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?
An externally hosted image should be here but it was not working when we last tested it.
Jan E Veiset
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.
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.
Last edited:
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
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
"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). ?
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). ?
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.