#40
gm = Rp / (µ+1) = 80Ω / (2,7+1) = 21,6Ω
There is 0% difference.
I got gm from the data book.
What do you want to suggest here?
No comments to the other text, this is a moderators job.
D.
tubetvr said:... However if you use the same formula for a low µ power triode as a 6C33C the result is very inaccurate. A 6C33C has a µ of typically 2.7 and a Rp of 80 ohm which gives a gm of 33.75mA/V, with the faulty formula this would give an output impedance of 1/gm which in this case is 29.6 ©. If we instead use the correct formula Zo = 1/[(1/Rp)+gm] we get 21.6 © a 37% difference which can't be seen as insignificant. ... Regards Hans
gm = Rp / (µ+1) = 80Ω / (2,7+1) = 21,6Ω
There is 0% difference.
I got gm from the data book.
What do you want to suggest here?
No comments to the other text, this is a moderators job.
D.
gm = Rp / (µ+1) = 80Ω/ (2,7+1) = 21,6Ω;
Gm for a triode is µ/Rp, not Rp/µ+1
Regards Hans
Re: #40
1/gm = Rp / (µ+1) = 80Ω / (2,7+1) = 21,6Ω
There is 0% difference.
Darius
Originally #41 posted by oldeurope
gm = Rp / (µ+1) = 80Ω / (2,7+1) = 21,6Ω
There is 0% difference.
I got gm from the data book.
What do you want to suggest here?
No comments to the other text, this is a moderators job.
D.
1/gm = Rp / (µ+1) = 80Ω / (2,7+1) = 21,6Ω
There is 0% difference.
Darius
Re: Re: #40
I think you missed the point. The problem is that 1/gm is NOT the correct formula for output impedance, it is only an approximation in the limit of mu>>1. If you look at the output impedance, this formula gives a 30% error for Hans's example.
And, once again, by definition, gm = mu/rp, NOT (mu + 1)/rp.
oldeurope said:
1/gm = Rp / (µ+1) = 80Ω / (2,7+1) = 21,6Ω
There is 0% difference.
Darius
I think you missed the point. The problem is that 1/gm is NOT the correct formula for output impedance, it is only an approximation in the limit of mu>>1. If you look at the output impedance, this formula gives a 30% error for Hans's example.
And, once again, by definition, gm = mu/rp, NOT (mu + 1)/rp.
no problem ...
Hi SY,
now I am totally confused. I'll think about it.
Most important is this :
Back to topic please, thanks!
Kind regards,
Darius
Originally #46 posted by SY
I think you missed the point. ... If you look at the output impedance, this formula gives a 30% error for Hans's example.
And, once again, by definition, gm = mu/rp, NOT (mu + 1)/rp.
Hi SY,
now I am totally confused. I'll think about it.
Most important is this :
The ECC83 12AX7 has definitely µ>>1. Thus my calculation is correct.Originally #46 posted by SY
... The problem is that 1/gm is NOT the correct formula for output impedance, it is only an approximation in the limit of mu>>1. ...
Back to topic please, thanks!
Kind regards,
Darius
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