Following the sub-dedate on this thread;
http://music-electronics-forum.com/s...ad.php?t=10431
with Sir Cuitous about output impedances for the 6G15 in order to come up with a suitable impedance for a theoretical 1:1 isolation transformer for the 6G15 output (is this a trick question? - I thought 1:1 meant that the reflected load was the same as the input impedance of whatever was following the earlier device? So I could be barking WAAAYYYY up the wrong tree - please tell me if this is so), I decided to start another sub thread too help me work out the answer. So..... (big breath in - )
To my simple mind, the 6G15 has two stages that provide source impedance leading into the mixer (I think that much is correct - but Ah Foohey - I could be wrong about that too). These stages are in parallel and one of them is a cathode follower. Does this mean that output impedance is the product of paralleling those last two source impedances? And since the cathode follower has really low output impedance, does that make the total output impedance at the mixer really low too? Or does that depend on how the mixer is set?
Formula for bypassed gain stage; Zout = Ra || ra
Anode resistance for a 12AX7 triode is 70k
Therefore, for the recovery stage Zout = 100k x 70k/(100k + 70k) = 41k (I think that's right)
Formula for a CF stage Zout = ra/mu (I'm not sure about this tho')
Assuming the gain is about 0.93 for a typical CF stage, therefore for the 12AX7 CF stage; Zout 70k/0.93 = 752k (That doesn’t sound right - what have I done wrong? The formula?)
Anyway even if I didn’t get the CF Zout impedance correct, if the two output stages were paralleled, then does that make the output impedance:
Zout = normal stage Zout x CF stage Zout/(normal stage Zout + Cf stage Zout)? (= really low?, or does that depend on how the mixer is set, as I pondered earlier)
http://music-electronics-forum.com/s...ad.php?t=10431
with Sir Cuitous about output impedances for the 6G15 in order to come up with a suitable impedance for a theoretical 1:1 isolation transformer for the 6G15 output (is this a trick question? - I thought 1:1 meant that the reflected load was the same as the input impedance of whatever was following the earlier device? So I could be barking WAAAYYYY up the wrong tree - please tell me if this is so), I decided to start another sub thread too help me work out the answer. So..... (big breath in - )
To my simple mind, the 6G15 has two stages that provide source impedance leading into the mixer (I think that much is correct - but Ah Foohey - I could be wrong about that too). These stages are in parallel and one of them is a cathode follower. Does this mean that output impedance is the product of paralleling those last two source impedances? And since the cathode follower has really low output impedance, does that make the total output impedance at the mixer really low too? Or does that depend on how the mixer is set?
Formula for bypassed gain stage; Zout = Ra || ra
Anode resistance for a 12AX7 triode is 70k
Therefore, for the recovery stage Zout = 100k x 70k/(100k + 70k) = 41k (I think that's right)
Formula for a CF stage Zout = ra/mu (I'm not sure about this tho')
Assuming the gain is about 0.93 for a typical CF stage, therefore for the 12AX7 CF stage; Zout 70k/0.93 = 752k (That doesn’t sound right - what have I done wrong? The formula?)
Anyway even if I didn’t get the CF Zout impedance correct, if the two output stages were paralleled, then does that make the output impedance:
Zout = normal stage Zout x CF stage Zout/(normal stage Zout + Cf stage Zout)? (= really low?, or does that depend on how the mixer is set, as I pondered earlier)
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