ECC83S/12AX7 Omitting grid leak, input impedance and preamps

[ruffrecords]

There is a limit on how much max the load impedance can be.
If you load the secondary with a 1 mega ohm resistor, does not mean the primary will have the same input impedance.
Then there is no meaning saying 600 ohm transformer. They will say only 1:! transformer.

See post #17 for a more detailed explanation

Matching the impedance to 600 ohms is only for maximum power transfer, which is not applicable in this case.

Not necessarily true. it was in the old days but not so much today. Depsite that, people do still insist on designing with transformer that work best with a 600 ohm secondary load which means the primary looks like 600 ohms to whatever is driving it. They really should be using 10K;10K bridging transformers which have much higher primary inductances.

Cheers

Ian

 

[rayma]

Crazy thought I am sure but could one add a series resistor (with or without the existing grid stopper) in each grid leg and a pot between the legs right after (on the grid side) to form a two sided voltage divider essentially partially shorting the input signal between the phases?

Yes, that's a standard balanced L pad attenuator, sometimes called a U pad.

 

[mandu]

See post #17 for a more detailed explanation

Not necessarily true. it was in the old days but not so much today. Depsite that, people do still insist on designing with transformer that work best with a 600 ohm secondary load which means the primary looks like 600 ohms to whatever is driving it. They really should be using 10K;10K bridging transformers which have much higher primary inductances.

Cheers

Ian

If the source has 50 ohms impedance (case 1) or 600 ohms impedance (case 2) connected to the primary of a 600 ohm impedance transformer, in which case will there be more voltage output at the secondary of the transformer? All the while I thought case 1. Can you clarify (in this case of driving the grid). Regards.

 

[ruffrecords]

If the source has 50 ohms impedance (case 1) or 600 ohms impedance (case 2) connected to the primary of a 600 ohm impedance transformer, in which case will there be more voltage output at the secondary of the transformer? All the while I thought case 1. Can you clarify (in this case of driving the grid). Regards.

It depends. When people talk about source impedance, what they really mean is small signal source impedance. However, this gives no indication of the driving capability of the the source. Most semiconductor outputs are designed to drive loads in excess of 2K and preferably 10K. A 50 ohms output impedance is extraordinarily low. Most, if not all, semiconductor output stages have build out reistors in series with the output to prevent instability when driving capacitive loads; 75 to 150 ohms is typical. This largely defines their output impedance.

Lastly, there is no such thing as a 600 ohms impedance transformer. Transformers do not have an intrinsic impedance. The load they present to the source is (mostly) a combination of the reflected load attached to the secondary and the inductive reactance of the windings. The latter varies with frequency. So if the secondary is effectively open circuit (driving a grid) the inductive reactance dominates. In your example, at 20Hz, the 50 ohm source would create a higher secondary voltage but at 1KHz both sources would create almost identical secondary voltages (how identical depends on the design of the transfromer).

Cheers

Ian

 

[57786]

Yep, and the circuit as drawn (as before) will have an input impedance which is essentially the inductance of the primary winding. There is no reason to assume it is 'low' - define 'low'; if it's 20 times your source impedance at your lowest usable frequency - it is not 'low'.

So, really, you need to find the primary inductance to work out what you can and cannot drive the circuit with, then you will know where to place the attenuator, and which type is the best fit.

With no data outside of '600 ohm', drive it with 600 ohm signal generator, with open secondary and vary the frequency until voltage across the secondary is equal to the generator output voltage divided by 2, note the frequency and then work it back from there.

You might be surprised, or not, but this would be the best way to proceed IME.