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I'm trying to understand the concept of driver impedence and how interstage transformers affect this parameter. For now, consider the working design to be single-ended, class-A. The output tube is an 845.I understand that some output tubes (especially transmitting tubes) want a low impedence driver. This can be achieved by using a driver tube with a low plate resistance (6H30P, 6BX7, 300B etc.). Some designs use a kathode follower or a Mu-follower to get even lower drive impedence. (Someone please stop me if I'm getting anything wrong here).
So how does an interstage transformer affect the drive impedence seen by the output tube? Does the plate resistance of the driver tube get "reflected" through the transformer? Does the turns ratio affect this?For example, would a driver tube plate resistance of say, 1200 ohms, be seen by the output tube as 2400 ohms if the transformer were wired for 1:2 step up? How about step-down? Would the 1200 ohm plate resistance be seen by the output tube as 600 ohms if the interstage were connected 2:1 step-down?
How would these connections affect the primary side of the interstage? In other words, would a 2:1 step-down connection change the plate load seen by the driver tube? Is there an impedence on the secondary side of the transformer that gets reflected through the interstage to affect the primary load on the driver tube--or is the driver's load established by the Henry's of the primary?
I'm guessing that with enough signal voltage to start with, a step-down transformer will allow me to get a low drive impedence without resorting to kathode followers. Obviously, a step-down interstage will throw away signal voltage so there must be enough to start with, (from preceding stages of gain).
Although perhaps an odd topology, connecting an interstage as Push-Pull to Single-Ended (and step-down 2:1) might produce enough voltage gain (from the push-pull pair) and still lower the drive impedence seen by the output tube-----or does a pair of push-pull tubes on the primary side result in "twice the effective plate resistance, compared to a single tube" and nullify the lower Z achieved by the 2:1 step-down connection?
Any help with these questions would be much appreciated.
Follow Ups:
Lots of good questions... Let me see if I can do justice to them!Basically, the interstage transformer (IT) transforms the voltage as the turns ratio and the impedance as the square of the turns ratio. Let's set up an example. Let's imagine a SE driver with an Rp of 1.2K driving a triode output tube through an IT with a turns ratio of 1:2. At the simple level, for every volt of signal developed at the plate of the driver, 2 volts appear at the grid of the output tube. However, the grid of the output tube "sees" a source impedance of 1.2K x (2 squared) or 4.8K. The result for a step down connection is just the reverse: 1 volt becomes 0.5 volt and the reflected impedance is 300 ohms. Big difference!
With transmitting tubes like the 845 there are two challenges. The first is an adequately low Rp driver to ensure that the grid capacitance can be driven up to an adequately high frequency, and the second is that the driver can provide enough voltage to the grid to get the full power out of an 845, which requires a pretty massive grid swing. Unfortunately, while a step down IT solves the first problem, it aggravates the second. Your idea of a P-P driver, while heresy to the SE devotee, is not such a dumb idea, because it largely cancels even order harmonics in the driver stage, while potentially providing a lower frequency cut-off than a paralleled SE driver stage with the same tube. However, it doesn't buy you lower drive impedance than a paralleled driver.
My advice is to choose a low Rp tube with a high power supply voltage maximum, so that you can swing the required grid voltage and have a low driving impedance, which takes us right back to the short list of tubes in your second paragraph.
Kevin Carter
K&K Audio
www.kandkaudio.com
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Thanks!Two questions I'm still not sure of.
First, in the PP to SE interstage connection (stepped down 2:1), is the impedence seen by the output tube one-quarter of the Rp of a single driver tube (as in SE to SE) or do both PP driver tubes get added together to make a combined impedence?
Second, my question about the load seen by the driver tube(s) is still unclear. What factors contribute to this load? Is there something on the secondary side (associated with the 845) that contributes?
My interest in using a PP to SE topology lies in lower Z drive but it also would seem to increase the drive voltage generated compared to a single driver tube while reducing the "excursion" each tube would swing. This should help lower distortion as well. Any thoughts?
Lastly, what possible connections might a LL-1660s be used in for a design of this type? (it wouldn't have to be 2:1 necessarily, just something similar).
In the case of the P-P to SE IT, the grid of the output tubes "sees" the plate-to-plate impedance divided by the square of the turns ratio. So the grid sees roughly 2 x Rp x 0.25.The driver tube(s) "sees" the grid impedance, which at low frequencies is predominantly the reactive impedance of the IT secondary and at high frequencies is the parallel sum of this and the reactive impedance of the output tube grid capacitance.
As I wrote in the first post (almost), the advantages of Class A P-P to SE IT drive (compared to the same two tubes used in parallel SE) that are apparent to me are: for a given size of transformer, a lower low frequency response limit, lower even order harmonic distortion, and half the drive impedance for the same driver tube bias point.
With the caveat of little or no thought given to the frequency response consequences of my choice, you could try something similar to connection Alternative N with the secondaries wired up to give 2.25+2.25:2.
Kevin Carter
K&K Audio
www.kandkaudio.com
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