**EMRFD Message Archive 9794**

MessageDateFromSubject9794 2014-03-03 10:07:23 ford@highmarks.co... Impedance question fig 6.56 I have a practical question about coupling stage-to-stage and impedance transformation.

Focus on Figure 6.56 of EMRFD Rev 1

^{st}. This is an IF amp with AGC based on dual gate mosfets. I’m adapting it to a similar design and have questions.

The transformer T2 in the drain lead of Q3 has a 1K resistor in parallel with the primary. I must presume this acts to terminate the stage into 1K ohm.

The description calls for T2 to be 28T : 4T of #28 wire on a type 43 core. Ok, simple enough. I am led to believe that the impedance transformation ratio is the ratio of turns squared primary over turns squared secondary. 28x28=784 and 4x4=16 and 784 / 16 = 49. The 1K termination on the primary is then 1000/49= 20 ohms or so. Does this suggest that Q4 and Q5 present a characteristic impedance of 20 ohms? My success in getting good transformation ratios beyond 16:1 or so is not good. Is it reasonable to expect good results at 49:1?

The secondary then terminates into Q4 / Q5. These PNP devices appear to have a characteristic impedance in the 510 ohm range—assuming the 510 ohm resistors dominate that impedance. Then we see Q5 with a collector resistor of 680 instead of 510. Just looking at the symmetry of the circuit, why is it 680 instead of 510?

Then we see the final transformer tapped off the 510 on one side of the differential amp, terminated into a filter (of presumably 510 ohm impedance) and then transformed down to (presumably) 50 ohms with another transformer with a turns count of 15 : 5. 15x15=225 and 5x5=25 and 225 / 25 = 9 and 510 / 9 = 57 ohms. The one side has a 680 and terminates directly into Q6, which appears to be an impedance of 10 ohms or so.

Why is T2 designed as it is? Why isn’t the 680 actually 510? What is magic about the 1K in parallel with the primary of T2? Why is a transformer necessary to transform good power transfer from 510 to 50 but not 680 to 10? I need some practical perspective on when to be concerned about impedance and when to just press on…

Any help would be informative as I’m adapting a similar amp and am confused as to the 49x impedance ratio.

Ford-N0FP

9810 2014-03-06 15:02:16 w7zoi Re: Impedance question fig 6.56 Hi Ford, and group,Sorry I did not respond earlier. We were out of town on a vacation trip and I didn't look at any on line electronics while away.The IF amplifier of Fig 6.56 is an adaptation of the one used in the Progressive Receiver (PR) from QST,November, 1981. The early stages are dual gate mosfets with AGC applied to the gate-2 bias line. Theoriginal PR used two stages, but this one uses three. Gain is kept low by loading the drains withfairly low load resistance. The dual gate mosfets are followed by a differential pair. I happened to use PNP parts here, for that was handy in driving a detector. A NPN pair would work as well. The transformer serves to decouple common mode signals from the FET chain from the high gain differential pair. One could probably get by without a transformer here. The transformer is not intended to offer a careful impedance match. Rather, it provides a low impedance (essentially, a voltage source) drive for the diff pair. The differential pair is powered with a single divider to bias the two bases at the same DC voltage, and a "long tail" resistor for the emitters. Two identical resistors bias the two emitters that are then tied together with a capacitor, a scheme I picked up from K7HFD many years ago.The input impedance of the diff pair is going to be a few hundred Ohms or more at the frequencies we areusing here. This is essentially established by the DC emitter current in the diff pair. The outputimpedance at each of the two collectors is going to be quite high. We load the two with resistor loads. The output of Q4 is terminated in a 510 Ohm resistor, thus providing a 500 Ohm drive for the crystal filter. The other side of this filter goes through a 3:1 turns ratio transformer and is then loaded with a 50 Ohm circuit. This provides the 500 Ohm termination required by the crystal filter. Other values could be used for other crystal filters. Indeed, if you are building your own crystal filters, you even had that option available to you. The collector of Q5 happens to have a 680 Ohm load. There is no good reason for this and you could certainly use a 510 Ohm load. It merely drives Q6, which is the detector. The slightly higher load resistance will alter the dc bias which ripples through to bias the op-amp.A differential pair amplifier has good reverse isolation and even better isolation between the two outputs when they are separated. I measured this when working with the hybrid cascode IF circuit and the isolation between the two collectors was close to the limit of what I could measure with ease.Most of the design of the Fig 6.56 amplifier happened at the bench. The turns ratio of the transformer is not really a critical parameter. Some experimentation may provide results that are more in line with your needs. Adding turns to the secondary would increase the drive impedance applied to the bases and would increase the gain, perhaps by quite a bit. The reverse isolation may be compromised. This is a servo loop, so reverse isolation is always a consideration.You can certainly do some computer modeling of the circuit, although it's hard to find good models for dual gate FETs. Back in earlier times we used a FET cascode circuit as a model for a dual gate GaAsFET and were always surprised to find that the model worked as well as it did. The hybrid cascode topology with a JFET cascode connected to a NPN is readily modeled in SPICE though.Incidentally, the amplifier of Fig 6.56 is still a part of my home station CW receiver and continues tooperate today.Good luck with your experiments.73, Wesw7zoi9812 2014-03-06 16:26:23 Nick Kennedy Re: Impedance question fig 6.56 OK, you got an answer from the designer himself.I will point out one error in your post:The description calls for T2 to be 28T : 4T of #28 wire on a type 43 core.

The primary is actually 20 turns, not 28. Looks like you took the wire size as the number of turns.73-Nick, WA5BDU