After these two photos look for a working conjecture as to the funcitonality of this box. This was excerpted from an email exchange with Dan Sheingold. It includes interesting insights about the trenary counting system.
Finally, Charles Dodds lent his KT-4 for schematic tracing. You can download it in PDF format, or in LTSpice format. LTSpice is available for free at www.Linear.com
While I had the box, I measured a few resistance values as a sanity check for my traced schematic. There still unresolved discrepancies in the switch wiring.
Date: Sun, 11 Jan 2009 16:07:10 -0500
To: Joe Sousa
From: Daniel Sheingold
Subject: Re: K4 T Philbrick Transconductor
Joe:
I can't vouch for the contents of that box; it was just a guess. Perhaps if the circuit were traced out and the connecting points appropriately shown in a real circuit, we might be closer to the solution.
Best regards,
Dan
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Dan:
Interesting info, I really like the economy of the trenary system. It is the bridge between octal and BCD.
I have shared some of your thoughts on the K4-T with the owner of the unit.
As you may expect, I would like to post your thoughts on this unit and on the trenary system that seems to be very economically employed in this box.
...snip...
Best Regards,
-Joe
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At 09:35 PM 1/10/2009, Daniel Sheingold wrote:
Joe:
The K4 series in the late '40s and early '50s consisted of large boxes (multiplier, function fitter,dynamics simulator, etc.) the width of essentially 4 K3 boxes, practically rack width.
This one doesn't seem to be a bona fide K4 of that generation. I don't know why anyone would give it a K4- prefix. Perhaps Bob remembers it.
Superficially, it looks as though it could be a separate component based on one of the gain-setting inputs of an SK5-U--to allow someone to build a single-input SK5-U, with a loose pair of op amps, for example. However, the front-panel diagram is not very helpful.
By the way, the gain settings of the SK5-U are based on ternary-coded decimal, rather than typically BCD, switching logic. That is, each input resistor can be switched to the direct summing point, the inverting summing point, or to ground. Thus gains are built up like so (taking a single term) with fewer resistors:
-9 = -9 (Binary would be -8 -1)
-8 = -9 +1
-7 = -9 + 3 - 1
-6 = -9 + 3
-5 = -9 + 3 + 1
-4 = -3 - 1
-3 = -3
-2 = -3 + 1
-1 = -1
0 = 0
+1 = + 1
+2 = +3 - 1
+3 = +3
+4 = +3 + 1
+5 = +9 -3 -1
etc.
Dan
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The following contribution came from the owner of the box, Charles Dodds:
Joe,
I am having a friend help me with the ohm meter testing (it has been years since I did anything like this). He is a computer sales and engineering pro.
Also, I have cut and pasted from my research, the opinion of a seasoned electronics engineer (a friend of my friend). His vintage opinion, is based on the photos I sent you, is as follows:
"This would be a resistance bridge, aka dual precision voltage divider. In a passive set up it could provide two (nearly) equal voltages on the C & D outputs. A first typical use would be measuring an unknown resistor (or more general impedance containing Capacitor or Inductive elements), by comparison with a known one. For example the excitation voltage would be applied to A, and B would be grounded. Then C and D could be set up to have equal voltages. Or they could be set up to have another precisely-known ratio, by using the click switches that allow adjustments in 1% steps.
To make a precise measurement, one would want to make a setup where two almost equal values would be compared by subtraction -- but this device doesn't provide sign inversion function. Suppose we put the signal on A, with D connected to the inverting input of an operational amplifier. The noninverting amplifier input is grounded, and the opamp output is put onto B. Now A has +V on it from our external source, and B has -V* on it, due to the opamp inverter action. Now as we click the resistor switches point C can be varied from reaching the top of the input (A) so pt C gives output(C) = +V, down to zero when the resistors on the left side of the bridge have been set to be equal, exactly balancing the current from the + V of A with the current from -V* of B. If the smaller resistance on the left side would be on the bottom of the chain, the output would be closer to the bottom potential, ie it would be negative. So this is a handy-dandy bipolar voltage source. To make it a "stiff" voltage source, so it didn't get loaded down by your load, another opamp would be used as a follower with +unity gain on output C.
In general terms, the connection between A and C would be called a transimpedance and, because of the opamp's inversion of the voltage to B, could vary between + and - some large value. The actual resistance to the assumed ground (defined by the noninverting opamp input level) would be a variable and relatively small resistance. But the electrical model of this box with input A and an assumed ground, would have output current from C to the same ground which can vary from a lot +, to essentially zero when the system is balanced, to a lot - when one is maximally unbalanced on the left side toward the bottom of the left arm. Due to tolerance of the parts, the nominal zero voltage at C probably wont be exactly zero, so some current would come out of C. This can be cancelled by a direct leakage of a little current from A to C - this is the purpose of the 2.7 Meg and 2x 22 Meg resistors in parallel across this terminal pair. Perhaps the residual unbalance went the other way - then the connection would have the bridging resistors going from B to C, to leak in a little current of the opposite sign. (It may be that the assignment of the connectors is intended to be different from what I said, because the trim resistors seem to be between A and B. Or maybe these parts were added afterward - they are cheap carbon resistors. Added incorrectly, and the owner became frustrated by the absence of a balance point.)
To get the stable precision, they have used wirewound resistors, which will also have inductance. So this system probably begins to have frequency-dependent errors above a few 10's of kHz. Nowadays we have metal film resistors with 4 or 5 digit accuracy and few $ cost, so this same strategy could be implemented up to a few 10's of MHz. For rf, into the GHz range, it is more convenient to use wideband inverting transformers to get the voltage sign inversion, via a center-tapped winding, with the centertap grounded. Then one has a sample of + and - signals to combine as desired."
Thanks,
Chuck