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This is a high voltage supply from a Data Precision 8200 voltage reference source. In this simulation, the output voltage on the far right will be 10x what is input plus or minus in between the 130 volt rails. For example if -1v is fed into the far left PNP base, about -10v will appear on the far right output (trimming required in real life). My analysis shows that the output voltage is proportional to the current flowing through the amp meter in the center of the schematic. For example, +100uA flowing to the left (to ground) will produce +10v on the out. -1000uA (flowing to the right, out of ground) will produce -100v on the output. In essence, the far right two transistors are biased on and drop voltage proportional to the current passed through their collector to emitter. For example, as less current flows through the upper right NPN transistor the transistor drops more voltage from collector to emitter, the lower right transistor drops less and the difference is seen at the output. Hence, if you set the input supply to zero volts, you get zero out as the voltage is balanced between the two transistors.
If anyone has a comment to add on the functionality, please add it! I am analyzing this to perform a repair on a Data Precision 8200 voltage reference source.
UPDATE 6-3-16: I was able to successfully repair the DP 8200 high voltage circuit. This simulation helped a lot to understand it. I believe that the previous owner placed a load too large on the output and the circuit drew too much current and destroyed the output transistors. I replaced the D40Ns with TIP50s. The pins don't line up but a little bodge wire and were in business!
Data sheet:
http://www.ko4bb.com/manuals/97.91.48.154/Data_Precision__8200_VoltageCurrent_Calibrator_Service_Manual.pdf
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