🚀 go-pugleaf

Hopefully this is a simple question, but I forgot my transistor theory
since it's been so long. 

This is the circuit : 
                                                                
                                   <--- Ic                           
                              __________     Vcc                             
                             |                                   
                             >    300 ohm                               
                             <                                   
                             |                                   
                             |                                   
                1K ohm      /    C                                
                         |/           NPN                            
            --- ^^^^^--- |\           MPSA42                            
                            \                                    
           input             |   E                                
                             |                                   
                            ____                                  
                             --                                   
                                                                


Vcc is about 100V; the NPN transistor MPSA42 is in a conventional 
common emitter circuit, collector resistance = 300 ohm, input 
resistor (base resistor) is 1K ohm, emitter is grounded. 

My question is : when I turn off the transistor, what is the leakage
current drawn from the + 100V supply ?

I have a Sprague data book, which says, for MPSA42, Icbo is 100nA
at Vcb of 200V. Is Icbo the same thing as the leakage Ic in the
above circuit ? Sprague conveniently ignored to show their measurement
circuit, or define their terms. 

(To turn off the transistor, I have the choice of applying either 0V
or -12V through the 1K resistor. -12V will break down the base-emitter
junction, but the 1K prevents damage. I'll do whichever gets me
lower leakage current from the 100V Supply) 

Thanks much, 

/ Jon Sreekanth

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