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enter image description here

what happen if the load is a high resistance ( let's say 100K )

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    \$\begingroup\$ Then you get less than 1ma through it. \$\endgroup\$ – pjc50 Feb 4 '18 at 18:56
  • \$\begingroup\$ What do you think? What's your hypothesis? \$\endgroup\$ – Harry Svensson Feb 4 '18 at 18:56
  • \$\begingroup\$ Good question! It illustrates the need for specifying compliance when designing a current sink or source. \$\endgroup\$ – AlmostDone Feb 4 '18 at 19:17
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – clabacchio Feb 5 '18 at 7:45
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The more the load resistante increases, the less voltage you´ll have in the 1k resistor, due to the fact you won´t be able to source 1mA as load resistance has increased and power supply is 10V. That 1.6V will reduce as well, and the transistor will be as saturated as possible. Together with all of this, base current will increase as less voltage will be dropped across the 1.6K resistor. And finally, alongside with that, your 1mA current source won´t hold and the current will decrease.

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Will you accept a rough approximation?
With a 100k load resistor, the transistor saturates. Saturation occurs when the collector-to-base voltage drops to near zero...

schematic

simulate this circuit – Schematic created using CircuitLab

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The circuit, with exactly those component values comes from Art of Electronics, Horowitz & Hill. Read what they have to say about \$V_{CEsat}\$... the BJT cannot drag the collector down to \$-90V\$ in order to draw 1mA through 100k!

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