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Consider the voltage level at the circled node,
if that voltage level is calculated by the
Kirchhoffs voltage law (summing from the 5V rail to the 0V rail)
, but if the node is also connected to the chip, we don't know
what components are available inside that chip, can we assume
the voltage...
I start peeking the file content, and discover its a
Personal CAD Schematic file.
Is there any free viewers for those file?
Schematic FILE
It has a header like this
Or is it interchangable, the file was created in 1994.
Okay, I remember now. The current of a transistor circuit doesn't depend on the DC voltage, but the base current/voltage. So the Ic doesn't depend on Rc.
Right?
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Ok, get it, the potential difference all over is 18V with a 0.65V transistor difference. divided by Rc...
Thanks, I understand some of the parts now. The transistor is acting like a diode, but 18V - 0.65V, if the 0.65V is he drop across the transistor, but why not considered Rc which might have a resistance of several Kilo ohms or Mega on its own. if so, the voltage drop across it might be ~17V...
No voltage drop across T1? Sorry, I am quite new to current mirror circuit
Why not IcRc + Ic(Resistance of transistor 1) = 18V
18V / Rc + Resistance of transistor 1) = Ic
https://en.wikipedia.org/wiki/Current_mirror
I don't understand if the positive rail is +9V and negative rail is -9V
Why the current is 18V - 0.65V / Rc?
Anyone explain this?
Thanks
Jack
Sorry, I have come back.
In physics, I remember there is one thing called displacement current, where energy stored in magnetic fields are transformed into electric fields in between the cap and transformed back into magnetic fields on the other side of the cap for continuous currents. So my...
I have a 520W power supply for my personal computer appliances.
I used to have a 500W power supply for the the same computer set. If the mains supplies 13A of current, would the 520W power supply draw more energy than the 500W one from the mains to the supply, meanwhile the current is constant...
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