The power supply needs to be negative for an NPN.You could make that with, say, and NPN transistor, by grounding the base and running the collector and emitter to a positive power supply through separate resistors.
Hmmm... maybe it will be saturated because the emitter and collector will be a roughly the same voltage. The exact voltage depends on the design of the particular transistor.
Your experience has led you astray.Alright lets logically think this thru.
A transistor is properly biased (NPN) when the base is more positive than the emitter,
and the collector is more positive than the base.
Under these conditions you have the base emitter junction acting as a diode in the forward biased way, the base collector junction is acting as a reversed biased diode, due to the purposely designed thickness of the emitter collector junction a barrier is formed and bla bla bla..
It works as a current amplifier.
Now if the base is made more positive than the collector, then there is a collector base forward bias, which causes the transitor to leave its linear amplifier curve.
One way this happens is if the collector resistor is to large and the base voltage is to large, causing a significant voltage drop across the collector resistor, thereby lowering its collector voltage with respect to the base voltage.
Under these conditions, the base emitter junction acts as a forward biased diode, with the collector almost being open, which means is, very little current will flow into the collector, and most of the current from ground will flow out the base, this is not saturation, but a wrongly biased transistor.
Saturation is a transistor that is at the top of its linear curve, with maximum current flowing, where any increase in base current has little to none effect of the collector current.
So a forward biased base emitter, and a forward biased collector base junction can not enter saturation, but only act as a base emitter diode, with the collector almost opened.
This is just my hobby experiance (NON professional) opinion.
Actually, most of the current (typically more than 90%, according to simulations) will flow through the collector. The b-c junction is typically larger than the b-e junction, so b-c will hog most of the current.As a by-the-by if you short the source and drain of a JFET you get a very low leakage diode. See a 'PAD 5' for example. If you do the same with an NPN tansistor the emitter will have a lower Vf and most current will flow through the emitter, but it will still lokk like a diode.
Given everything you described, I think the transistor was saturated. In a saturated transistor with an emitter resistor, Vc will not be near ground, but Vce will still go to near zero volts. The key here is where you connect your black probe. It needs to be on the emitter, not on ground.<snip>but the transistor was not saturated, because the CE drop was no where near ground, but just very low for the expected calculated values.
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by Jake Hertz
by Jake Hertz
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