I have learned in electronics courses about an NPN transistor is that current Ib controls Ic, the current Ic flows from c to e, and Ib flows from b to e.. how could the current pass from b to c (in NPN)?? I read this in an analysis for this circuit http://sub.allaboutcircuits.com/images/04090.png which is located here http://www.allaboutcircuits.com/vol_4/chpt_3/3.html and I'm taking about Q1

Hazim

 

Current can pass through the base-collector junction if the base is about 0.6V more positive than the collector. In normal operation of a NPN transistor you don't need to pass current through the b-c junction.
If you use collector as the emitter and the emitter as the collector, then current will pass through the c-b junction like as the b-e junction as normally. However, due to the different doping levels of the emitter and the collector (collector doping lower than the emitter) the characteristics of the transistor, like its gain, change.

 

Hi hazim,

At first glance, the circuit does seem strange, but if you follow the analysis from the 2nd link in your post, it's explained fairly well. I'll try to expand on that here to make it more understandable.

First, lets look at the case where the input is low. This is straightforward and the Q1 NPN get bias current thru the base-emitter junction in the normal manner, which sets Q2's base close to ground and much less than a diode drop above ground.

Now, back to the other case where the input is high. Note that Q1 is an NPN bipolar transistor. Therefore, as you would expect, there are three junctions: B to E, B to C, and C to E. Since Q1-E is above Q1-B, albeit not by much, but less than a diode drop, and Q1-C is a bit more than a diode drop above ground (thru the Q2 B-E junction) and much lower in voltage than Q1-B, and since the B-C junction is a P-N, just like the B-E junction, Q1-B to Q1-C will be forward biased and conduct the current from VCC thru R1 thru the Q1 Base to Collector junction and into Q2's Base to emitter junction, turning on Q2. The drop across Q1 Base to Collector should be in the area of 0.65V to 0.7V. In this case, with the current flowing from Q1-Base to Q1-Collector, Q1 is operating solely as a diode and not as a transistor. But when the input switch is grounded, Q1 is operating as a transistor.

I hope that explained it. Try this. Take an NPN transistor with a beta of about 100, perhaps a TO-92 size, and take a DMM that has a diode checker and put the plus lead to base and the minus lead to emitter and you'll read about 0.65 to 0.7V. Now, keeping the plus lead of the DMM on the base, move the DMM's minus lead to the Collector and you'll also read about 0.65V to 0.7V. It is a diode junction and operates this way in the circuit we've been talking about.

Good luck,
Kamran Kazem,
kkazem

 

When an NPN is saturated (a normal mode of operation), current must flow from base to collector.

 

Ok thank you all, I understood this issue well

 

When an NPN is saturated (a normal mode of operation), current must flow from base to collector.

Why would current flow from base to collector when saturated? I've read that before, and it doesn't seem to make sense to me. The only way that current flows from B to C is if B has a higher voltage that C. When the transistor is in saturation C is(usually) much higher than B.

 

If any current flowed from collector to base in normal operation, the transistor would latch on and need the base actively pulled low to turn off.

That only happens under thermal runaway conditions, definitely NOT in normal use.

 

You might find this section of the Ebook enlightening - http://www.allaboutcircuits.com/vol_3/chpt_4/3.html

The meter check of a transistor shows two PN junctions - BE & BC. If the BC junction is forward biased, then there will be current flow.

 

Why would current flow from base to collector when saturated? I've read that before, and it doesn't seem to make sense to me. The only way that current flows from B to C is if B has a higher voltage that C. When the transistor is in saturation C is(usually) much higher than B.

The base voltage is higher than the collector voltage during saturation of an NPN.
When a BJT is in saturation, Vce is typically less than 200mV, sometimes much less. Perhaps you are thinking of MOSFET saturation, where, unfortunately, the definition of saturation is basically the opposite for that of a BJT.

 

The base voltage is higher than the collector voltage during saturation of an NPN.
When a BJT is in saturation, Vce is typically less than 200mV, sometimes much less. Perhaps you are thinking of MOSFET saturation, where, unfortunately, the definition of saturation is basically the opposite for that of a BJT.

Oh, sorry I didn't know you were talking about Vce, I thought you were talking about voltage at the collector, not the voltage dropped across it.

 

Oh, sorry I didn't know you were talking about Vce, I thought you were talking about voltage at the collector, not the voltage dropped across it.

Do you see any difference between Vc and Vce in this circuit?

 

Oh, sorry I didn't know you were talking about Vce, I thought you were talking about voltage at the collector, not the voltage dropped across it.

If the emitter is grounded, and you are measuring wirh respect to GND, Vc=Vce.