When I try to prove that Vce for saturation is below .2V and I take npn transistor I simply consider emitter to be at 0V and thus base emitter junction being forward biased base gets a voltage of (.7 V + x) where x is positive constant then to forward bias base collector junction I will have to have forward Voltage of at least .8V so collector voltage becomes .7v+x-.8v+z where z is a positive constant
and x is greater then z so collector voltage should become -.1V+(x+z) thus What I am getting is Vce should be greater then -.1V
When follow same procedure for PNP I get Vce greater then .1V
which is totally in coherent with <.2V What I am doing wrong please help
Electronic – How to prove that Vce saturation lies below .2V
bjtnpnpnpsemiconductors
Related Solutions
In general, it is the states of the PN junctions inside the transistor which will determine what operation region it is in. However, after gathering some experience, one can deduce the states of the above junctions by inspecting the circuit itself without actually measuring the voltages at the terminals.
An example:
Lets analyze the circuit you've referenced.
Once the switch is closed a current of approximately \$1mA\$ will flow into the base, which will cause:
$$V_{BE} \approx 2V$$
Since this is higher than the minimum of \$0.6V-0.7V\$ for being out of cut-off - the transistor is in one of its operational modes. In reality, the Base-to-Emitter voltage will not rise much beyond \$0.6V-0.7V\$ (due to presence of protection resistor R1), which means that the Base current will be a bit higher than \$1mA\$.
Knowing that the motor is \$12V, 100mA\$, and that the transistor is capable of handling \$100mA\$ Collector-to-Emitter current, we can deduce that:
$$I_C = I_{Motor} \approx 100mA$$
Given that we know (from motor's specs) that the motor will consume \$100 mA\$ at \$12V\$, the voltage on the motor:
$$V_{Motor} \approx 12V$$
Which leads to:
$$V_C \approx 0V$$
But this means that Collector-to-Base junction is forward biased which implies that the transistor in saturation.
The above analysis is quite general for this configuration (full voltage rated motor switched by matching BJT), therefore, in majority of circuits like this one, the transistor will be in saturation.
Experienced engineers perform the analysis above at a glance, knowing that the transistor in saturation a second after they see the schematics.
In a nutshell, bipolar junction transistors work because of the physical geometry of the two junctions. The base layer is very thin, and the charge carriers that are flowing from the emitter to the base do not recombine right away — most of them pass right through the base altogether and enter the depletion region of the reverse-biased base-collector junction. Once this happens, the strong field in this region quickly sweeps them the rest of the way to the collector terminal, becoming the collector current.
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Best Answer
**if you want to estimate Vce with a part number, it is helpful to use Vce(sat) @ Ic=10*Ib then go from there.
The hFE declines to about 10% of the max linear value at the rated Vce(sat). The Vce(sat)/Ic = Rce is a computed term that is often useful to know for a switch characteristic when choosing a device or a collector load. Rce decreases with increasing base current and collector current when saturated.
The actual Vbe depends on these values and could be 0.6V if Ic is 1mA and Vce=0.06V.
You can also see Rce (typ) @ 25'C at full saturation with Ic/Ib=10 for this part;
Rce=60 Ohms @ Ic=1 mA Vce= 60mV.
Rce= 5 Ohms @ Ic=10 mA Vce= 50mV.
Rce= 1.3 Ohms @ 100mA Vce = 130mV.
Rce = 0.8 Ohms @ 500 mA Vce= 400mV.
This trend is common but the ohmic scale reduces for higher power switching transistors.
2N4401 example General Purpose Small signal TO92
When operating in the linear region the collector is a high impedance current sink.