# How can I prove that Vce saturation lies below .2V?

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

• if you want to estimate Vce with a part number, it is helpful to use Vce(sat) @ Ic=10*Ib then go from there if you know the actual ratio and actual load current as the bulk resistance causes a linear rise in Vce/Ic The actual Vbe depends on these values and could be 0.6V if Ic is 1mA and Vce=50mV – Tony Stewart Sunnyskyguy EE75 Jun 10 '19 at 1:58
• |Vce(sat)| is not necessarily less than 200mV. For example, for 2N4401 it will be >200mV typically if Ic >150mA for Ic/Ib = 10 and it will never get less than about 400mV at Ic = 500mA regardless of Ib. – Spehro Pefhany Jun 10 '19 at 2:20
• @Buzz bee - you are treating this like an algebra problem. BJTs are much more complicated than that. See the Wikipedia article: en.wikipedia.org/wiki/Bipolar_junction_transistor – Mattman944 Jun 10 '19 at 2:23
• This assumption is incorrect: ".7v+x-.8v+z where z is a positive constant". This expression incorrectly models the voltage and current relationships within an NPN BJT, and is the reason your analysis isn't working. (Unrelated comment: For numeric values less than 1, always put a leading zero to the left of the decimal point: CORRECT 0.7, INCORRECT, .7). (Unrelated comment: Please use punctuation marks like commas and periods in your written prose. Your message is very difficult to read and understand because it has no punctuation marks.) – Jim Fischer Jun 10 '19 at 7:50

**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.