1
\$\begingroup\$

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

\$\endgroup\$
  • \$\begingroup\$ 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 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 10 '19 at 1:58
  • \$\begingroup\$ |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. \$\endgroup\$ – Spehro Pefhany Jun 10 '19 at 2:20
  • 2
    \$\begingroup\$ @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 \$\endgroup\$ – Mattman944 Jun 10 '19 at 2:23
  • \$\begingroup\$ 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.) \$\endgroup\$ – Jim Fischer Jun 10 '19 at 7:50
1
\$\begingroup\$

**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

enter image description here

When operating in the linear region the collector is a high impedance current sink.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.