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Here is my task:

Find input voltage Vul and base voltage of T1, so that T1 enters full reverse active mode. Explain. Voltage base-emitter of BJT (when it conducts) is 0.7V, voltage base-emitter in saturation is 0.8V and voltage collector-emitter in saturation equals 0.1V.

enter image description here

For BJT to work in forward active mode, EBJ should be forward biased (Vbe = 0.7V) and CBJ should be reverse biased (Vbc < 0.4V).

For BJT to work in inverse active mode, EBJ should be reverse biased and CBJ should be forward biased. Does it mean that in this case Vbe = -0.7V and Vbc > 0.4V, for BJT on left side on image below? Does BJT in this mode behave like one on right side on same image? I know that in this mode, we have small gain Beta_reverse.

enter image description here

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    \$\begingroup\$ The inverse gain is not always small. Some transistors are made to have similar forward and reverse beta. Your diagram on the right is confusing - the symbol emitter is marked C... at the risk of sounding Lincolnesque, calling it a C does not make it a C. \$\endgroup\$ – Spehro Pefhany Jun 3 '17 at 19:45
  • \$\begingroup\$ I didn't see I made mistake. I will fix it. \$\endgroup\$ – Zdenko Paldum Jun 3 '17 at 19:48
  • \$\begingroup\$ I'm currently learning TTL implementations of logic gates and Beta_reverse is very small there... \$\endgroup\$ – Zdenko Paldum Jun 3 '17 at 19:50
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    \$\begingroup\$ @ZdenkoPaldum Then you might want to read the following link from here, which deals with the TTL NAND gate: electronics.stackexchange.com/questions/304642/… \$\endgroup\$ – jonk Jun 3 '17 at 19:51
  • \$\begingroup\$ You redrew the same schematic on the right as the left... On the Left E should be positive and current flows form E to C (for Inv Active) \$\endgroup\$ – sstobbe Jun 3 '17 at 19:53

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