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Drawing #1

At this first picture i have a NPN transistor with Beta=75, 2N222A transistor, both Inputs are 5V. it was very simple to calculate the current that will flow through the LED. as it is simply I1 * Beta = I2, and I3 = I1+I2.

enter image description here

But if i replace it with a resistor of value 5k for example the I1*B = I2 relationship is not valid anymore.

My question is, I1*Beta = Maximum I2 current, would that be a better explanation of a relationship or i don't understand something ?

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1 Answer 1

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The problem here is that your transistor is no longer in forward-active mode, but in saturation. The equation \$I_C = β·I_B\$ is only valid in forward-active mode.

When the transistor is in saturation mode, the equation to use becomes \$V_{CE} = V_{CE,sat}\$, and the transistor acts like (to first-order approximation) a voltage source. If all you care about is \$I_C\$, you can ignore the base entirely since you don't have anything in series with the emitter.

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  • \$\begingroup\$ so how would i calculate the Ic current in the example with resistor ? \$\endgroup\$ Oct 30, 2018 at 13:07
  • \$\begingroup\$ @AntonStafeyev See my edit. \$\endgroup\$
    – Hearth
    Oct 30, 2018 at 13:11
  • \$\begingroup\$ As mentioned, that transistor is in saturation mode. That means that the Vce of the transistor is small, if we assume that it is 0.3 V (in the datasheet of the 2n2222 Vce = 0.3 V when Ic = 150 mA), what will Ic be? Think of how much voltage drops across the 5k resistor. \$\endgroup\$ Oct 30, 2018 at 13:12

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