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I understand that a transistor will amplify current by a certain gain factor if it is below saturation.

But what if there is a resistor that reduces current?

So let's say there is 1mA between base and emitter and a gain of 100x that means there will be 100mA between collector and emitter?

But what if there is a 5v supply voltage and a 1000ohm resistor between supply and collector, it will be impossible for this much current to flow.

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    \$\begingroup\$ That mode of operation is : saturation. \$\endgroup\$ – analogsystemsrf May 5 at 23:23
  • \$\begingroup\$ You typically duplicate the setup a few million times, invent a whole new branch of using electronics and port Doom to the result. \$\endgroup\$ – DonFusili May 6 at 12:32
  • \$\begingroup\$ For bipolar transistors (BJT) it is called saturation mode, but just be aware that for field-effect transistors (FET) the saturation mode means almost exactly the opposite. \$\endgroup\$ – jpa May 6 at 13:06
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I understand that a transistor will amplify current by a certain gain factor if it is below saturation.

Good.

But what if there is a resistor that reduces current?

Then the transistor will drive into saturation.

There are many times when this is useful - most notably in the NPN low-side switch. Here we want the transistor to behave like a switch so we inject a base current high enough to ensure that the transistor saturates so that the collector-emitter voltage, VCE is as low as possible. This eliminates variation in load current due to variations in the transistor gain and also ensures that power dissipation in the transistor is at a minimum.

enter image description here

Figure 1. A typical NPN low-side switch can be driven into saturation by making R1 low enough. Source: LEDnique by the author.

To ensure saturation it is normal to assume a much lower current gain than the hfe paramater might suggest. 10 to 20 is typical. See the linked article for a worked calculation.

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  • \$\begingroup\$ Will the b -> e current be reduced if the c -> e current is reduced? \$\endgroup\$ – user221241 May 6 at 9:25
  • \$\begingroup\$ No. The \$ I_{be} \$ current is determined by R1 and \$ V_{be} \$ of the transistor. It doesn't "know" about the collector current. \$\endgroup\$ – Transistor May 6 at 9:29
  • \$\begingroup\$ So the "beta" for the transistor only holds when R1 and R2 are within certain values? \$\endgroup\$ – user221241 May 6 at 9:34
  • \$\begingroup\$ See if the discussion on Physics Forums helps. \$\endgroup\$ – Transistor May 6 at 9:44
  • \$\begingroup\$ @user221241 β only applies when the transistor is in forward-active mode. The relevant characteristic for saturation mode is called the saturation voltage, Vce,sat. \$\endgroup\$ – Hearth May 6 at 20:09
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The purpose of biasing a transistor amplifier is to make sure that this problem does not occur. The resistor values are selected so that without any input signal (other than the dc bias) the output voltage will be about half way between ground and Vcc. That setup allows for the largest possible linear amplification.

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