# Why do we use one bias (VCC) for transistor amplifier circuits and 2 supplies for opamp circuits? [closed]

Opamps are said to have 2 supplies, so that the output can rail between these two points, then why do transistors have only one bias point.

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• power rail and bias point are not the same. Also, there is plenty of transistor circuits out there that swing between a positive and negative rail, not just above ground. – Linkyyy Apr 23 at 9:23

A supply rail and a biasing point are different things.

Some opamp circuits use a negative and a positive supply rail because sometimes that makes the circuit simpler.

Here are four basic opamp circuit, the inverting and non-inverting amplifier for both single and dual supply configurations:

Note how extra resistors and capacitors are needed in the "Single supply" circuits. The resistors are needed to create a 6 V DC voltage (to bias the opamp properly) and the capacitors are needed to block that same 6 V from reaching the input and output.

Using a negative and positive supply rails can also be done with transistor circuits. Many audio amplifiers do this, for example, this circuit contains no opamps but uses a +40 V and - 40 V supply: Source

Basically this circuit is a "power opamp" with feedback.

A biasing point is the current (or voltage) at which we bias a transistor. It is an operating point, for example we make 10 mA flow through a transistor and allow the signal to make that current vary between 5 mA and 15 mA.

This can be done with a single (usually positive) supply but also with a negative and positive supply. Which one is used is irrelevant to the biasing of the transistor.

Because an opamp is DC coupled and a common emitter amplifier is AC coupled. For the transistors to keep working the collector-emitter voltage must be positive. So the output cannot swing below the lower voltage rail.

The common emitter amplifier filters out the DC voltage. The C1 capacitor together with the R1 and R2 voltage divider network work like an adder. R1 and R2 specify the bias voltage $$\V_\mathrm{bias}\$$. Hence $$V_\mathrm{base} = V_\mathrm{bias} + V_\mathrm{in}.$$ This way you can have a lower input voltage than the lower rail and keep the transistor operational.

Image taken from here.

• Could you please provide a link or citation to the original source of your graphic? – Elliot Alderson Apr 23 at 11:47
• @ElliotAlderson Sure. – user110971 Apr 23 at 12:00