Contents
How do we calculate the output voltage of differential amplifier with active load?
I want to compute the value of VO. I would have guessed that VO = VCC – VEB. Asuming VCC = 5V and VEB = 0.6 V, that would render VO = 4.4 V, as in the left branch of the circuit.
What is the voltage gain of differential amplifier with active load?
1. What is the voltage gain of a differential amplifier with active load? Explanation: If all the resistor values are of the equal value, then the circuit will have a voltage gain equal to exactly one or unity. Hence it is also called as a Unity Gain Differential amplifier with active load.
What is current mirror in Mosfet?
A current mirror is a circuit designed to copy a current through one active device by controlling the current in another active device of a circuit, keeping the output current constant regardless of loading. The current being “copied” can be, and sometimes is, a varying signal current.
What are advantages of actively loaded MOSFET differential pair?
In the two previous articles, we introduced the actively loaded MOSFET differential pair and discussed two prominent advantages of this configuration—namely, improved biasing (compared to the use of drain resistors) and differential-to-single-ended conversion without loss of gain.
What is the gain of an active load MOSFET?
As discussed in the first section of The MOSFET Differential Pair with Active Load, the magnitude of this amplifier’s gain is the MOSFET’s transconductance multiplied by the drain resistance: Now let’s incorporate the finite output resistance:
What’s the differential gain of a short channel MOSFET?
Keep in mind, though, that state-of-the-art short-channel MOSFETs will have higher lambda (and thus lower gain). Now that we have explored small-signal output resistance, we are ready to analyze the differential gain of our actively loaded MOSFET differential pair.
What is the Overdrive voltage of a MOSFET amplifier?
Let’s start by looking at a passively loaded, non-differential MOSFET amplifier: The current source biases the FET so that it can operate in the saturation region. The overdrive voltage (V OV )—i.e., the gate-to-source voltage (V GS) minus the threshold voltage (V TH )—will be whatever value corresponds to a drain current of I BIAS.