Question

For the amplifier shown determine the value of the bias current (Ib) and input offset current (Io) respectively.

(a) Ib = IB1 + IB2

Answer 1

To determine the values of the bias current (Ib) and input offset current (Io) for the amplifier, let's break down the given expression and equations:

1. Bias Current (Ib):

For a differential amplifier, the bias current is the total current provided to the input transistors.

The equation for the bias current is:

Ib=IB1+IB2I_b = I_{B1} + I_{B2}Ib​=IB1​+IB2​

Where:

• IB1I_{B1}IB1​ is the bias current for transistor Q1.
• IB2I_{B2}IB2​ is the bias current for transistor Q2.

2. Input Offset Current (Io):

The input offset current is the difference between the bias currents of the two input transistors:

Io=∣IB1−IB2∣I_o = |I_{B1} - I_{B2}|Io​=∣IB1​−IB2​∣

Where:

• IB1I_{B1}IB1​ is the bias current for the first transistor (Q1).
• IB2I_{B2}IB2​ is the bias current for the second transistor (Q2).

Given:

• If you have specific values for IB1I_{B1}IB1​ and IB2I_{B2}IB2​, you can directly substitute them into the equations above to calculate IbI_bIb​ and IoI_oIo​.

Example:

If the bias currents are as follows:

• IB1=100 μAI_{B1} = 100 \, \mu AIB1​=100μA
• IB2=90 μAI_{B2} = 90 \, \mu AIB2​=90μA

Then:

1. The total bias current:

   Ib=IB1+IB2=100 μA+90 μA=190 μAI_b = I_{B1} + I_{B2} = 100 \, \mu A + 90 \, \mu A = 190 \, \mu AIb​=IB1​+IB2​=100μA+90μA=190μA

2. The input offset current:

   Io=∣IB1−IB2∣=∣100 μA−90 μA∣=10 μAI_o = |I_{B1} - I_{B2}| = |100 \, \mu A - 90 \, \mu A| = 10 \, \mu AIo​=∣IB1​−IB2​∣=∣100μA−90μA∣=10μA

Summary:

• Bias current (Ib) = IB1+IB2I_{B1} + I_{B2}IB1​+IB2​
• Input offset current (Io) = ∣IB1−IB2∣|I_{B1} - I_{B2}|∣IB1​−IB2​∣

Let me know if you have the specific values for IB1I_{B1}IB1​ and IB2I_{B2}IB2​ or if you need further assistance.