Q-point of a diode circuit
ECE
John Stahl
Western Michigan University
Copyright: John Stahl. For educational use.
Do not distribute or reproduce without
permission
Solve for the Q-point of the
circuit shown if the diode has a
current of 1mA at 0.6v. Assume
nVT = 25.8mV.
Take a few minutes and try to
approach the problem. Start
with nodal analysis.
Copyright: John Stahl. For educational use.
Do not distribute or reproduce without
permission
Solution
Start by tracing the currents in the circuit at
the node. Write the current equation for
the node.
Iin
𝐼𝑖𝑛 = 𝐼𝐷 + 𝐼2
The diode’s current behavior follows:
𝐼𝑄
𝑉𝑄 = 𝑉1 + 𝑛𝑉𝑇 ∙ ln⁡
𝐼1
Where V1 and I1 are the priming points
given in the problem description.
Copyright: John Stahl. For educational use.
Do not distribute or reproduce without
permission
ID
I2
For this example we will use the method which allows us to solve for
the Q-point in one equation using a numerical solver. We need to set
the current equation in terms of VQ and IQ.
𝐼𝑖𝑛 = 𝐼𝐷 + 𝐼2
𝐼𝐷 = 𝐼𝑖𝑛 − 𝐼2
𝐼𝐷 =
For a resistor V=IR.
Replace resistor currents.
5 − 𝑉𝐷
𝑉𝐷
−
750Ω 1.5𝑘Ω
We are looking for the Q-point current in the Diode. ID and VD are the Q-point values
for the diode. There is room to reduce the above fraction.
5 − 𝑉𝑄
𝑉𝑄
𝐼𝑄 =
−
750Ω 1.5𝑘Ω
Copyright: John Stahl. For educational use.
Do not distribute or reproduce without
permission
Place the current equation into the diode voltage equation. Where V1
= 0.6v, I1 = 1mA, and nVT = 25.8mV
5 − 𝑉𝑄
𝑉𝑄
−
750Ω
1.5𝑘Ω
𝑉𝑄 = 0.6𝑣 + 25.8𝑚𝑉 ∙ ln⁡
1𝑚𝐴
This method works best with a numerical solver.
𝑉𝑄 = 0.6434𝑉
5 − 0.6434 0.6434
𝐼𝑄 =
−
750Ω
1.5𝑘Ω
𝐼𝑄 = 5.380𝑚𝐴
The Q-point of the diode is:
𝑉𝑄 = 0.6434𝑉
𝐼𝑄 = 5.380𝑚𝐴
Copyright: John Stahl. For educational use.
Do not distribute or reproduce without
permission