Question

In the circuit below, the voltage $V_L$ is ______________ V (rounded off to two decimal places).} 

Hint:

In ideal current mirrors, the output current scales directly with the W/L ratio. Determine each mirrored branch current, apply KCL at the node, then use Ohm’s law across the load resistor to get $V_L$.

Answer 1

Assumptions: All MOSFETs are matched, operate in saturation, and channel-length modulation is neglected (ideal current mirrors).
Step 1: Reference PMOS mirror.
The left PMOS (W/L$=1$) is diode-connected and forced to carry $1\,\text{mA}$ by the current sink. A PMOS with W/L$=10$ sharing this $V_{SG}$ therefore sources \[ I_{P,\text{set}} = \frac{10}{1}. 1\,\text{mA} = 10\,\text{mA} \] into the diode-connected NMOS (W/L$=10$), setting $V_{GS,n}$ for $10\,\text{mA}$.
Step 2: NMOS mirror to the load node.
An NMOS with W/L$=5$ using the same $V_{GS,n}$ sinks \[ I_N = \frac{5}{10}. 10\,\text{mA} = 5\,\text{mA} \] from node $V_L$ to ground.
Step 3: PMOS mirror to the load node.
A PMOS with W/L$=7$ sharing the $V_{SG}$ of the $1\,\text{mA}$ reference sources \[ I_P = \frac{7}{1}. 1\,\text{mA} = 7\,\text{mA} \] from $V_{DD}$ into node $V_L$.
Step 4: Node current and $V_L$.
Net current through the $1\,\text{k}\Omega$ resistor (to ground) is \[ I_R = I_P - I_N = 7\,\text{mA} - 5\,\text{mA} = 2\,\text{mA}. \] Hence \[ V_L = I_R . 1\,\text{k}\Omega = 2\,\text{mA} \times 1000\,\Omega = 2.00\,\text{V}. \] \[ \boxed{V_L = 2.00\ \text{V}} \]