C-X4 · Two junctions around one loop

SQUID: Flux into Voltage

How can a magnetic flux change a superconducting circuit’s voltage? Flux changes the phase difference between the two junctions. Their supercurrents reinforce or cancel, changing the largest current the loop can carry at zero voltage. A fixed bias above that limit produces a voltage that varies periodically with flux.

Send current through a superconducting ring with two weak links. Sweep the flux and watch interference move the switching threshold and the voltage response.

Critical current $\text{Loading}$

DC voltage $\text{Loading}$

Applied flux $\text{Loading}$

Circuit state $\text{Loading}$

A small change, a different state

Change the physical setting

Applied flux

0.32

\Phi/\Phi_0

Bias current

1.7

I_b/I_0

Junction asymmetry

0.08

d

The default experiment is recorded automatically.

Model and scale

Two sinusoidal Josephson junctions have negligible loop inductance. Their critical currents can differ. A noiseless overdamped effective-junction model gives the dc voltage; capacitance, self-field screening, hysteresis and feedback electronics are omitted. Each mean junction critical current is ten microamperes and the effective resistance is ten ohms. The chip, flux marks and moving carriers are schematic; carrier motion illustrates current, not individual electron trajectories.

From the experiment

Observe, then compare

Cyan: recorded model sweep. Amber marker: present setting. Current in microamperes, voltage in microvolts.

Recorded flux sweep

Voltage response

Start with the essentials

Focus question: How can a magnetic flux change a superconducting circuit’s voltage?
One-sentence intuition: The phase difference determines how much the two junction supercurrents reinforce.

Core mathematical model

Critical current without loop inductance

I_c(\Phi)=2I_0\sqrt{\cos^2(\pi\Phi/\Phi_0)+d^2\sin^2(\pi\Phi/\Phi_0)}

The mean single-junction critical current sets the scale. The dimensionless imbalance is the difference divided by the sum of the two junction critical currents.

Flux quantum

\Phi_0=\frac{h}{2e}

One flux quantum advances the relative phase by one full turn.

Overdamped effective-junction voltage

\overline V=R\,\operatorname{sgn}(I_b)\sqrt{\max(0,I_b^2-I_c^2)}

This noiseless dc approximation neglects junction capacitance and loop self-inductance.

Common difficulties

Reading the model correctly

Typical misconceptionA flux quantum is a tiny magnetic particle moving through the hole.

Better mental modelIt is a unit of magnetic flux that sets the phase periodicity. The marks inside the ring indicate applied flux schematically.

Run the experiment

01
Cancel the two paths
Compare Constructive interference and Destructive interference.
What to observe: At half a flux quantum the critical current reaches its minimum.
02
Choose a voltage-sensitive bias
Select Flux-sensitive voltage, then move Applied flux.
What to observe: The fixed current crosses the critical-current envelope and generates a periodic voltage.
03
Unbalance the junctions
Select Unequal junctions and repeat the flux sweep.
What to observe: The cancellation minimum rises because the supercurrent amplitudes differ.