Calculating the current through a single-electron transistor

When a bias voltage great enough to overcome the Coulomb blockade is applied, current will flow through a single-electron transistor as electrons tunnel on and off the island. For low bias voltages and temperatures, electrons tunnel onto the island from the source electrode then off the island to the drain electrode one at a time. This occurs in any of the diamonds in Fig. 3 labeled by two charge states n, n+1. The greatest modulation of the current can be achieved when a single-electron transistor is biased in this regime so this is the most important case to consider for applications.

When a single-electron transistor is biased in the 0,1 diamond at low temperature, all charge states n will decay by tunneling until either the charge state n = 0 or n = 1 is achieved. Charge state n = 0 decays to charge state n = 1 as an electron tunnels left through junction 2. This occurs at a tunnel rate Γ_2L(0) = -ΔE_2L(0)/(e²R₂). Here R₂ is the tunnel resistance of tunnel junction 2. After this tunnel event, the transistor is in charge state n =1 and it decays back to charge state zero at the rate Γ_1L(1) = -ΔE_1L(1)/(e²R₁). In steady state, the probability P(0) that charge state n = 0 times the rate that it tunnels 0→1 must be equal to the probability P(1) that it is charge state n = 1 times the rate that it tunnels 1→0.

Γ_2L(0)P(0) = Γ_1L(1)P(1)

This equation simply states that the average current through the two tunnel junctions has to be the same. By combining this equation with the condition, P(0) + P(1) = 1, the probabilities can be calculated.
P(0) = Γ_1L(1)
Γ_1L(1) + Γ_2L(0)

P(1) = Γ_2L(0)
Γ_1L(1) + Γ_2L(0)

The current is e times the probability that the transistor is in charge state n = 0 times the rate it tunnels to charge state n = 1. Equivalently the current is e times the probability that the transistor is in charge state n = 1 times the rate it tunnels to charge state n = 0.

I =	eΓ_1L(1)Γ_2L(0)
	Γ_1L(1) + Γ_2L(0)

This can be written as,

I =	ab	,
	C_Σ(R₂a + R₁b)

where a = C_ΣV₁ + ne - Q₀ - C₁V₁ - C₂V₂ - C_g1V_g1 - C_g2V_g2 + e/2 and b = -C_ΣV₂ - ne + Q₀ + C₁V₁ + C₂V₂ + C_g1V_g1 + C_g2V_g2 - e/2. This equation is valid at low temperature when for positive bias voltages for 0 < a < e and 0 < b < e. For negative bias voltages, the equation is valid for 0 > a > -e and 0 > b > -e. The light blue regions in the following figure show where this expression for the current is valid.

The current in the light blue diamonds is zero along the boundry of the gray Coulomb blockade regions where a = 0 or b = 0. The maximum current in this diamond occurs at the point a = e, b = e; where the current is I = e/(C_Σ(R₁ + R₂)).

The voltage gain at constant current (dV_b/dV_g1) and the transconductance (dI/dV_g1) can be determined by differentiating the expression for the current. The maximum voltage gain is -C_g1/C₁ if junction 2 is grounded and -C_g1/C₂ if junction 1 is grounded. The maximum transconductance is achieved near the boundaries of the Coulomb blockade. At a = 0, dI/dV_g1 = -C_g/(R₁C_Σ) and at b = 0, dI/dV_g1 = C_g/(R₂C_Σ). The individual resistances of the two junctions in a SET are typically determined by measuring the transconductance.

The current can be calculated as a function of the gate voltage using the following form.

As more charge states are included, the formula for the current gets more complicated. The formula for the current at low temperature in the diamonds labeled (n-1,n,n+1) is,

I =	R₁(a+(n-1)e)(b-ne)(b-(n-1)e) + R₂(a+(n-1)e)(a+ne)(b-ne)	.
	C_Σ(R₂²(a+(n-1)e)(a+ne) + R₁R₂(a+(n-1)e)(b-ne) + R₁²(b-ne)(b-(n-1)e))