 Mismatch remaining after common RC gain cancellation

The mismatch remaining after common RC gain cancellation is the timing error caused by unequal gains in the amplitude and threshold paths after their common multiplicative part has canceled. It depends on their ratio, so improving a shared gain alone does not remove it.

For the zero-baseline RC threshold-crossing time, let $a=\theta /H$ with $0<a<1$. If the amplitude is multiplied by $G_H>0$ and the threshold by $G_{\theta }>0$, define the residual relative error by $G_{\theta }/G_H=1+ϵ$. The exact result from ratiometric cancellation of RC timing gain errors isHere $\tau$ is the time constant. As $a(1+ϵ)$ approaches one, the crossing is arbitrarily late; at or above one there is no finite rising crossing. An error budget must check this existence condition as well as a small timing residual.

For a new example, set $H=1\mathrm{V}$, $\theta =0.8\mathrm{V}$ and $\tau =1\mu \mathrm{s}$. A one-percent positive mismatch gives $\Delta t=-\log (0.96)\mu \mathrm{s}\approx 40.822\mathrm{ns}$; a one-percent negative mismatch gives $-\log (1.04)\mu \mathrm{s}\approx -39.221\mathrm{ns}$. The first-order estimate is $\Delta t\simeq \tau aϵ/(1-a)$, or $\pm 40\mathrm{ns}$ here. The unequal exact magnitudes are a consequence of the logarithm, not an asymmetric tolerance specification.

If $|ϵ|\le e<(1-a)/a$, monotonicity gives the exact worst absolute errorThus a $10\mathrm{ns}$ bound in this example requires $e\le (1-a)(1-e^{-0.01})/a\approx 0.00248754$, about $0.249\%$. This is a bound on the gain ratio, not on either gain separately.

For small fractional path errors ${\eta }_H,{\eta }_{\theta }$, $ϵ\simeq {\eta }_{\theta }-{\eta }_H$ andA perfectly shared component has zero contribution to that difference. This is the law of propagation of uncertainty applied to the two gains; it does not cancel a change in $\tau$, an additive threshold error or a nonzero unscaled baseline.^[ref]

A proposed check varies the common gain while holding its ratio fixed, then varies only one path. Recording the two electric potential differences and the crossing on an oscilloscope distinguishes the exact common-gain invariance from the surviving mismatch.