To be honest, I don't see any problems, though I have read the question twice now.

I don't even see any problems soldering the RTC, as an SO8 package really isn't that small, even for a hand made PCB. What's your PCB production method?

To maintain good accuracy, you might consider keeping the circuit at a controlled temperature. You could use a Peltier element plus a thermistor to regulate the temperature.

There are such things as temperature compensated oscillators (TCXO) and Oven Controlled Oscillators (OCXO), but these don't seem to be available at 32.768kHz.

I'm going to have a stab at some maths :)

The DC resistance of a conductor - any conductor - is calculated as:

\$R_{DC} = \frac{{\rho}l}{A}\$

Where \$\rho\$ is the resistivity of the conductor in \$\Omega/m\$, \$l\$ is the length in meters, and \$A\$ is the cross-sectional area in m².

The thickness of 1oz copper is \$0.000034798m\$. Say you have a 3mm (or 0.003m) wide trace. The cross-sectional area is (approximately, assuming a perfectly rectilinear cross-section) \$0.000034798 × 0.003 = 0.000000104m^2\$. Resistivity of copper is \$1.68×10^{−8}\$ at 20C, and your trace is 100mm long (0.1m).

\$R_{DC} = \frac{1.68×10^{−8} × 0.1}{0.000000104} = 0.016153846\Omega\$ at 20C.

Ok, now for the tricky bit. The temperature co-efficient (\$\alpha\$) for copper is 0.003862.

\$R(T) = R(T_o)(1+\alpha{\Delta}T)\$

So for a temperature of 30C we have a \${\Delta}T\$ of 10C, or 10K (30 - 20 = 10, K = C + 272.15).

So \$R(30) = R(20)(1+0.003862×10) = 0.016153846×1.03862 = 0.016777708\Omega\$

So now solve Ohm's Law for voltage. Say you have 100mA flowing through the trace. That's \$V=RI\$, so \$0.016777708×0.1 = 0.001677771\$ or \$1.678mV\$ dropped across the trace at 30C.

Who says you need online calculators?

(Now, it's been about 20 years since I did this kind of thing at college, so I may be completely wrong ;) )

## Best Answer

Try both directions.

Very likely only one of both values is member of the particular E-series.

Sometimes the tolerance ring is somehwhat thicker than the others or its distance to the end of the resistor is smaller, but AFAIK this is not reliable.

If in doubt, measure.