In the last post we worked on some implementation details to consolidate the bill of material while improving the power rating of the decade resistor. However, we didn’t actually analyze the design regarding power dissipation. So let’s get some definitions out of the way (goto Conclusion for tldr):

nnDecade nn providing resistance values 0Ω0\Omega, 110nΩ1 \cdot 10^{n}\Omega to 210nΩ2 \cdot 10^{n}\Omega
RnR_{n}Total resistance of decade nn as set by the switches
R0nR_{0n}Resistance of an individual resistor of decade nn (in the picture above: R00=2ΩR_{00} = 2 \Omega)
P0nP_{0n}Power rating of any individual resistor of decade nn.
I0nI_{0n}Current through resistor of decade nn @ power rating
Imax,nI_{max,n}Maximum allowable current for decade nn
Pmax,nP_{max,n}Maximum allowable power dissipation decade nn
Umax,nU_{max,n}Maximum allowable voltage for decade nn
Imax,totI_{max,tot}Maximum allowable current for the programmable resistor (all decades)
Pmax,totP_{max,tot}Maximum allowable power dissipation of the programmable resistor (all decades)
Umax,totU_{max,tot}Maximum allowable voltage for the programmable resistor (all decades)

The definitions imply that we use identical resistors throughout a decade. For this post we also assume that the power rating of all resistors are the same. In reality this is not the case. That’s one of the reasons for part 2 of this post…

Exemplary calculation for the first decade

First decade n=0
First decade n=0

Let Imax,0I_{max,0} flow into the first decade (n=0n=0, R00=2ΩR_{00}=2\Omega) with all switches open except for SW12. The resulting resistance is R0=5Ω=52R00R_{0}=5 \Omega = \frac{5}{2} R_{00}. Then:

  • a current of Imax,0/2I_{max,0} / 2 will flow through the resistors with the reference designators R1, R2, R7, R9, R8, R10 (see schematic above)
  • a current of Imax,0I_{max,0} will flow through the resistor with the reference designator R13

Hence Imax,0I_{max,0} is defined by the current that flows through R13 at its power rating:

Imax,0=I00=P00R00I_{max,0} = I_{00} = \sqrt{\frac{P_{00}}{R_{00}}}

The maximum total power dissipation of the decade is the sum of the power dissipation of the individual resistors (at their respective current). Accordingly:

Pmax,0=2R00(Imax,02)2+4R00(Imax,02)2+R00Imax,02P_{max,0} = 2\cdot R_{00} \cdot \left(\frac{I_{max,0}}{2}\right)^2 + 4\cdot R_{00} \cdot \left(\frac{I_{max,0}}{2}\right)^2 + R_{00} \cdot {I_{max,0}}^2=52R00Imax,02=52P00=\frac{5}{2} R_{00} \cdot {I_{max,0}}^2=\frac{5}{2} P_{00}

The maximum allowable voltage is:

Umax,0=R0Imax,0=52R00I00=52P00R00U_{max,0} = R_0 \cdot I_{max,0} = \frac{5}{2} R_{00} \cdot I_{00}=\frac{5}{2} \sqrt{P_{00}\cdot R_{00}}

Table for decade nn

To generalize the example shown above, here’s a table showing the results for all resistance values in tabular form.

R0R0n\frac{R_0}{R_{0n}}Imax,nI_{max,n}Pmax,nP_{max,n}Umax,nU_{max,n}
00limited by relayslimited by relayslimited by relays
112I0n2 I_{0n}2P0n2 P_{0n}P0nR0n\sqrt{P_{0n} \cdot R_{0n}}
222I0n2 I_{0n}4P0n4 P_{0n}2P0nR0n2 \sqrt{P_{0n} \cdot R_{0n}}
332I0n2 I_{0n}6P0n6 P_{0n}3P0nR0n3 \sqrt{P_{0n} \cdot R_{0n}}
44I0nI_{0n}2P0n2 P_{0n}2P0nR0n2 \sqrt{P_{0n} \cdot R_{0n}}
55I0nI_{0n}52P0n\frac{5}{2} P_{0n}52P0nR0n\frac{5}{2} \sqrt{P_{0n} \cdot R_{0n}}
66I0nI_{0n}3P0n3 P_{0n}3P0nR0n3 \sqrt{P_{0n} \cdot R_{0n}}
77I0nI_{0n}72P0n\frac{7}{2} P_{0n}72P0nR0n\frac{7}{2} \sqrt{P_{0n} \cdot R_{0n}}
88I0nI_{0n}4P0n4 P_{0n}4P0nR0n4 \sqrt{P_{0n} \cdot R_{0n}}
99I0nI_{0n}92P0n\frac{9}{2} P_{0n}92P0nR0n\frac{9}{2} \sqrt{P_{0n} \cdot R_{0n}}

Considerations for multi-decade resistors

If the power rating of the individual resistors in the programmable resistor are identical, the maximum current Imax,totI_ {max,tot} is limited by the decade with the highest resistance. To prove this, we show that the largest maximum permissible current of a higher decade mm is still less than the smallest maximum permissible current of a lower decade n<mn < m.

1. Determine the smallest maximum permissible current for the lower decade nn

With the total resistance of a decade of Rn=lnR0nR_n = l_n \cdot R_{0n}, the smallest maximum current for decade nn results for ln>3l_n > 3 (see table for sigle decade). Therefore:

min(Imax,n)=P0nR0nmin(I_{max,n}) = \sqrt{\frac{P_{0n}}{R_{0n}}}

2. Determine the largest maximum permissible current for the higher decade mm

With the total resistance of a decade of Rm=lmR0mR_m = l_m \cdot R_{0m}, the largest maximum current for decade mm results for lm{1,2,3}l_m \in \lbrace{1, 2, 3\rbrace} (see table for sigle decade). Therefore:

max(Imax,m)=2P0mR0mmax(I_{max,m}) = 2 \cdot \sqrt{\frac{P_{0m}}{R_{0m}}}

3. Utilize the assumptions

If mm is the next-higher decade (i. e. m=n+1m = n + 1), the resistor value for decade mm is ten times larger than for decade nn

R0m=10R0nR_{0m} = 10 \cdot R_{0n}

Also, we assumed that the power ratings of all resistors all equal:

P0m=P0nP_{0m} = P_{0n}

4. Bringing it all together

max(Imax,m)=2P0mR0m=2P0m10R0n=210P0nR0nmax(I_{max,m}) = 2 \cdot \sqrt{\frac{P_{0m}}{R_{0m}}} = 2 \cdot \sqrt{{\frac{P_{0m}}{10\cdot R_{0n}}}} = \frac{2}{\sqrt{10}} \cdot \sqrt{\frac{P_{0n}}{R_{0n}}}0.63min(Imax,n)\approx 0.63 \cdot min(I_{max,n})max(Imax,m)<min(Imax,n)\Longrightarrow max(I_{max,m}) < min(I_{max,n})

Therefore, under these assumptions the power rating of the programmable resistor is limited by the highest non-zero decade.

Conclusion

The main takeaways are:

  • The maximum power rating of a decade can vary depending on the selected resistance value. Therefore, it makes sense to calculate and display the applicable power rating for the resistance value set by the user
  • Using the topology and applying the assumption (in particular: same power rating for all resistors) shown above, the power rating is at least two times the power rating of the individual resistors
  • Then the power rating of the programmable resistor is limited by the highest non-zero decade

The next post in this series will focus on the actual component selection and the implications on the subject of the power rating of the programmable decade resistor.