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The arithmetic used to define Divider-Mox resistors can be confusing to those who do not work with them on a routine basis. Below the form are some helpful notes, if necessary.
The first thing to understand is that there are two resistors on a Divider-Mox device. The shorthand used refers to one resistor R1 and the other resistor as R2. R1 will have one resistance value and R2 will have its own value. When we add the two resistance values together, we have the total value of the device (RT). So, R1+R2=RT.
The overwhelming application for Divider-Mox resistors is a voltage divider circuit. The effect of the resistors is to lower a high voltage potential to a much lower level. The amount of this change is proportional to the ratio between the total resistance value of the divider (RT) and R2. The following example should clarify this explanation.
Suppose that the need exists to reduce the high voltage in a circuit by a factor of 1,000. This means that the voltage out of the circuit would equal the voltage into the circuit, divided by 1,000. It does not mean to lower the voltage 1,000 volts. If the circuit were operating at 10,000 volts, we would be looking for 10 volts. This defines the need for the ratio to be 1,000:1.
Other considerations define how much "load" the resistor can have on the circuit. Let us assume that it is determined that the total resistance value of the device is to be 500M ohm. This becomes the total resistance value (RT). We know RT, and we know the ratio. The steps that are taken to determine the values of R1 and R2 are as follows. R2 is found by dividing the total resistance value by ratio. In our example this is dividing 500M ohm by 1,000. This shows the value of R2 to be 500K ohm. We know from before the RT=R1+R2. We know RT=500M ohm and R2=500K ohm. Therefore, R1=500M ohm minus 500K ohm or 499.5M ohm.
If the tolerance on the device were defined to be 1%, the complete description of the values would be:
RT = 500M ohm ±1%
R1 = 499.5M ohm ±1%
R2 =500K ohm ±1%
Ratio =RT/R2 = 1,000:1 ±1%