PCA9615 Termination

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PCA9615 Termination

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henryy
Contributor IV

According to the datasheet, the output impedance is 100 Ohm differential. On page 9, it mentioned 'R2 is 120, R1 is 600, The parallel combination yields a termination of 100'. How this was calculated? Two R1s should be in series then parallel with R2. Then R1 should be 300 Ohm to yield a combination 100 Ohm. Did I miss anything? May I use R2=100Ohm and R1=10k to reduce the current?

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guoweisun
NXP TechSupport
NXP TechSupport

Hi

No you can't use R1=10K. In the calculation you have to keep the voltage drop of R2=200mV to keep system working normally.

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henryy
Contributor IV

Thanks for the reply. You are right to keep the voltage drop of R2 = 200mV for working normally. But the voltage over R2 depends on the differential output, not the R1. the Two R1s only provide bias.

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guoweisun
NXP TechSupport
NXP TechSupport

Hi

Same current flow into R2/R1s/R1s,there is voltage drop in R2 200mV,so also there are voltage drop in R1s.

If R1s value is large the voltage drop of R2 can't reach 200mV.

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