Stk412 750 Datasheet 16.pdf
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Stk412 750 Datasheet 16.pdf
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Si alguien más necesita, pude resolver el problema gracias a la ayuda de los colegas que me ayudaron a ver el problema, no estoy seguro de lo que pensó pero creo que el problema era que no tenÃa el ide para arduino en el equipo y también no tenÃa instalado el Ãndice de arduino.
gracias por la colaboración y saludos
Biodegradation of a novel herbicide (Bcls) in soils: kinetic, degradome and related microbial mechanism analyses.
Bcls [4-amino-2-[4-methoxy-3-(trifluoromethyl)phenyl]-6-trifluoromethylpyrimidin-5-yl-(6-methoxybenzoxazol-2-yl)methanone], a novel herbicide, was applied to a grassland paddy soil under a tillage system to examine its degradation and microbial transformation in soil. Incubation with un-sterilized soils under both aerobic and anaerobic conditions showed that Bcls degraded readily in soils, with a half-life (t 1/2) of 8.9-23.9 days, and that the degradation rates increased with time. The mineralization of Bcls was estimated to be in the range of 11.2-22.2% during 180 days of incubation in soil. The degradation rates of Bcls spiked to soils under no-till and mid-tillage were 1.6-2.2 times faster than those under plow-tillage. At relatively high soil concentrations (>1 mg/kg), Bcls was more rapidly degraded in soil than the soil POM (particulate organic matter) adsorbed portion, but it was more slowly than the soil organic carbon (C) adsorbed portion. In the study of the related microbial mechanism, Bcls-degrading denitrifying bacteria were enriched and identified as Paracoccus sp. strain ILM 2-3, which degrades Bcls by reductive dehalogenation pathway. Another
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Here I have written my problem to the best of my knowledge, can some one please tell me where am I wrong. Thank you for your time A: Your understanding is close, but incorrect. If the open-circuit voltage of D1 is Vout, you are observing the correct thing. If D1 doesn’t turn on (because it only conducts when it’s configured for high-side switching), R2+R1 will conduct up to a maximum of Vout = 2 x Iout + (Vout – Vin). If Vin = 5 V, then the maximum Vout that R2+R1 will conduct is 5 V, and the most positive current that R1+R2 can conduct is 2 x 0.5 = 1 ampere. That leaves the possibility of \$I_R = 2 \frac{Vin-Vout}{R1+R2} \$ amperes flowing through D2, so D2 conducts up to at least (2 / 1) = 2 V. The real issue is that the D1+D2 junction cannot conduct enough current to light up the LED when you have a 2 V input. That’s because the D1+D2 junction will always conduct the maximum amount of current needed to drive the LED. The LED will be conducting the rest. You can get \$V_{LED}=I_C \times R_C\$ where \$I_C\$ is the collector current of the transistor. So, with \$I_C = I_{D1+D2} \approx 3.0 – 3.5 mA\$ and a 0.01 V LED, you’ll be getting 0.5 V or less at the LED, which is barely enough for it to barely glow. Furthermore, while the LED can turn on with \$I_{D1+D2} < I_{R2} \le I_{D2}\$ (so \$V_{LED} = I_{D1+D2} \times R_C \le \frac{V_{LED}}{R_C} \le \frac{V_{LED}}{2R_2} = \frac{1}{2} \frac{V_{LED}}{R_2} < I_{R2} \times R_C\$), it can't turn on
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