Recent content by gazp1988

this is what i have. PD= (Vin-Vout) x (Iout) = (18 - 15) x 1 = 3w TJ = TA + PD (θHS + θCHS + θJC) TJ= 75 + 3 (θHS + 0.2 + 0.6) to work out the heat sink value PD = ΔT/(θJC + θHS) = 3 = 50/(0.6 + θHS) = θHS = 50/(0.6 + 3) = 13.89 C/W TJ= 75 + 3(13.89 + 0.2 + 0.6) = 119.07°C <125°C this shows...

i see what i have done, sorry. I = 2... Vpp = 2 / (2 x 50 x 0.047) = 0.426

i see i have made a mistake now. using the second formula above: Vpp= 4.5/(2 x 50 x 0.047)

i was assuming that 2A stated in the question was the load resistance or as quoted in the question resistive load

9 = dc output 2 = double the frequency 50 = frequency 0.047 = reservoir capacitor 2 = load resistance given in the question, is this information you wanted

thank you for your responses. yes berkeman, its suppose to be like that. i have used both above formula and gave me the same answer of 0.957 Vpp = 9/(2 x 50 x 0.047 x 2) = 0.957 Vpp = 4.5/(2 x 50 x 0.047) = 0.957

Thank you for your quick reply. Vpp=/ frac{1} {2FCRl} {Vc} This equation came from my notes that were attached to the assignment equation

Homework Statement A 9 V power supply delivers 2 A to a resistive load. The a.c. supply is 230 V, 50 Hz and a bridge rectifier is used in conjunction with a 0.047 farad reservoir capacitor. Homework Equations Vpp = 1/(2FCRL) * VC The Attempt at a Solution 1/(2*50*0.047*2) * 9 =0.957v after...

it seems that there is some initial output interference could it be caused because of output noise??

ive restructured it now

changing the voltage to +/-12 has no difference to the output?

i see what what i ahev done now the original question stated "Calculate the current I to give an output voltage of 0.1 V if Rf = 10 MΩ, R1 = 90 kΩ, R2 = 10 kΩ." I've taken 0.1v as the input voltage.

because that is the input we were given for the circuit