Recent content by shseo0315

Sorry, I still didn't get this. First, I have set Ic=20mA R3 = Vcc / Ic = 3.0V / 20mA = 150ohm I have set 20mA for Ic. Since B=200, I used Ic/Ib = B to get Ib. Ib = 0.1mA. then I used Ib = Vb /R2 = 1.4v/R2 = 0.1mA. Then, R2 = 14k Finally, Vcc * (R2 / R1+R2) = 3.0V * (14k/R1+14K) = 1.4V...

It is sort of working now. Actually the LED brights up fully at 3.7V and as you decrease power supply, LED light weakens. But I thought it should be turned off when the battery is full. As we decrease voltage, I thought the LED lights up. But why does LED act as some kind of indicater that uses...

So I have used R1 = 1.8k R2 = 4.7k and R1 = 56ohm When Vcc is 3.7v, Vb would be around 2.67V, which makes BJT to turn on. As Vcc decreases, Vb will reach 1.4V and BJT turns off and current only goes through RED LED and light it up. When Vb is around 1.4V Vcc is around 2V. The current through the...

That's understandable. Vb = Vcc * (R2/ R1+R2). For the BJT to turn on, you need at least 1.4v(Vdiode + Vbeon). To increase Vb, I will try to use some higher resistance for R2 around 4.7K with R1 fixed as 2k. Also, as you have mentioned I will lower the resistance of R3 to have enough...

Hi, Thanks for the replies. I appreciate it. So I understand what you all have said. So I tried to replace the green LED with a diode. That makes Vb to be aroudn 1.4v to turn on BJT as the reply says. I used R2=2k, R1=1K2, R3=150ohm. It still is not working. Since 3.3v cannot be changed...

For my project, the Lithium Ion Polymer Battery - 3.7v 2600mAh powers arduino, transmitter and a few other parts. My whole design works on 3.3v. I'm trying to build a circuit to indicate the status of the battery using LED. When the voltage of the battery is consumed below 3.0v, I want to...

using the property above, I get (1/12)ln((2x+3)(2x-3)) + c = t then, e^12t = (2x+3)(2x-3) + e^c e^12t - e^c = (2x+3)(2x-3) here how can I go further to have x equals to whatever. thanks a lot. it really helps.

Homework Statement dx/dt = 9-4x^2 , x(0) = 0 when I integrate, am I supposed to use the property below? int du / (a^2 - u^2) = 1/2a ln(u+a / u-a) + c or how do I integrate this by using partial fraction? tips anyone? Homework Equations The Attempt at a Solution

now I get int (x^2)(e^(1/x^3)) then? stuck here

Homework Statement int((x^2)(e^-3lnx)) anyone tips for this?Homework Equations The Attempt at a Solution

Homework Statement x dy/dx -y = 2x^2 y, y(1)=1 Homework Equations The Attempt at a Solution x dy/dx = y(2x^2 + 1) int 1/y dy = int (2x^2+1)/x dx int 1/y dy = int 2x dx + int 1/x dx ln y = x^2 + lnx + c y = exp(x^2 + c) + x and in this case initial condition doesn't work. please help. I...

Homework Statement I haven't done much math for years. just a little reminder would be very appreciated. when you take integral for (2x^2+1)/x dx I believe this can break down into integral 2x dx + integral 1/x dx What I wondered about was is this summation of those two as...

from 1/(x+1) +c = 1/(y+1) that is y+1+c = x+1 right y(x) = x+c this is what I get, but the answer is quite different which is y = (1+x)/[1+c(1+x)] -1

Thanks. But on the way, 1/(1+x^2)dx = 1/(1+y^2)dy if I take an intergral, I get -1/(x+1) + c = -1/(y+1) + c This is 1/(x+1) + c = 1/(y+1) right? The answer states that y = (1+x)/[1+c(1+x)] -1 I don't know how to get there.

Homework Statement (1+x)^2 dy/dx = (1+y)^2 Homework Equations The Attempt at a Solution The post I put up a while ago actually turns out to be the one above. So far I'm not getting the right answer, please help.