Hey guys I don't have a scanner so bear with me a second. I was given a circuit including a NPN transistor. The collected is connected to a resistor Rc and a Vcc = 5 volts above it. The base is connected to an AC signal Vs and DC source VBB in series with one another. There is a resistor Rb = 25k between the base and those sources. Finally the emitter is connected to ground. Given B = 120 and r(pi) = 5.4k Find Vbb and Rc such that Q-Point is center of load line. Vo is also equal to Vc

So here's what I did.

DC Analysis:

r(pi) = B/gm => gm = B/r(pi) = 22.22 mA/V

gm = Ic/Vt => Vt*gm = Ic = .578 ma

Ic = BIb = 4.82uA

-Vbb + RbIb +. 7 = 0 Vbb = .8205V



So where I get stuck is trying to solve for the load line Vc=Vo versus Ic

I know ic = gmVeb where Veb = (r(pi)/r(pi)*rb) * Vs

I also knew Vc = Vo = RcIc

So I tried solving using those and got a load line equation of Vo = constant*Vs. Not what I was looking for. Your help is appreciated!

 

Hey guys I don't have a scanner so bear with me a second. I was given a circuit including a NPN transistor. The collected is connected to a resistor Rc and a Vcc = 5 volts above it. The base is connected to an AC signal Vs and DC source VBB in series with one another. There is a resistor Rb = 25k between the base and those sources. Finally the emitter is connected to ground. Given B = 120 and r(pi) = 5.4k Find Vbb and Rc such that Q-Point is center of load line. Vo is also equal to Vc

So here's what I did.

DC Analysis:

r(pi) = B/gm => gm = B/r(pi) = 22.22 mA/V

gm = Ic/Vt => Vt*gm = Ic = .578 ma

Ic = BIb = 4.82uA

-Vbb + RbIb +. 7 = 0 Vbb = .8205V



So where I get stuck is trying to solve for the load line Vc=Vo versus Ic

I know ic = gmVeb where Veb = (r(pi)/r(pi)*rb) * Vs

I also knew Vc = Vo = RcIc

So I tried solving using those and got a load line equation of Vo = constant*Vs. Not what I was looking for. Your help is appreciated!

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Your math looks good to me, except where you calculated Ib and called it Ic, but you seem to know that. You found Vbb, what about Rc?

Since the emitter is at ground and Vcc s 5volts your output will swing between almost 5 volts peak to peak across a 2.5 volt Q point. So you want Vc at 2.5 volts Using your calculation for Ic=.578mA you can find Rc by Vc=5-.578mA*Rc=2.5

To draw your load line you need Is, that is the saturation current. This is found as Is= Vcc/Rc which looks like a little over 1mA

On the vertical axis put Ic, on the horizontal Vce. Draw a line from Is to Vce=5 volts. Plot your Q point at 2.5 volts and .578mA. In all this I used your numbers and I rounded off calculations. You do the math.

 

Thanks so much! Yea I got the same numbers. Have a wonderful day.

 

Ok, another fast question. I just tried calculating the voltage gain of the circuit. I did A = vo/vs = Rc*gm*r(pi) / Rb + r(pi) and got a gain of 17,070!

That seems ridiculously high....

 

I got an Rc of 4.3kohms. The gain, Av=-gm*Rc which should be about -95v/v Where - means signal inversion. By the way the slope of your Is-Vce is -1/Rc.

 

Yea I got 4.325 kOhms. However, my professor gave us the above equation for solving for voltage gain. I am confused as to why it doesnt work.

 

Wait. Since my Gm is in units of mA/V, do I have to divide my answer by 1000 to get it back to amps?

 

I learned that for a CE amp with totally bypassed Re via a capacitor, the gain is -gmRc. But by your professors equation the answer is -17v/v. Let me know how it comes out, please. I'm interested. I may just go ahead and breadboard this circuit and find out myself! At any rate, I've told you what I think I've been taught and amps I've designed seem to work as planned -- most of the time! Again, let me know what your professor says, please?