Why does this transistor amplifier fail if Rc is greater than 50 ohms?

Question

This simple circuit amplifies a 1 µV 50 ohm 7 MHz signal (e.g. RF off an antenna) quite nicely (31 dB):

However, I'm stumped as to why R2 must be so low (50 ohms), giving a Ic of close to 50 mA. Raising R2 to 1k still provides 11 mA of Ic, yet ruins all amplification (note the scale is now nanovolts). Instead, it attenuates:

Why? 11 mA of bias current should be more than enough. I see, however, that as I raise R2, the voltage at the base goes up, and the voltage at the collector goes down. Along with that, gain goes down, until R2 is about 500R, at which point VCB is about 0.6 V and gain is about 0 dB. Why?

I conjecture that the cause is along the lines of:

1. Raising R2 causes less Ic
2. which lessens Ie
3. which raises Ve
4. which lowers Vbe
5. which lowers Ib
6. which, multiplied by a factor of beta, lowers Ic
7. causing the cycle to repeat.

But I am not sure if that is correct, if so how to quantify it, or if so, how it ends. Furthermore: Why am I seeing the need for such a low Rc on this particular circuit but not in any other standard transistor amplifier.

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I've studied the linked questions, and, as Tim requested, am clarifying what's uniquely being asked here:

I understand why we need all 4 resistors: We need R3 and R4 to bias the base. We need R1 to provide a DC path for bias current. (We could use a short circuit instead, but this would just waste power while degrading quality.) And of course the gain comes entirely from having R3.

Likewise, I understand that biasing is important for many reasons and not necessarily best done at halfway between rails. The question isn't about a particular bias point.

Rather, this is a circuit analysis question: How do I analyze the effect of raising R2? It would seem to me that raising R2 should increase gain. A small change in Ib results in a larger change in Ic = β * Ib, which results in a change -R2 * Ic in output voltage. Higher R2 should increase the gain: as long as output voltage (collector) doesn't crash into input voltage (base).

But practically, I'm seeing very different results: We need to keep R2 very low for the circuit to work. I can't determine the cause, or even determine how to analyze it. That is this question.

Answer 1

I wonder about accurate simulation of RF, using your current understanding of bipolar circuits. But I'll drop that and move on.

A couple of bad things done in that circuit.

• Trying to bias the base at \$\frac12 V_{_\text{CC}}\$
• Using a very tiny biasing-pair resistor current, as a ratio of the expected collector current.

Those two things stick out like sore thumbs.

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For your circuit:

• \$V_{_\text{TH}}=V_{_\text{CC}}\cdot\frac{R_4}{R_3 + R_4}\$
• \$R_{_\text{TH}}=\frac{R_3\,\cdot\, R_4}{R_3 + R_4}\$
• \$I_{_\text{B}}=\frac{V_{_\text{CC}}-V_{_\text{TH}}}{R_{_\text{TH}}+\left(\beta+1\right)\cdot R_1}\$
• \$I_{_\text{C}}=\beta\cdot I_{_\text{B}}\$
• \$I_{_\text{E}}=I_{_\text{C}}+I_{_\text{B}}\$
• \$V_{_\text{B}}=V_{_\text{TH}}-I_{_\text{B}}\cdot R_{_\text{TH}}\$
• \$V_{_\text{C}}=V_{_\text{CC}}-I_{_\text{C}}\cdot R_2\$
• \$V_{_\text{E}}=V_{_\text{B}}-V_{_\text{BE}}\$

Saturation just starts to happen when the collector voltage is the same as the base voltage or else is between the base voltage and the emitter voltage. We can simplify that and just say \$V_{_\text{C}}=V_{_\text{B}}\$ is the moment of initial saturation (to avoid \$\le\$ and \$\ge\$ in the equation.) Until that moment, the bipolar is in active mode. After that moment it is in saturation mode.

So:

vth = vcc*r4/(r3+r4)                # Thevenin voltage of R3 & R4
rth = r3*r4/(r3+r4)                 # Thevenin resistance of R3 & R4
ib = (vth-vbe)/(rth+(beta+1)*r1)    # Use KVL to get this
ic = beta*ib                        # A given, by definition
ie = ic+ib                          # Taking the magnitude
vb = vth-ib*rth                     # Thevenin source, less drop due to base current
vc = vcc-ic*r2                      # Power source, less R2 drop
ve = vb-vbe                         # By definition
R2 = solve(Eq(vc,vb),r2)[0]         # Per condition, Vc = Vb, for saturation
R2
r3*(-beta*r1*vcc - r1*vcc + r4*vbe - r4*vcc)/(beta*(r3*vbe + r4*vbe - r4*vcc))

That allows you to work out the maximum value of \$R_2\$ to avoid saturation (or the minimum value needed to guarantee saturation.) The only assumptions you must apply here are an estimate of \$V_{_\text{BE}}\$ and an estimate for \$\beta\$. These will be assumptions. The Spice modeling will do much better at working out those two values. But if you use Spice modeling to capture those two values, then the above equation will yield a value that matches up.

Assuming \$\beta=190\$ and \$V_{_\text{BE}}=750\:\text{mV}\$ (I'm clearly pulling these out of a hat):

R2.subs({r1:50,r3:33e3,r4:33e3,vcc:12,beta:190,vbe:.75})
243.533834586466

So that's the prediction. Here's LTspice:

From those figures, I find \$\beta=179\$ and \$V_{_\text{BE}}=761\:\text{mV}\$:

R2.subs({r1:50,r3:33e3,r4:33e3,vcc:12,beta:179,vbe:.761})
255.329869127227

LTspice provides:

And it appears that now \$\beta=178\$, instead. But it's narrowing down very quickly.

The only remaining question you have is about estimating what happens upon raising \$R_{_\text{C}}\$'s value in your own circuit from \$50\:\Omega\$. That moment you use a word like raising you should immediately think derivative. And you are asking about the sensitivity of the collector voltage to changes in \$R_2\$.

I think you should find a good web page on the topic of sensitivity equations, as I'm not going to teach it here.

But, let's look at what LTspice says:

So \$\beta=192\$ and \$V_{_\text{BE}}=760\:\text{mV}\$. I can now make an estimate of what happens if I make a 10% change:

rho = simplify(derivative(vc,r2)/vc*r2)
(rho*10).subs({r1:50,r2:50,r3:33e3,r4:33e3,vcc:12,beta:192,vbe:.76})
-1.90909918936151

It says that a 10% increase will lower the collector voltage by a magnitude of about 1.91%. Let's test this:

Yup. The sensitivity equation works!