High Gain Amplifier using BJT

I need some assistance on building a high gain amplifier using only BJTs, capacitors and resistors. I have made some before using op amps, but for this assignment I am not allowed to use Op Amps at all.

The amplifier has to be capable of providing a voltage gain of at least 1.5k V/V. We can use any value we want for the resistors, capacitors and input.

Here I attach a Multisim simulation of what I have so far:

If you can see at the end, I am able to get a gain of 1.5k, but my instructor says that I am using too many parts for it to achieve that. He said that I have to take away a lot of resistors and capacitors for it to be ok, or to use current mirrors or a differential amplifier and cut the number of components I am using. And this is the part I need help with because I do not know how to achieve that by my own.

• If I were told "too many parts" and nothing else I'd ask for the boundary conditions to go from "unstated" to "stated clearly." How many BJTs? How many resistors? How many parts total? Etc. I'd try and nail it down. If refusing to add anything else, I'd take that as them taking the power to jerk my chain for as long as it pleased them and probably find someone else. (I'm assuming this is undergrad.) – jonk May 3 '17 at 22:17
• It is possible to get Av =1000 or 60dB if your output impedance is higher than source from a single transistor. so Av=1.5k is easy to do, but what are your other dozen specs typical for any OA? – Tony Stewart Sunnyskyguy EE75 May 3 '17 at 22:50
• Cripes. I just noticed you are depending on $r_e$ for your gains. No managed global feedback to linearize or set gain. Kind of open loop and very temperature dependent, isn't it? Or am I missing something? – jonk May 4 '17 at 1:37
• I think his issue is not #of components but number of passive components. You may want to refresh on differential amps and current mirrors. Maybe study the good old 741 circuit you can find here en.wikipedia.org/wiki/Operational_amplifier for ideas and pointers. You may also get on better if you use +- rails instead of going one sided. – Trevor_G May 4 '17 at 2:21
• Make the whole thing DC-coupled. That will eliminate 3 parts per stage except the first. – Marquis of Lorne May 5 '17 at 21:50

Here's an answer that will achieve at least a gain of 1500 from just one common BJT. Details about the design below will, I think, make a clear example out of why you would not design one like this. But your instructor wants just a few parts and doesn't seem to care about anything else. So:

simulate this circuit – Schematic created using CircuitLab

I've used a 2N5550 because it can tolerate the required higher $V_{CEO}$ needed here. I've used a single $90\:\textrm{V}$ power supply so that I could get the necessary gain. I've assumed that $Q_1$ operates at room temperature and since it will only experience less than $50\:\textrm{mW}$, it probably won't be any higher than perhaps $10\:^\circ\textrm{C}$ above ambient, anyway. Not enough difference to matter here.

The design follows this logic:

1. I assumed temperature variation is modest enough to ignore.
2. Voltage gain will be $A_V=\frac{r_o\vert\vert R_C}{r_e}=\frac{\left(r_o\vert\vert R_C\right)\cdot I_C}{V_T}$.
3. Since $A_V\ge 1500$, it follows that: $\left(r_o\vert\vert R_C\right)\cdot I_C\ge 39\:\textrm{V}$
4. I chose to more than double this and set $V_{CC}=90\:\textrm{V}$ on the usual basis of setting the quiescent collector voltage "midway" and to allow room for the Early effect, too.
5. To further increase room for gain, I decided to lower the quiescent collector voltage to $20\:\textrm{V}$
6. I wanted to minimize distortion (which means reducing variation around the quiescent collector current), and taking into account #5 above, I decided the maximum output swing would be $v_{pk}=5\:\textrm{V}$
7. Decision #6 suggests that the maximum input swing is $v_{pk}=\frac{5\:\textrm{V}}{1500}\approx 3\:\textrm{mV}$

First thing is to pull up 2N5550 specs from an ORCAD model I found: $V_A=100\:\textrm{V}$, $I_{SAT}=2.511\:\textrm{fA}$, $n_e=1.241$, $I_{KF}=.3495$, and $\beta_F=213.4$.

Don't think that the $\beta_F$ is accurate, though. The $n_e=1.241$ and $I_{KF}=.3495$ are nominally $n_e=1.5$, $I_{KF}=.4$, which yields an adjustment factor of very close to 1 for typical collector currents. But given these values, and a rough idea of the collector current I'm expecting, the modified $\beta$ for purposes of estimating the base current will be about $\beta_{ADJ}=\frac{213.4}{2.04}\approx 105$ for our purposes below.

I probably should have used a different BJT to keep this simpler, but this illustrates an issue regarding some BJTs. These two values may vary a lot, but it usually works out that they don't combine to create such a difference.

The 2N5550 is one of the weird ones in this regard. It's adjusted $\beta$ varies more substantially over collector current than is common and its effective $\beta$ isn't what you'd imagine if only looking at one of its spice parameters.

But using this device actually makes an exaggerated example about why one does NOT depend upon $\beta$ for biasing a BJT like this. Not only does the circuit overly depend upon $\beta$ for the quiescent collector bias point, but the $\beta$ value of any one given physical unit of this part family can vary a lot depending on even modest changes in collector current.

From the above, I know that:

\begin{align*}R_C& \ge \frac{\left(V_A+V_{C_Q}\right)\cdot 1500}{V_A+V_{C_Q}-1500\cdot V_T}\cdot\frac{V_T}{I_C}\\\\&\ge 38.5\:\textrm{k}\Omega\end{align*}

In calculating the above, I decided to estimate $I_C\approx 1.5\:\textrm{mA}$. However, once it computed out a value for $R_C$, I then selected a higher standard value of $R_C=45\:\textrm{k}\Omega$ for it.

This choice now means I'm predicting a quiescent collector current of $I_{C_Q}=\frac{90\:\textrm{V}-20\:\textrm{V}}{45\:\textrm{k}\Omega}=1\frac{5}{9}\:\textrm{mA}$. I also therefore get $r_o= \frac{V_A+V_{C_Q}}{I_C}=\frac{100\:\textrm{V}+20\:\textrm{V}}{I_C}\approx 77\:\textrm{k}\Omega$, suggesting that my gain will be $A_V=\frac{\left(r_o\vert\vert R_C\right)\cdot I_C}{V_T}\approx 1700$.

The base current calculation was complicated by those factors. But we can now compute an estimated value as $\frac{I_C}{\beta_{ADJ}}\approx 15\:\mu\textrm{A}$. Using a rough estimate that $V_{BE}\approx 700\:\textrm{mV}$, I figure $R_1=\frac{90\:\textrm{V}-700\:\textrm{mV}}{14\:\mu\textrm{A}}\approx 6\:\textrm{M}\Omega$. I used a standard value close to that, but larger because I already used a lowish collector voltage and I don't want to push it any lower than that.

That completes the design.

Test it out in spice using an input source with no more than $3\:\textrm{mV}$ peak. Unloaded, the output should be about right, I think. Here's a model:

.model 2N5550 NPN(Is=2.511f Xti=3 Eg=1.11 Vaf=100 Bf=213.4 Ne=1.241 Ise=2.511f Ikf=.3495 Xtb=1.5 Br=3.24 Nc=2 Isc=0 Ikr=0 Rc=1 Cjc=4.883p Mjc=.3047 Vjc=.75 Fc=.5 Cje=18.79p Mje=.3416 Vje=.75 Tr=1.212n Tf=560.1p Itf=50m Vtf=5 Xtf=8 Rb=10)

(For high gain arrangements like this, by the way, there is also a small effect called the "Late effect" which might impinge. But I think the effect is likely small enough not to seriously impair the predicted gain.)

So that's what you get without any specifications to speak of.

The maximum gain you can get from a single bipolar using resistive load is

$$VDD/0.026$$

hence you need 1,500 * 0.026 ~~ 40 volts for single-stage resistive load.

Given current source loads, you can reach these gains in 5 or 10 volts. The only challenge will be stability of operating point. Opamps use these methods, achieving Open Loop Gains of 10^5 or 10^7 or higher, and depend on external feedback to precisely define the gain.

Early integrated circuits used this topology:

simulate this circuit – Schematic created using CircuitLab

• Thanks! If I do this circuit in Multisim, should it work? I am really lost with high gain multipliers. – MrGamez22 May 4 '17 at 13:34
• You must insert DC-blocking capacitors at both input and output. Start with 1uF. – analogsystemsrf Apr 4 '18 at 16:34