Electronic – How to build BJT amplifier with Vin and Vout requirements (peak to peak values)

acamplifierbjtdcgain

I have been working on a project but I'm not sure if I'm complying with the requirements that I was given.

  1. Design a BJT amplifier such that VIN=0.1 V and VOUT=5 V (peak to peak values). Justify the gain of the amplifier by performing DC and AC analyses for the design.

  2. Increase the value of VIN in order to reach the saturation point in the DC load line. Justify the value obtained experimentally using the theoretical analysis of the Q-point.

Here is a screenshot of the BJT amp I have done in iCircuit. I adjusted the values until I was able to get a 5 V output but I'm not sure if my circuit meets the requirements. Can anyone help me with this project?

Screenshot of the BJT amp
enter image description here
enter image description here

Here are the images of the diagram Joink Suggested. When it initial y displays a 5 V P-P value and then it switches to alternate current with a peak to peak value of 107mV

But the output volate is 5 Peak to peak from the capacitor as you suggested in my opinion.
5 P-P
enter image description here

Then it displaysenter image description here the following

Best Answer

I'll provide an approach. There are many such, not just one. But I want to write this out quickly, so I'll just plow through with some short-cuts.

  1. The maximum voltage gain is about 40 times the quiescent collector current (in millamps.) You want a voltage gain (supposedly, from what I can read out of what you have written) of 50. So to be safe I'd set the quiescent collector current to \$2.5\:\text{mA}\$. Should be fine.
  2. From this quiescent current, it is reasonable to conclude that the quiescent base-emitter voltage is about \$700\:\text{mV}\$.
  3. I like to reserve about \$2\:\text{V}\$ for the minimum \$V_\text{CE}\$ of the BJT, in order to keep it well away from saturation.
  4. I like to reserve at least \$1\:\text{V}\$ for the quiescent emitter voltage for a variety of reasons, but importantly because I would like to place temperature and part variation issues under management.
  5. With \$9\:\text{V}\$ total (assuming your battery is fresh), this means there is about \$6\:\text{V}\$ left over for the collector. Since you need a range of only \$5\:\text{V}\$, this means I can (and I want to) leave about \$1\:\text{V}\$ margin at the top end of the collector swing. In short, I don't want the collector to move any higher than \$8\:\text{V}\$.
  6. Therefore, the quiescent collector voltage will be \$8\:\text{V}-2.5\:\text{V}=5.5\:\text{V}\$.
  7. From (1) and (6), I can compute a collector resistor of \$\frac{9\:\text{V}-5.5\:\text{V}}{2.5\:\text{mA}}=1.4\:\text{k}\Omega\$.
  8. From (1) and (4), I can compute a DC emitter resistor of \$\frac{1\:\text{V}}{2.5\:\text{mA}}=400\:\Omega\$.
  9. From (2) and (4), I know that the quiescent DC base voltage should be \$1\:\text{V}+700\:\text{mV}=1.7\:\text{V}\$.
  10. To be conservative, I'll assume that the base current of the BJT will be no more than about \$\frac{2.5\:\text{mA}}{\beta=100}=25\:\mu\text{A}\$.
  11. To make a "stiff" resistor divider (in the sense that it is relatively unaffected by variations in the required base current), I know that the current through the two base divider resistors should be about \$\frac1{10}\$th the quiescent collector current (or 10 times the current calculated in (10) above. So this means about \$250\:\mu\text{A}\$.
  12. The divider resistor, from base to ground, is then \$\frac{1.7\:\text{V}}{250\:\mu\text{A}}=6.8\:\text{k}\Omega\$.
  13. The divider resistor, from base to the supply rail, is then \$\frac{9\:\text{V}-1.7\:\text{V}}{250\:\mu\text{A}+25\:\mu\text{A}}=26.545\:\text{k}\Omega\$.
  14. To get the gain, I need the total AC emitter resistance to be \$\frac{1400\:\Omega}{50}-\frac{V_T=26\:\text{mV}}{I_Q=2.5\:\text{mA}}\approx 18\:\Omega\$. I should probably take into account the value computed from (8) above, but it's effect is minor here. So I'll ignore it in the resulting circuit, today.

So here is the resulting design after taking into account nearby standard resistor values:

schematic

simulate this circuit – Schematic created using CircuitLab

The above should take a \$100\:\text{mV}_\text{PP}\$ input signal and generate a \$5\:\text{V}_\text{PP}\$ output signal.

Feel free to ask questions, now. But hopefully that provides one possible approach to solving your question.

Note

This assumes audio frequencies. This means it will not work correctly for a \$1\:\text{Hz}\$ signal source. My recommendation is to try it with \$1\:\text{kHz}\$. But if you increase \$C_e\$ to \$470\:\mu\text{F}\$, then it will work okay down to perhaps a little less than \$100\:\text{Hz}\$.

[There's another issue with the design. It probably needs something to reduce its gain at higher frequencies. A cheap "fix" for this is a small-valued capacitor (perhaps \$2.2\:\text{nF}\$, for example) placed in parallel to \$R_c\$ (or the same small-valued capacitor used in series with \$2.2\:\Omega\$, where this series combo is placed in parallel to \$R_c\$.) That will roll off the gain at higher frequencies.]

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