# How can I build BJT amplifier with Vin and Vout requirements (peak to peak values)

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?

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 Then it displays the following

• Looks pretty random. Do you have any calculations you made here? Or was it more "hunt and peck," so to speak? Can you explain any reasoning at all for the design you have above? Is the $10\:\text{V}$ supply rail required? Or is that something you can adjust per your desires for the design? Also, are you really supposed to use $1\:\text{Hz}$? And finally, do you notice your RMS and your P-P for the blue line? Do you know how those values are calculated?
– jonk
Nov 22 '18 at 6:24
• A quick calculation tells me that your BJT is saturated, so the collector isn't doing much. I also see that your purple P-P value shows as $200\:\text{mV}$ and not $100\:\text{mV}$ as your text says. So I think $100\:\text{mV}$ is the peak value, not the peak to peak value. Just another note to add to others. But no, this design won't meet any reasonable requirements I can imagine.
– jonk
Nov 22 '18 at 6:34
• Jese, I'm willing to help out. I just need to know where you are at in terms of your education here. If you have no clues at all, it's fine. It just means I start somewhere else. If you have some clues, I need to know which ones you have so I can build on those. I can try and help. But it would be nice if you could help me understand where you are at, first.
– jonk
Nov 22 '18 at 17:39
• Thank you for your answer. I am currently trying to solve this by working backwards. I am currently taking an introductory class for this project and know how to solve problems that i am given with initial conditions and resistors but dont know how to solve this particular project with Vin and Vout as my conditions. My logic tells me that i can solve for VC(2.5V) = 9V(my battery) - Rc(Ic). Im using 2.5v since the requirements say that its 5 v peak to peak values, as you suggested. Thank you very much for your help. Nov 22 '18 at 19:45
• I appreciate getting a response. Your writing doesn't suggest much of an approach, though. It's almost entirely hand-waving, so to speak, without any approach in mind. (It seems, to me, anyway.) Besides this, your schematic shows $10\:\text{V}$ but your comment says you are using a $9\:\text{V}$ battery for this. In total, I feel the whole thing is rather slippery and muddled in your mind and you don't have any starting approach. This worries me about the education you are supposedly getting (or your presence of mind when in class.)
– jonk
Nov 22 '18 at 20:17

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: 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.]

• My word!!!! Never in the nearest dream would I have come up with that. Let me figure out if I understand the process. Thank you very much for your honest help. I will make sure to recommend you as a helping lad in this community. Thank you!!! Nov 22 '18 at 22:50
• I have posted my observations on the main text of the question, everything seems to be in order as you suggested, but let me know what you think. Nov 22 '18 at 23:39
• @JeseReyes Just a note. I'm not sure if you are really wanting to use 1 Hz. But the circuit is designed for audio. You should either use REALLY BIG capacitors if you intend on keeping the 1 Hz or else you should use 1 kHz or something like that.
– jonk
Nov 23 '18 at 0:38