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I have been exploring BJT transistors and implemented a common base amplifier. I connected a signal generator to the voltage source labelled V2 in the below schematic. The oscilloscope image shows the input above the amplified waveform. It appears that it is in fact working, 10.8/.288 amplification. I wish to possibly use it to amplify a dynamic mic. I plan to reduce the amplification and/or control it by a potentiometer. Anyways..... my question is this

Why is the amplified signal "squareish" and not a decent replication of the input signal. It looks more like an amplified square wave. I created the circuit in LTSpice and simulated it. The simulation is shown below. I am reading materian on bjt transisters and especially common base amps,how is this fixable? Which resistor values should i look to change Rc, Re, or Both. Should the current level be changed through the emitter. With no signal generator, it is 205 micro amps on the real circuit. Do I need to adjust the voltage biase on the base. Thanks for nay advice.

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  • \$\begingroup\$ You are seriously overdriving the amplifier. Reduce the input signal level and/or reduce the gain of the amplifier. \$\endgroup\$ – Peter Bennett Mar 30 '20 at 0:02
  • \$\begingroup\$ Jeffrey, what are you trying to achieve? How did you design this? What were the goals? We can't just go fix something, or suggest changes, without a clue about what qualifies as "success" to you. \$\endgroup\$ – jonk Mar 30 '20 at 0:08
  • \$\begingroup\$ @jonk I am trying to make an amplifier to take the small signal from a dynamic mic, which should be in the range of a few millivolts, and feed it to a ham radio transmitter. Once fed, I wish to experiment with the level of amplification to see what "sounds best" on another radio. The other radio will of course receive the transmitted signal with its volumn appropriately set. \$\endgroup\$ – Jeffrey Edward Messikian Mar 30 '20 at 0:33
  • \$\begingroup\$ @jonkI would like to use a potentiometer and "adjust the volumn" on the tx side until I find the optimal circuit. I copied this circuit from the internet and understood how the gain and especially how the emitter resistance (.025 mv) works. I took this circuit with these parameters from an internet circuit. Success is when the amplified waveform does not look like a square wave. I will try and adjust the amplification. \$\endgroup\$ – Jeffrey Edward Messikian Mar 30 '20 at 0:33
  • \$\begingroup\$ @Peter Bennett Thank you, I will try that. I copied this from the internet. It was meant to take a dynamic mic as input. I fed it point 4 volts peak to peak from my signal generator. I will reduce the gain/modify it or try and lower my signal generator (unlikely). My main issue is the replicated wave the fact that the top and bottom are not shaped the same. I do not know how to adjust it so that the top and bottom look like they were mirrored, but I will try and lower the gain. \$\endgroup\$ – Jeffrey Edward Messikian Mar 30 '20 at 0:43
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The current through the base-emitter junction of a transistor is governed by the diode equation: \$i_e = I_S \left(e^\frac{v_{be}}{n\ V_T} - 1\right)\$. At room temperature, \$V_T \simeq 26\mathrm{mV}\$. So unless your peak-peak input voltage is significantly smaller than \$26\mathrm{mV}\$, you'll get distortion.

On the bright side, you have lots of excess gain in that circuit. The quickest way to get more linearity is to put a resistor in series with the emitter, between the emitter and the intersection of R4 and C2. I think I'd start with a value somewhere between \$200\Omega\$ and \$1000\Omega\$ and see how things look.

And -- don't think for a second that your resistor values are realistic. 1% resistors are standard these days, yet still a bit absurd for a transistor stage like that because any real transistor that you plug in there will add far more variation than a 5% resistor will.

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  • \$\begingroup\$ Well, the 1% resistors just mean that the design doesn't have to be compromised still further to handle large unknown resistor variations as well as everything else the design otherwise has to compromise in order to manage (such as part parameter variations, temperature...) One less thing to deal with is a nice-to-have, I think. \$\endgroup\$ – jonk Mar 30 '20 at 3:49
  • \$\begingroup\$ I implemented this circuit from a schematic. The resistor values were described in standard sizes such as 10k, 47k. I saw what I attached on the oscilloscope pic and decided to implement it in LTSpice. I used a 20000 count LCR meter to obtain the actual resistor values in my circuit. I used those values in LTSpice. I skipped the caps. Sadly, the LTSpice sim did not exactly match the oscilloscope, though I would say it is close. I still need to look more at LTSpice and its sim. I never saw the diode equation (current) related to BJT's so I will investigate that and its application hre \$\endgroup\$ – Jeffrey Edward Messikian Mar 30 '20 at 13:31
  • \$\begingroup\$ What I read/know about BJT's was mostly from the web site allaboutcircuits.com I will dig more to understand the excess gain problem you say I have. I will especially try and understand how the diode equation relates at low input values \$\endgroup\$ – Jeffrey Edward Messikian Mar 30 '20 at 13:37
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    \$\begingroup\$ If you have more than enough gain and lots of distortion, the gain isn't a problem -- it's a life saver. \$\endgroup\$ – TimWescott Mar 30 '20 at 15:18
  • \$\begingroup\$ The second-biggest reason your LTSpice simulation wouldn't match your circuit exactly is because transistors vary a lot from unit to unit. The biggest is that it's a simulator, not the real thing -- you're supposed to understand that, and understand transistor theory, and understand what you can and cannot learn from simulation. At your level, just remember that it's kinda approximate. \$\endgroup\$ – TimWescott Mar 30 '20 at 15:19

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