Amplifier Gain and Frequency Response Using AI

電子學問題ＡＩ求解方式  

Solve for the gain. Plot frequency response for the following  amplifier circuit.

source

share grok 4 (able to consider parasitic capacitance) Bode plot   (188 kHz @Cπ ≈ 50 pF, Cμ ≈ 2 pF)

Claude Bode plot gives 188 kHz

ChatGPT 5 gives 187 kHz

Gemini 2.5 Pro gives 188 kHz

Lower cut off comes from coupling capacitors (not shown here)

我用Claude計算電晶體放大器，得出頻率響應理論值 2.4 MHz
然後我給他一組Grok 算出來的數據是 125 kHz
兩者差了20倍

我沒有跟Claude 說這組數據哪裡來的，他稱讚我的數據是應該是用Spice 專業軟體跑出來的，而且提醒我他的計算是根據純理論，因為Spice 可以考慮更多細節去電路模擬，所以Claude認為我的數據比較值得參考。

接下來我跟他講這組數據是 Grok 4 算的，
Claude 馬上變換語氣，說頻率響應不可能是125 kHz
業界標準至少在1 MHz 以上，
說Grok 
1. Does not make any sense (完全沒有道理）
2. Against common sense (違背常識）
3. Distrust (不值得信賴）

可是剛剛不是還在稱讚「我的模擬」做得好？

以前Claude 還會稱讚別家的AI有什麼優勢，現在看起來不演了。

Solve for the gain. Plot frequency response for the following two stage amplifier circuit.

source

share rough guess

share (Cπ = 22.1 pF, Cμ = 3.0 pF) RE is bypassed; Bode plot, fH= 10.46 MHz, RE not bypassed

Solve for the gain. Plot frequency response for the following two stage amplifier circuit.

source

schematic

solve the gain (artifact)

source

frequency response (share)

freq. response (capacitor value settings changed)

artifact, share Cπ2=15 pF, Cμ2=3pF, 

freq response, gain=8513, low cut off=46.5 Hz, high cut off=312kHz

Gemini gain=8509, low cut off=46.5 Hz, high cut off=296 kHz @Cπ2=15 pF, Cμ2=3pF (good match!)

artifact, share gain=8559, 

Hybrid-π Model, With bypass capacitors, without ignoring base current loading effects on bias networks

artifact gain=9460

Ignored base current loading effects on bias networks

artifact

Without bypass capacitors, Ignored base current loading effects on bias networks

solve for the gain, plot freq response

share (source)

-194 for Vc vs. Vbase, Ibase ignored

share

Hybrid-π Model, With bypass capacitors, without ignoring base current loading effects on bias networks, Av=-165
share including frequency response

Gemini 2.5 Pro gain & lower cut off freq match Claude 4 Sonnet, 4kHz

Design a two stage amplifier using BJT.

the first stage is class A. the second is class AB. freq response cut off at 10Hz and 25 KHz

Alternatively use two diodes in the second stage, textbook, theoretical

History

share grok 4 (able to consider parasitic capacitance) Bode plot   (188 kHz @Cπ ≈ 50 pF, Cμ ≈ 2 pF, 

but Claude Bode plot gives 2.4 MHz @Cπ ≈ 50 pF, Cμ ≈ 2 pF)

Claude with corrected topology gives 193 kHz 

Claude gives 188 kHz no correction.

ChatGPT 5 gives 187 kHz no topology correction required as Grok 4

Low cut off comes from coupling capacitors (not shown here)

Claude 4 Sonnet (Extended mode) Bode plot   (assuming different Cπ ≈ 70.4 pF, Cμ ≈ 7.0 pF)

topology uncorrected