# Simulation result of bjt cascode amplifier

I use square wave as the input source to my BJT cascode amplifier. However, I got my output voltage as

I used transient analysis in PSPICE to obtain the simulation result. Can someone tell me whether the result is correct or the output voltage should be a square wave as well? I think this graph due to capacitor but I am not so sure about it.

• Hi, could you add a schematic? It seems to me that the square wave input signal you are feeding to the amplifier has a so low fundamental frequency respect to the minimum input lower cutoff frequency that the circuit behaves as an approximate differentiator. The spikes you show seem the approximate $\delta$-pulses resulting from the derivation of the rising edge and the falling edge of the input square waves. – Daniele Tampieri Oct 7 '18 at 6:29
• Thank you for replying. added schematic. Can I know is it possible to get complete square wave at the output voltage? – sterstar Oct 7 '18 at 6:33
• I try to write down an answer even at the risk of not be a nice one, since the space for comments is too restricted. – Daniele Tampieri Oct 7 '18 at 8:17
• Change all the pF to nF for starters then optimize bias voltages for max gain and swing. – Tony Stewart EE75 Oct 7 '18 at 9:11
• I took the liberty of changing a few values. dV/dt=V/RC may help you understand your results. tinyurl.com/ybbnxdo9 – Tony Stewart EE75 Oct 7 '18 at 9:25

As stated in my comment above, it seems to me that the problem of your circuit is that, at the fundamental frequency of the test input signal $$\V_5\$$, it behaves as a differentiator. I suggest to follow the steps below in order to see if it is really this the problem of the circuit, and if so solve it

1. A first check you could do is to increase the fundamental frequency $$\f_{V_5}\$$ of your input signal. From the analysis of your output waveform, it appears that $$\f_{V_5}\simeq 130\mathrm{kHz}\$$: try to raise progressively your frequency up to $$\f_{V_5}= 500\mathrm{kHz}\$$ and see what happens. If the output waveform resembles more and more a square wave, we have identified the problem and you can go to the second step

2. It seem to me that the biggest problem of your circuit is the low value of $$\C_E\$$ this implies an associated lower cutoff frequency of $$f_{c_\mathrm{Low}}=\frac{1}{2\pi R_E C_E}\simeq 59.0\mathrm{kHz}$$ Try $$\C_E=2700\mathrm{nF}\$$ and see what happens.

Further notes.

• While analyzing the influence of $$\C_L\$$, Harry Svensson simulated the response in the time domain of the high pass RC circuit that this capacitor makes with the load resistor $$\R_L\$$, to a square wave at $$\125.0\mathrm{kHz}\$$. The output waveform, shown in the comments to this answer, is distorted but resembles the shape of the input square wave: this confirms my intuition, supported intuitively by the fact that $$f_{R_LC_L}=\frac{1}{2\pi R_L C_L}\simeq 15.9\mathrm{kHz},$$ that the problem lies in the value of $$\C_E\$$.
• Tony EE rocketscientist, in his comments to the question, showed the simulation in the time domain of the response to a square wave at $$\1.00\mathrm{MHz}\$$ of an optimized cascode circuit: the shape of the output waveform looks very close to the input square wave, confirming again that a proper response of the circuit relies on a accurate design of the decoupling time constants.
• What about the size of $C_L$? – Harry Svensson Oct 7 '18 at 9:46
• The cutoff frequency associated to $C_L$ and $R_L$ is nearly 15kHz so I think it is not the problem. Also the very low level of output voltages makes me think tat is the gain stage that has a voltage gain less than 1. – Daniele Tampieri Oct 7 '18 at 10:05
• Hmm, it barely resembles a square wave. Oh well. – Harry Svensson Oct 7 '18 at 10:37