I'm trying to design a small amplifier for AM signals to use before the usual diode-C-R detection (the reception in my area is very poor). I devised a QUCS simulation as below: QUCS simulation model

The diode and detection circuit is shorted or deactivated because I wanted to analyze first just the transistor output.The DC simulation shows approx the values I calculated for the biasing (1 mA quiscent current, 0,7V bias on the base and 4,5V on the collector) but when I run the AC simulation I get very strange results: [![Simulation results][2]][2]

Up left is Ve and Vc, up right input (measured at the "antenna") and output (measured at the 2kOhm resistor), low left Vb (base voltage) and low right the input.

I appreciate inputs on how to solve the problem as well as how to design a regenerative feedback for the circuit.

UPDATE: further experimentation made me realize that the antenna can't be floating, it has to be connected to ground. Point is: if I connect it through a capacitor (let's call it stray capacitance) the output gets smaller depending on the size of the capacitor (the bigger the smaller the reduction). If it's tied directly (or the "stray" capacitor is big enough) the input gets correctly amplified but the "positive" part gets squished (negative feedback?): artefacts

Can anybody help?


That lop-sided output is correct for large inputs to a bipolar. You are injecting ~~ 26 milliVolts peakpeak, or +-13mvppeak.

A positive input 13mV will cause e^13/26 increase in current, so vout drops

A negative input 13mV will cause e^-13/26 decrease in current, so vout rises.

Note the increase and the decrease are not symmetric.

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  • \$\begingroup\$ I still fail to understand why the output is not symmetric \$\endgroup\$ – Luca Jul 9 '17 at 17:50
  • \$\begingroup\$ e^0.5 is 1.64; 100% * 1.64 is 164 (increase of 64); 100%/1.64 is 60 (decrease of only 40). Thus a 64/40 = 1.6:1 imbalance, which about what your plot shows. \$\endgroup\$ – analogsystemsrf Jul 9 '17 at 22:34
  • \$\begingroup\$ I'm still not sure I'm following but so I should always expect non symmetric output for relatively large inputs? \$\endgroup\$ – Luca Jul 10 '17 at 9:56
  • \$\begingroup\$ The math I used is an exact description of the voltage-in and current-out for diodes and for bipolar emitter-base diodes. For low distortion (1% THD) the V_in needs to be 1mVpp. For 0.1% THD, V_in needs to be 0.1mVpp. For 0.01% THD, V_in needs to be another factor of 10 smaller. The intercept points for bipolar (IP1 and IP3) are near 0.1voltPeakPeak, and are not identical. \$\endgroup\$ – analogsystemsrf Jul 10 '17 at 15:50

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