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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?

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2 Answers 2

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The value of the 5M resistor is way too high and does not provide enough base current. Since the emitter voltage is zero then the transistor is not turned on and its collector voltage should be almost 9V, not 4.2V. The emitter resistor should be providing plenty of negative feedback to reduce the distortion you see. What is the transistor part number?

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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, 2017 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, 2017 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, 2017 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, 2017 at 15:50

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