I'm reading at the moment the schematic of the Baofen UV5R radio (Baofeng UV5R schematic). If I understand the circuit correctly, this circuit is used to enable/disable the microphone (Q16) and to detect VOX (Voltage at C14). The question is, why do I need such a complicated circuit with Q17, D21... instead of a simple lowpass filter with recifier diode as in the second image below connected at the same node as C137? VOX detection

Simple detection circuit

  • \$\begingroup\$ At the C137 node, voice amplitude from MIC1 is too low...the AC amplifier Q17 is required so that D21 peak-to-peak rectifier works efficiently. In any case, your LTspice circuit won't do a similar function...C13 might change its DC voltage when you drive it with a large AC signal, but C16 will remain at 0V DC. \$\endgroup\$ – glen_geek Jun 6 at 16:08

Andy is correct that you need gain but his estimate of gain is incorrect.

The gain is only 30 +/-5 with such a low bias current from 1M5 to amplify say a 50mVp to 300mVpp. The dual Sch. Diodes clamp and rectify the signal but drop about 50mV each at this xx nA current level leaving about 200 mVdc for the ADC to decide when to enable and disable mic for fast attack and slower cutoff with software hysteresis.

Also 104 caps are 100nF and the p suffix is redundant but a reminder of the printed value on the part vs the logical value usually shown in schematics. So your schematic has the wrong cap value. (Which is why actual values are used on schematics) {rookie designer;} but a pro compared to questions on this site.

Your design barely exceeds the conduction threshold at 50mV and would need to be at least 150mVp to trigger the same threshold. So the VOX would cutout your first utterance.

So you need gain, although his circuit could be improved with better values to raise input impedance, add 500k pullup for more Ic and thus lower re =26/Ic with Ic @ 1uA , raise Rb and Rc and thus more Av, without loading the audio much.

  • \$\begingroup\$ Hi Tony, thank you! Never seen this p suffix. Is it used for capacitors only and do have resistors a similar suffix? From where did you get the value 26 without knowing the exact partnumber of the npn? \$\endgroup\$ – Murmi Jun 6 at 19:28
  • \$\begingroup\$ Yes the gain is a lot lower than hundreds; more like 50 to 100. \$\endgroup\$ – Andy aka Jun 6 at 19:50
  • \$\begingroup\$ That’s a BJT characteristic that is temperature sensitive , some use 25 or 26 and the gain is more like 25 after loading from clamp diodes and cap. Never 50 or 100 with 4k7 load and 1k input unless it was biased with much more current a greater supply voltage with a larger Rc/Rb ratio. \$\endgroup\$ – Tony Stewart EE75 Jun 6 at 19:56
  • \$\begingroup\$ @Murmi this is another example electronics.stackexchange.com/questions/553388/… \$\endgroup\$ – Tony Stewart EE75 Jun 6 at 20:16

why do I need such a complicated circuit with Q17, D21

Q17 has a voltage gain of of several hundred so, if your input signal is not tiny you can dispense with that amplifier stage. But you can't get rid of the 2nd diode in D21 to ground after C109 because it (and the other diode) are needed to properly rectify a signal and sustain an output DC level that is representative of the amplified microphone audio signal.

Put it another way - C139 needs to pass an AC current and the diode to ground passes negative current whilst the series diode passes the positive current to the output capacitor C140. If you didn't have the diode to ground, C139 would charge up and you'd lose the output signal after a short while.

enter image description here

  • \$\begingroup\$ Thank you Andy for the explanation. Makes sense with the amplifier. So I made wrong assumptions in my simulation for the microphone to not see that. Yes I've seen in the simulation that the diode to ground is needed to discharge the capacitor. Thank you! \$\endgroup\$ – Murmi Jun 6 at 19:21
  • \$\begingroup\$ @murmi take the 2 minute tour to see what to do next. \$\endgroup\$ – Andy aka Jun 6 at 19:51

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