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In this non-contact AC voltage detector, I understand the working principle of taking a tiny induced current in the antenna and amplifying it using transistors. However, I do not understand why the resistors have been used and why the particular values have been selected.

My understanding is that a transistor will give a DC current gain based upon the particular transistor used of say 100. So why has the 1 MegaOhm resistor been used? Surely your induced current will be so small you want to maximise the gain of that signal? I'm just struggling to understand why the resistors have been used.

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  • \$\begingroup\$ these impose Miller Effect, to what avail? \$\endgroup\$ – analogsystemsrf Aug 13 at 3:06
  • \$\begingroup\$ @analogsystemsrf I'm sorry could you explain what you mean by that? \$\endgroup\$ – Greyhurst Aug 13 at 17:12
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The base-emitter current is a fraction of the collector-emitter current, but a non-zero fraction. For a larger collector-emitter current you need a large base-emitter current. The RF energy might be too weak to supply the larger base current required if you were to try and maximize your collector-emitter current in a single stage.

The 1 megaohm resistor is sized to be small enough so that T1's collector-emitter current can supply reasonable about of base current to T2 is, but large enough so that the base current of T1 does not load down the signal source too much.

The collector resistor for T2 is smaller to allow it to produce more collector-emitter current by taking advantage of the fact that there's a lot more power driving its base.

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  • \$\begingroup\$ I thought that the voltage drop would be relatively consistent across the transistor regardless of the current? I would assume that a 1 megaohm resistor would limit the current flowing through T1 so you're putting a smaller current into T2 to apply the gain too. \$\endgroup\$ – Greyhurst Aug 12 at 21:17
  • \$\begingroup\$ @Greyhurst There's the assumption. Yes but not in the places you're thinking, The voltage drop across the base-emitter junction will be relatively consistent, but the base-emitter current will modulate the collector-emitter current. As a result, the collector-emitter is NOT constant. This in turn modulates the current through the resistor which modulates the voltage drop across the resistor. The transistor gain in this case (hfe) is poorly controlled so you're only really relying on the transistor as a switch, not as an actual linear amplifier. \$\endgroup\$ – DKNguyen Aug 12 at 21:19
  • \$\begingroup\$ Note that there's no signal picked off the collector resistor. It's passing via the base-emitter junction, instead. I suspect there's a better way of explaining why the largest value resistor is used in stage 1, a lower value in stage 2, etc. For example, does your explanation answer why the same collector current for T1 (which becomes emitter current) wouldn't occur regardless of its collector resistor value, if the base current is taken as determined by the antenna's signal reception? \$\endgroup\$ – jonk Aug 12 at 21:20
  • \$\begingroup\$ @jonk Whoops. Totally missed that no signal was being taken off the collector. In which case the 1Megaohm resistor is not to produce a large voltage drop but just because only a little current is needed to drive the base of the next transistor. \$\endgroup\$ – DKNguyen Aug 12 at 21:21
  • \$\begingroup\$ @DKNguyen There's an even more interesting answer than that, I think. \$\endgroup\$ – jonk Aug 12 at 21:22

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