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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 MΩ 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.

enter image description here

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  • \$\begingroup\$ these impose Miller Effect, to what avail? \$\endgroup\$ Aug 13, 2019 at 3:06
  • \$\begingroup\$ @analogsystemsrf I'm sorry could you explain what you mean by that? \$\endgroup\$
    – Greyhurst
    Aug 13, 2019 at 17:12

3 Answers 3

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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, 2019 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, 2019 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, 2019 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, 2019 at 21:21
  • \$\begingroup\$ @DKNguyen There's an even more interesting answer than that, I think. \$\endgroup\$
    – jonk
    Aug 12, 2019 at 21:22
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However, I do not understand why the resistors have been used...

The resistors are there to limit the current through the collectors. The primary advantage of this is the conservation of battery power.

The circuit needs to work with a wide range of effective capacitance values between the "antenna" and mains voltage. (There is also a capacitance between the person holding the device and mains ground, and between the person holding the device and the detector circuit ground. But to simplify things we can think of this as a combined capacitance between the "antenna" and the mains.)

As a result of the wide range of in effective capacitance values between the "antenna" and mains voltage, there is a wide range of current into and out of the base of T1. Note that the current flows in both directions, and must do so, because the current through a capacitor must average to 0. The reverse current through the transistor is primarily the result of base-emitter junction reverse break-down. (If you attempt to model this circuit in a simulator, be aware that some simulators do not model reverse breakdown current. You may have to add an "ideal" diode in parallel with the base-emitter junction, with a \$V_{fwd}=2V\$ and a \$V_{rev}\$ of say 5 V (or -5 V depending on whether the parameter is supposed to be positive or negative). If "one-sided" ideal diodes are available, as is the case with CircuitLab, one could simply use a reverse oriented ideal diode with a \$V_{fwd}=5\$. Don't use a Zener, because the leakage current will (or may) render the model inoperable.

Here is a CircuitLab model if one wants to play around with it.

schematic

simulate this circuit – Schematic created using CircuitLab

And for no particular reason than to help understand the circuit, here are the base to ground voltages for the three transistors, and the collector current for T3.

enter image description here enter image description here

Regardless of the current into and out of the base of T3, the current through the collector of T3 is limited by the voltage drop across the load components and T3's collector resistor. In some similar circuits, the collector resistor is 1k and there is only a LED for load. In that case, the current through the LED is limited to a little less than 9 mA.

Together, the 1 M\$\Omega\$ and 100 k\$\Omega\$ resistors limit the current from the battery through T1 and T2 to about 100 uA. Much less than the 9 mA through the LED. But if the 1 M\$\Omega\$ and 100 k\$\Omega\$ resistors were not present, the current from the battery through T1 and T2 could be much more than the current through the LED. T3 could even operate in a saturation so deep that there is more base current than collector current.

If T1, T2 and T3 were all 2N3904 transistors, their typical current gain would be around 300 each. The combined current gain of T1 and T2 would be around 90000. With the 1 M\$\Omega\$ and 100 k\$\Omega\$ resistors absent, and an input current of only 1 uA, the current through the emitter of T2 would be 90 mA! This would wear down the battery much quicker than necessary. With even more "antenna" current, one might even destroy a small transistor.

However, I do not understand ... why the particular values have been selected.

Assuming a current gain of 300, T3 only needs 30 uA base current to give 9 mA collector current. With a 9V battery, and 100 k\$\Omega\$ collector resistor, T2 can give 90 uA emitter current. So, the 100 k resistor is big enough, but not too big.

The 1 M\$\Omega\$ resistor attached to T1 could actually be a 100 k\$\Omega\$ resistor as well without significantly adding to the current draw of the circuit. However, since T2 only needs a small fraction of 30 uA, the designers decided to go with the 1 M\$\Omega\$ which is still not too large for T1.

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    \$\begingroup\$ Great answer written with much imagination and inspiration... I even wanted to try it, but I'm too lazy to do real experiments, and the simulator can't recreate the real situation. But who's to say what's going on in this mess of input capacitances? It seems to me that the reverse current goes through the shunt capacitance rather than through the reverse breakdown... \$\endgroup\$ Dec 20, 2023 at 20:51
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    \$\begingroup\$ Thanks for the praise. I modeled it in LTSpice, because I didn't trust CircuitLab's transistor Vbe-Ib model. But it turns OK CircuitLab model gives same results. I can add that. I don't think capacitance between base and ground can account for the current. It's average must also be 0, so one has only moved the problem. If current flows from "antenna" capacitor to ground through base-emitter junction, eventually, current needs to flow the other way. If leakage doesn't cut it, it seems it has to be reverse breakdown. (I'm using the Sherlock Holmes fallacy here. :-) ). \$\endgroup\$ Dec 20, 2023 at 21:35
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    \$\begingroup\$ @Circuitfantasist I've added a CircuitLab model if you want to play around with it :-) \$\endgroup\$ Dec 20, 2023 at 21:55
  • \$\begingroup\$ Oh, it's good that there are like-minded people here... I'm so tempted to jump out of bed (it's past midnight here) but I know I'll play around and it'll be too late. I don't know why I like this CircuitLab so much, maybe because it's so intuitive and clean of details? BTW looking at my phone in the dark the input voltage is some weird value of 168V. Is that the peak value of 110V? Here it is 315 V; so this indicator will work twice as well here :-)? \$\endgroup\$ Dec 20, 2023 at 22:31
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    \$\begingroup\$ @Circuitfantasist yes. 168 V = 120 V x 1.4 (approx sqrt(2)). Yes, the minimum capacitance to turn on the LED should be half as much if the voltage is doubled. \$\endgroup\$ Dec 20, 2023 at 23:04
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The two resistors are not needed if the collectors are connected together in the so-called "Darlington triplet".

I have used the @Math Keeps Me Busy's simulation as a basis for comparison.

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see, there is no significant difference between the results; only the LED current is slightly reduced (with 1 mA) due to the 1.5 V drop across the "triplet".

STEP 1a

STEP 1b

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    \$\begingroup\$ You may have just improved on a circuit that has been widely used across the globe! \$\endgroup\$ Dec 21, 2023 at 13:19
  • \$\begingroup\$ @Math Keeps Me Busy, Imagine what profit this is for Chinese manufacturers - 2 resistors x 1000000...000 :-) \$\endgroup\$ Dec 21, 2023 at 13:38

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