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We all know that the purpose of Bridge rectifier is to change the AC to DC,and the picture below is my assumption,and as i know ,the Q1, Q2, D2 and the first level RC circuit are the filter and voltage stable mechanism,so i think the circuit in the red square is the voltage stable mechanism.but i don't know how they work.

Can anyone give me some suggestions ?

By the way,i am not pretty sure about the first level RC ,is the first level RC R1 and C3?and can the resistance and capacitor value be calculated? enter image description here

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  • \$\begingroup\$ Q2's internal body diode will be conducting if Q2 isn't being activated by a gate voltage so it doesn't look like any form of regulator I've ever seen. \$\endgroup\$ – Andy aka Feb 28 '18 at 9:34
  • \$\begingroup\$ @Andyaka So do you think what is this?because it also connect with the DC/DC buck-converter \$\endgroup\$ – Shine Sun Feb 28 '18 at 10:31
  • \$\begingroup\$ I don't know. All I know is that Q2 will be a forward biased diode (0.7 volts) or a low impedance due to Q2 being turned on. So the voltage across it will range from 0 volts to 0.7 volts and not provide much regulation capability. Where does the circuit come from? \$\endgroup\$ – Andy aka Feb 28 '18 at 10:53
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    \$\begingroup\$ No, I can't explain more because I'm just guessing. \$\endgroup\$ – Andy aka Feb 28 '18 at 11:16
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    \$\begingroup\$ It doesn't look like the circuit performs any useful function. This means I can't agree with any theory! \$\endgroup\$ – Andy aka Feb 28 '18 at 11:29
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You posted a corrected version of the schematic in another question (which has now been deleted) with important differences (e.g. no battery). I reproduce it below:

enter image description here

Essentially, the circuit in the red box ensures that the circuit is only turned on when C1 contains sufficient charge:

  1. The piezo power source charges C1 through the bridge rectifier.
  2. During this initial phase, Q2 is off and the MAX666 and subsequent circuits are off.
  3. Zener D2 is biased through R4 and R3, limiting the base of Q1 to about 12V, while C1 will charge to a higher voltage.
  4. When C1 reaches about 12.6V, Q1 base-emitter starts conducting, bringing Q1 on.
  5. The current through Q1 collector goes through R2, giving a voltage drop which turns Q2 on.
  6. The "gnd" for the rest of the circuitry is now connected to C1's negative plate via Q2 and thus the circuit is powered (with MAX666 providing 5V regulation).

So Q1,D2,Q2 hold the system off while C1 charges to an appropriate voltage. Now what is C3 for?

  1. If C1 starts to lose charge/voltage, the MAX666 Vin and Vout will droop, causing improper circuit function. To prevent this, the MAX666 grounds the LBout pin (see What does "LB" out mean in a DC-DC buck converter?).
  2. The LBout low pulse is transmitted through C3 and momentarily turns Q1 off.
  3. This pulse is enough to break the "latch" action formed by Q1 and Q2, and Q2 turns off. The capacitive coupling (C3) ensures that Q1 is not held off forever.
  4. The rest of the circuit goes back to being unpowered, while C1 regains its charge, and we go back to step (1).

Edit in response to question:

The VN2222L NMOS has a Vgs threshold of about 2.5 V. This voltage will be created by a tiny current of 2.5 uA through R2.

So as Q1 turns on (allowing current from C1 to flow through R1,Q1,R2), Q2 will also start to turn on. This has a feedback effect because Q2's conduction lowers the voltage at Q1's base, thereby turning Q1 on strongly, which turns Q2 on more strongly. This is the "latch" effect which must be broken for the circuit to reset to step 1.

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  • \$\begingroup\$ I want to ask you some question. 1.About the "5."why does the Q1 collector goes through R2, giving a voltage drop which turns Q2 on? Vgs>Vth(assume Vth almost=0),and the Q2 will turn on,but how to prove that Vgs will bigger than 0 \$\endgroup\$ – Shine Sun Mar 2 '18 at 13:32
  • \$\begingroup\$ Why does gnd connected to C1's negative plate via Q2 and thus the circuit begin working? \$\endgroup\$ – Shine Sun Mar 2 '18 at 13:34
  • \$\begingroup\$ C1 is effectively the circuit's power supply. Before step 5, U1 gets a voltage from C1 through Vin (pin 8) but there is no return path back to C1, so no current flows through U1, so it gets no power. When Q2 turns on, the current can flow through it back to C1. This means U1 has power. Ignore the gnd symbol under C2 - it has no meaning in this case. \$\endgroup\$ – Mr Central Mar 2 '18 at 15:24

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