Misunderstanding from the author saying,about the current flowing path

The author show the working process of this schematic

1.The proposed system waits until the PZT device stores the energy vibrations produce during the positive half of vibration period $T_{VIB}$ before engaging switches $S_I$ and $S_N$ to discharge $C_{PZT}$ into off-chip harvesting inductor $L_H$

2.The inductor current $i_L$ rises to reach the peak, then the switches $M_{P1}$ and $S_I$ de-energize $L_H$ into the battery via $i_{BAT}^+$

3.The system again waits for the transducer to energize $C_{PZT}$ (in the negative direction) through the negative half of $T_{VIB}$ before prompting $S_I$ and $S_N$ to discharge $C_{PZT}$ into $L_H$, after which $M_{P2}$ and $S_N$ deenergize $L_H$ into the battery

I sort out what the author said:

1.the PZT will produce energy first,then $S_I$ and $S_N$ be "on" to let the current flow from $C_{PZT}$ to $L_H$

2.$M_{P1}$ and $S_I$ disconnect,and let the $i_{BAT}^+$ flow from the $L_H$ to the battery

3.when the PZT produce the negative part voltage,$S_I$ and $S_N$ are "on",and the negative current flow from the $C_{PZT}$ to $L_H$(this current direction is the oppsite of 1. current direction in the $L_H$ ),then $M_{P2}$ and $S_N$ disconnect to let the current flow from the $L_H$ to the battery

Hold on a second,i have a question about the 2. and 3.,if the $M_{P1}$ and $S_I$ disconnect,why can the $i_{BAT}^+$ flow from the $L_H$ to the battery?why can $M_{P2}$ and $S_N$ disconnect to let the current flow from the $L_H$ to the battery?It seems that i misunderstand something from the author saying,but i don't know which part do i misunderstand,can anyone correct my mistake?

1 Answer

After +ve swing of Cpzt: Dump charge into L current then release low side of +ve current as V+ into Vbat with CP1.

After -ve swing of Cpzt: Dump -ve charge into L current then release hi side of -ve current also as V+ into Vbat with CP2.

• Sir,i read the some part of the paper,the schematic is also from this paper,can you tell me your thinking or correct my mistake if i am wrong? – electronic component Mar 31 '18 at 3:16
• I corrected your thinking in my answer. It’s a full wave rectifier – Sunnyskyguy EE75 Mar 31 '18 at 4:11
• but sir,the topic of this paper is "A 2um BiCMOS Rectiﬁer-Free AC–DC Piezoelectric Energy Harvester-Charger IC " – electronic component Mar 31 '18 at 4:50
• I did not read topic, but I can read schematics. This is a boost flyback rectified current from a sinusoid piezoelectric. Charge up voltage is shunted to inductor L current then switched to charged Vbat, based on current polarity. – Sunnyskyguy EE75 Mar 31 '18 at 5:18
• CP1 and CP2 are "active rectifiers", not rectifiers, not pn junctions. Their Vf can be far below the 0.7V of a junction rectifier. The circuit is not rectifier-free. Instead it avoids use of pn junction rectifiers. – wbeaty Mar 31 '18 at 5:21