Is there any relation between the input and output of bridge rectifier that can explain this sentence?

Energy can be extracted independently of the excitation type and frequency, as long as $V_p$ exceeds the output buffer voltage $V_{buf}$,I am not understand what does this sentence mean,it seems that there is a relation,i don't know, between the input and output of bridge rectifier.Can anyone explain this to me?

By the way,i just saw the author said the $C_p$ limit the energy extraction when using an FBR(Full Bridge Rectifier),why?

As we know ,the remain voltage after passing the diode is the blue part,and V has relation with I,but i can't correspond it voltage to current,i mean,the second picture must have relation with the third picture,but i can't understand it. PEH=Piezoelectric Energy Harvesters

Energy can be extracted independently of the excitation type and frequency, as long as $V_p$ exceeds the output buffer voltage $V_{buf}$

Your load can extract energy as long as $C_{buf}$ has enough voltage. Using the energy stored in $C_{buf}$ of course depletes it and lowers its voltage.

The energy needed to charge $C_{buf}$ comes from the piezo element(s). These elements deliver an AC current, not DC. $C_{buf}$ needs DC so we need a rectifier. Here a bridge rectifier is used. It consists of 4 diodes.

At any time a maximum of two diodes will conduct, which ones depends on the direction of the current supplied by the piezo elements. These diodes are in series with the piezo elements and when operating in forward mode (conducting) diodes drop some voltage. For a silicon diode this is typically 0.6 to 1 V depending on the current.

The diodes will only conduct when the voltage across them is at least 0.6 V. As we have 2 diodes in series this adds up to 1.2 V. So $V_p$ has to be at least 1.2 V higher than $V_{buf}$ otherwise the diodes will not conduct and no charging current can flow.

Probably the text you quote assumes ideal diodes which do not exhibit the 0.6 V voltage drop but start to conduct as long as $V_{diode} > 0$. Then you could indeed day that $C_{buf}$ is charged whenever $V_p>V_{buf}$. Do note that ideal diodes do not exist!

• So if piezo element produce 1.5V AC,and as you said we have to have 1.2V to conduct two diode,so will the $V_{buf}$ become 0.3V?(1.5-1.2=0.3) – XM551 Apr 4 '18 at 13:16
• Yes that is correct, $V_{buf}$ will always be 1.2 V less than the highest value of $V_p$. – Bimpelrekkie Apr 4 '18 at 13:27
• I just realize that $V_{buf}$ is two diode drop,so to conduct two diode,so of course the input must bigger than $V_{buf}$ at least,right? – XM551 Apr 5 '18 at 0:12
• the author also said $C_p$ limit the energy extraction when using an FBR(full bridge rectifier),do you know why? – XM551 Apr 5 '18 at 1:31
• $C_p$ has to charged/discharged to reach the voltage $V_p$ that will allow the diodes to conduct. Suppose that $C_p$ has a very large value, then $V_p$ might never reach a usable voltage. – Bimpelrekkie Apr 5 '18 at 7:22

This means that the bridge will only pass current from the input to the output when the input voltage (whether positive or negative) exceeds Vbuf by two diode drops (1.5 to 2 V depending on how high the current is).

What the images try to represent (IMHO rather poorly) is that, given a sinusoidal input, the bridge will only charge Cbuf in the periods of the sinusoid where the voltage is high enough. In the other periods, when the input voltage is too low, the bridge will not conduct and the load will take it's current from Cbuf. That's why the voltage a Cbuf approximately resembles a saw tooth: going up during charge, going down during discharge.

• I just realize that $V_{buf}$ is two diode drop,so to conduct two diode,so of course the input must bigger than $V_{buf}$ at least,right? – XM551 Apr 5 '18 at 0:11
• the author also said $C_p$ limit the energy extraction when using an FBR(full bridge rectifier),do you know why? – XM551 Apr 5 '18 at 1:31