# How does this circuit automatically oscillate at the microblower's resonant frequency?

I'm in dire need of assistance with answering this question. I've exhaustively searched for hours only to find little. There's lots of answers here, but none that specifically answer this question. The question is: How does this circuit both find and automatically resonate at the microblower's resonant frequency? The closest answers I could find were in the patents of the driver circuit, and in this link (How to control the driving frequency when working with high power ultrasonic transducers). I understand that the op-amp is a rail-to-rail one, and both Q1A and Q1B make up a BJT transistor in a half-bridge type circuit; but I don't understand what the RC circuit does as well.

I'm a complete newbie with these things, so a thorough explanation would be greatly appreciated please!

• like this piezo element? ca.mouser.com/ProductDetail/Murata/MZB1001E00/… Your circuit has both negative and positive feedback. NEG is fixed ratio R7//C6 / R5. POS FB depends on load ratio to 15 Ohms may need adjustment. But if Positive ratio exceeds negative gain ratio, it becomes an oscillator determined by 1/(R15*Cload) for 1/2 cycle.. same as Schmitt trigger Astable osc. – Tony Stewart Sunnyskyguy EE75 Feb 2 '17 at 17:19
• @topher Did you understand anything I wrote? – Tony Stewart Sunnyskyguy EE75 Feb 3 '17 at 1:23
• @Tony Stewart You're explanation is amazing! I understand about half of what you wrote, but I'll read it a few more times and look up the definitions to a couple of the words used. I guess I'm still a bit lost, but I don't think it's possible to "dumb it down" any farther lol. So I'll do some learning on my own. – topher Feb 3 '17 at 21:26
• try either of the two Java simulations I made with (blue) linked text. Then point at the waveforms to see where they come from. , adjust speed to go in slow motion and change any value to see what happens. You can even increase the L value 100x and see how a series crystal resonates vs a relaxation oscillator. You can export and save links in your notepad. – Tony Stewart Sunnyskyguy EE75 Feb 3 '17 at 22:15

I assume the "microblower" (what is that?) is connected via the J1 connector? If so, then the the microblower is series resonant at it's operating frequency, and is providing positive feedback to the amplifier at that frequency. I assume that there is little or no DC continuity through the microblower.

The RC circuit (R7, C6) is providing negative feedback to stabilize the amplifier operating point. C6 lowers the gain at high frequencies to keep the amplifier from self-oscillating when the microblower is not connected. When it is connected, the positive feedback overwhelms the negative feedback, and the circuit oscillates.

• R7C6 has very little effect on the resonant frequency <1% with 10:1 range value change but suppresses stray noise and capacitance on R5 input. The dominant value is R6(22 ohm) and RsCs or ESR of piezo. R7/R5 ratio will affect frequency however as this defines the hysteresis threshold. So the explanation is not correct. – Tony Stewart Sunnyskyguy EE75 Feb 2 '17 at 19:20
• Tony, how much correctness do you think is needed here? The original question indicated that a graduate-level answer would probably be wasted. Plenty of things here affect the oscillation frequency. My answer provided the fundamentals (no pun intended) of operation. – Paul Elliott Feb 3 '17 at 0:44
• I think he is looking for the the critical factors that cause it to " resonate at the microblower's resonant frequency" Actually nothing forces it to find mechanical resonance but selection of R6 and R7/R5 ratio tunes the Relaxation effect which must be close to the actual resonance as it does not actually find it.. Your explanation defines how any resonance is possible but not the natural frequency of the piezo device. – Tony Stewart Sunnyskyguy EE75 Feb 3 '17 at 1:22
• Tony, I'm not sure that analyzing this as a relaxation oscillator is necessary. Consider R7/C6/R5 as providing negative feedback to set the DC operating point. We then look at the microblower as a series-resonant device. There will be in series some L, some C, and some R, with some overall parallel capacitance. There will be a series-resonant frequency where the phase and amplitude of the current through the microblower will provide positive feedback to sustain oscillation. R6 provides a load that sets the current through the resonant microblower, and feedback voltage for the amplifier. – Paul Elliott Feb 3 '17 at 1:41
• I concede it can work either way all depending on critical values of C,L,R – Tony Stewart Sunnyskyguy EE75 Feb 3 '17 at 1:53

I assume you know how a crystal oscillator works and how an Astable Multivibrator works. In this case it operates more like an RC Astable Osc with hysteresis.

If I understand your circuit, nothing "finds the natural resonance" but this circuit can be forced to oscillate with the positive feedback gain exceeding negative feedback gain with a piezo load in the +ve loop. You must tune it R5 and the other variables to adjust the frequency to match the mechanical resonance. This is unlike linear oscillators which do not use positive feedback and hysteresis like this design.

This microblower has similar characteristics of any crystal or resonator except bigger with the Series Motional Capacitance. I do not have any parameters for it. They all have an equivalent series and parallel resonant frequency due to a series motional capacitance which is very small (<1pF) in resonators used for stable clocks but very large C value when designed as a piezo pump. The series equivalent reactive elements in a crystal can be lumped as (Cs L Rs)//Cp equivalent circuit. The anti-resonant 180 deg phase shift for parallel resonance is not used here and is not discussed further but is common to CMOS clocks.

In order to pump high reactive power into the piezo device at a relatively low frequency to pump a fluid , we need an "Astable Multivibrator" equivalent circuit. This is often done with hysteresis and negative feedback for DC bias or positive feedback with AC coupling. In either mode the conditions for oscillation must be met and in this case the relaxation effect is the movement of the pump which has some interface series resistance and motional capacitance and leakage resistance. We can neglect the series equivalent inductance here.

The primary sources of relaxation oscillation are your R6 which effectively sense and limits current to the pump and the ReqCeq of the piezo pump.

Relaxation Effect.

The -ve feedback loop in this case sets the gain and the threshold level of hysteresis for the +-ve feedback signal to cross after decaying to >5T depending on gain ratio. When the pump current sense voltage drops below this fixed -ve input offset threshold, the output toggles along with the fix Vin- threshold.

I created a simulation using +/-10V where the R ratio 1.5k/270k= 55mV. This is the target threshold of hysteresis for this relaxation oscillator design.

This may give a longer stroke pump with 6T or more RC decay to 5.5% threshold of peak. We expect displacement to be the 2nd integral of motor current. I estimated the internal micropump ESR to be 1 Ohm, but it may be much higher, which of course affects the resonant frequency.

I hope you can figure out the rest on your own. My assumptions for the micro pump were 1 ohm and 1uF , but you can change these values in my simulation at any time. (left click edit in java simulation link above)

The formula for f period will be the exponential decay of T=(R5+Rc)C Blower on the positive feedback loop to the threshold defined by the negative feedback loop where it is shown by your example to be <1% Thus RfCf time constant must be much shorter than the oscillating period. <<10% to then become irrelevant. Op Amp must be capable of a large swing as well as high current, not common in Rail-Rail Op Amps , so the 470 Ohm output R of my simulation may be as high as 3k for some Op Amps and a CMOS or half bridge output may be better as this limits current pump thru high hFE transistors.

I derived this (verified by simulation), for Rs=ESR and C of piezo and R= R6+Rs then H=R5/R7 is the hysteresis level, which in linear circumstances (not here) would be Av-. and k ideally would be 1/2 but due to switch losses drops to k=0.27

$f = {H}^{\frac{1}{4}}\frac{k}{RC}$ H= 0.1%(noise sensitive) to 10% (more lossy pump) k=0.27

The conditions to be satisfied are;

• CM input range bias must satisfy single or dual supply to V/2
• Op Amp current capacity * hFE must exceed demand current of Piezo blower ( or output Z <1k ) or transistors may be replaced with Darlingtons
• choose C8 compensation such that C8R7 < 10% T of oscillator.

Here's another SIM with different values and a Gain pot. Note if Cf gets too large, it wont oscillate from above criteria.

• Since we neglected the physical mass resonant frequency, R5 will have to be tuned to the device to optimize the blower effect. – Tony Stewart Sunnyskyguy EE75 Feb 2 '17 at 21:25