LM3886 overshoot

[dr.power]

Hello guys,

I have an LM3886 driving a capacitive load piezo at say 40KHz by a bandwith of say 10KHz.
the pirzo load has a capacitance of almost 0.2uF. I have the problem of ringigng/overshoot at the output of the LM3886 when connected to the said load (the load has an impedance of say 5ohms).Can you guys tell me how to remove/reduce the overshoot problem plz?
the datasheet of the chip suggest using an LR circuit in series with the load (0.7uH & 10 ohms). I did s but the result was not promising. ANy idea plz? did I used the wrong values of lr for my chip at the said freq?
By the way what about the Q of the L used here? How to choice it? I do not know if it is better to have a highe or the lower q for the L here?

Thanks a lot

P.S As my load is capacitive do I need to use any zobbel netwoork too?

 

[dr.power]

Finall I could solve the problem of overshoot by putting a 70uH inductor in series with the piezo load.
the piezo as I said has a capacitance of 0.2uF and an impedance of 5ohms. But is it just right putting such a high inductor in series with the piezo load? The SPL of the piezo load seems to be higher by putting the said inductor. but I am afraid of damaging the piezo load by the inductor, Am I right? furthermore what will happen to the bandwidth of the load by putting such an inductor in series with the piezo load? I guess that the bandwidth would be reduced but I think I read somewhere that the bandwidth of the piezo load would be incrased by adding an inductor to the piezo loads. Any idea please?

Thanks

 

[Nigel Goodwin]

It's pretty standard practice to have a resistor/inductor (few turns of wire round a 10 ohm resistor) in series with the load, do you also have a zobel network?.

Why is overshoot a problem? - what are you trying to do?.

 

[ronsimpson]

Dr. Power,
Your 0.2uF and 70uH will resonate at 42khz.
You are trying to drive 40khz.
When you say 10khz bandwidth, do you want to get 30 to 50khz response?

 

[dr.power]

It's pretty standard practice to have a resistor/inductor (few turns of wire round a 10 ohm resistor) in series with the load, do you also have a zobel network?.

Why is overshoot a problem? - what are you trying to do?.

I tried to put a 10ohms in parallel with a 1uH inductor in series with the load but no affect on the overshoot. I tried to change the inductor and use the higher inductors, I could get an almost nice sine wave when I used a 60-70uH inductor in series with the load, The waveform started to change a bit from sine when I tried to add a 10ohms resistor in parallel with the said inductor but no overshoot. But Now I fear that I burn the load at higher frequenccies due to the 60-70uH inductor.

Regarding using a zobel network, as my load IS NOT INDUCTIVE but capacitive then I think I do not need a zobel (ie a resistor in series with a cap across the load) network.

 

[dr.power]

Dr. Power,
Your 0.2uF and 70uH will resonate at 42khz.
You are trying to drive 40khz.
When you say 10khz bandwidth, do you want to get 30 to 50khz response?

Right, the resonance freq of the LC as you told is almost 40KHz. the input of the LM3886 is drived by an square wave oscillator at just 40KHz, Do you think there would be any problem if I have an resonate circuit (The LC I mean) at the ouput of the LM3886 too?

Yes I am talking about 30 to 50kHz.
I am trying to make an audio generator out of the ultrasound,

 

[dr.power]

furthermore I would like to know how to calculte the power of the amplifier when the inductor is connected to it and makes resonance? is the power of the amplifier happens to be higher by adding the inductor in series to the capacitive load?

regarding to incrasing the bandwidth at resonance after adding the inductor I noticed that adding a damping resistor in series or in parallel to the inductor can causes the bandwidth to be high enough to support the bandwidth of say 10 or 20KHz, For instance if the inductor happens to be 67uH and the series/parallel resistor happens to be 10ohms then the bandwidth would be almost 23kHz or for R=20ohms the bandwidth would be 47kHz.

 

[unclejed613]

a couple of questions... what gain are you operating the LM3886 at? (the LM3886 is not unity gain stable) have you paid close attention to what the datasheet says about bypassing and lead dress?

one of the things you are likely running into, is that you don't have a lot of open loop gain at 40khz. at closed loop gains of 5 or less, the amp will also be unstable. some of the overshoot and ringing may be from the transducer, and not necessarily the fault of the amp.

also re-read this section:

REACTIVE LOADING
It is hard for most power amplifiers to drive highly capacitive
loads very effectively and normally results in oscillations or
ringing on the square wave response. If the output of the
LM3886 is connected directly to a capacitor with no series
resistance, the square wave response will exhibit ringing if
the capacitance is greater than about 0.2 μF. If highly capacitive
loads are expected due to long speaker cables, a
method commonly employed to protect amplifiers from low
impedances at high frequencies is to couple to the load
through a 10Ω resistor in parallel with a 0.7 μH inductor. The
inductor-resistor combination as shown in the Typical Application
Circuit isolates the feedback amplifier from the
load by providing high output impedance at high frequencies
thus allowing the 10Ω resistor to decouple the capacitive
load and reduce the Q of the series resonant circuit. The LR
combination also provides low output impedance at low
frequencies thus shorting out the 10Ω resistor and allowing
the amplifier to drive the series RC load (large capacitive
load due to long speaker cables) directly.

 

[dr.power]

a couple of questions... what gain are you operating the LM3886 at? (the LM3886 is not unity gain stable) have you paid close attention to what the datasheet says about bypassing and lead dress?

one of the things you are likely running into, is that you don't have a lot of open loop gain at 40khz. at closed loop gains of 5 or less, the amp will also be unstable. some of the overshoot and ringing may be from the transducer, and not necessarily the fault of the amp.

also re-read this section:

The gain of the LM3886 in my cause is 13.

I have used 100nF caps as bypass caps for the Chip as close as possible to the chip.
I am sure that the problem is due to CAPACITIVE load....

 

[unclejed613]

remember, op amps act as an active inductor, all by themselves. at 40khz, you only have 30db or so of negative feedback headroom left, so the amp is having some difficulty controlling it's output. with a 90 degree phase shift, this allows some "slop", which shows up as overshoot. if you were driving with sine waves, this wouldn't be so obvious, but you're driving with square waves and the amp takes a few microseconds to regain control of it's output after what might be a slew rate limited voltage swing, during which the feedback has no control whatsoever over the output. if you slew rate limit the input waveform so that the output is not slew rate limited (by the amplifier), the overshoot might clear up a bit.

 

[dr.power]

remember, op amps act as an active inductor, all by themselves. at 40khz

Why An op-amp acts like an inductor? I never heard something like that!

slew rate limit the input waveform so that the output is not slew rate limited (by the amplifier), the overshoot might clear up a bit.

How can I do a "slew rate limit" plaese?

Thanks

 

[unclejed613]

look at the Bode plot. the gain decreases linearly with frequency, which is very similar to the behavior of an inductor (except an inductor has no gain). an amplifier's output impedance increases at the same rate with frequency, which is another property of an inductor.

this app note is about driving capacitive loads, and can explain a lot of the things you are seeing.
https://www.electro-tech-online.com/custompdfs/2012/04/00884b.pdf

if you look at figure 4, you will see the op amp modeled as having an inductive output impedance.


slew rate limiting happens when the op amp (or in this case, a power amp, which is just a BIG op amp) can't keep up with changes in the input waveform. this is called the amplifier's "step" response. or slew rate. this figure is expressed in V/uS (volts per microsecond) in the data sheet. if the input step goes from -1V to +1V in 1 microsecond, and the gain of the amplifier is 10, then to accurately reproduce that voltage change, the amplifier needs to have a slew rate of 20V/us (input change=2Vin 1uS times gain). if the amplifier only has a slew rate of 10V/uS, then the output can only change that fast. the output becomes a ramp at the 10V/us rate, and the feedback has no control over the output during this time because the op amp can't keep up with the input. with square waves this is a problem. the rising and falling edges become more sloped than the input, and not always symmetrically either, since often the positive-going and negative-going slew rates are different. with sine waves, this doesn;t become a problem until you hit the edge of the amp's bandwidth.