how to make bpf oscillate at the center frequency

for a narow-band band pass filter(bpf), the bpf gain has the maximum gain at center frequency f0. now, i want to make the bpf oscillate at f0, how to do it? what i thought is that, if the loop becomes positive feedback, then it can oscillate at the center frequency f0. however, for a second-order bpf (active RC structure with feedback loop), how to make it a positive feedback loop? how to disconnect the bpf loop correctly? some advice will be appreciated, thanks.

Ihlblue, the answer is simple.
You must create a loop that fulfills Barkhausen`s oscillation condition, which consists of two parts:

1.) Loop phase: The phase at a non-inverting bandpasss at f=fo is zero. Thus there must be no additional phase shift in the loop to be realized (an inverting bandpass requires another inversion).
2.) Loop gain: If the gain at f=fo is Ao you must close the loop with a gain stage 1/Ao (this gain stage is a damping stage in case Ao>1).

The best design is a bandpass with Ao=1 (inverting or non-inverting). In this case, you create a loop with another unity gain amplifier (inverting or non-inverting, respectively).

- - - Updated - - -

Accidently, I have a circuit diagram available for a bandpass oscillator based on a multi-feedback topology (attachement).

The center frequency for Ao=-1 and R2=2R1 is

wo=(1/R1C)*sqrt[(Ro+R1)/2Ro].

The oscillation condition is fulfilled for RA=RB.

Practice: Make RB some percent larger than Ra and use gain stabilizing diodes (or any other gain stabilization method).

The key is to find a continual way that will:

(1) take a bit of energy from the bpf during each cycle,
and
(2) in turn to give it a slight 'kick' to keep it going.

Several topologies contain an LC tank loop, which has a particular resonant frequency.

Some topologies contain a coil and capacitor in series.

Sometimes the coil is tapped, or the capacitor is split into two series capacitors.

There is also a type based on the twin-T (even though it is typically used as a notch filter rather than bandpass).

The amplifying component is usually transistor or op amp.

The bandpass network might be connected in various ways to the amplifier. Again the key is to add resistors, etc., in the proper combination, so as to achieve just the right balance of give-and-take in the bandpass network.

If I understood it well, Bradthe Rad has presented a more or less interesting "philosophical" view of the oscillator problem.

As an example, the sentence

(Quote) Again the key is to add resistors, etc., in the proper combination, so as to achieve just the right balance of give-and-take in the bandpass network.

refers to the well-known Barkhausen condition, which is the mathematical form of the "right balance".

Ihlblue, you have asked for a circuit based on an active bandpass. This was the basis of my response.
Perhaps i should add that the circuit I have given with two opamps offers the possibility to have two outputs in quadrature at the same time (which sometimes is required).
Of course - as indicated by Bradtherad - you can use a passive bandpass and an amplifier which results in an one-opamp circuit.
There are many realizations of this concept. One of the best known alternative is the WIEN oscillator with a bandpass midband "gain" of 1/3 and an amplifier with a gain of (nominal) G=3.

Loop phase: The phase at a non-inverting bandpasss at f=fo is zero.
about the above comment, i have some doubts.
in bandpass-oscillator.pdf, when the input of multi-feedback filter has an increase, the output of inverting amplifier (also the input of multi-feedback filter) will increase also, then, finally, the output of inverting amplifier (or the input of multi-feedback filter) will be 'latched' to high level. so, oscillation will not occur.
similarly, for a three-inverter ring oscillator, the total loop phase is 540 (or 540-360=180), not 0. but it can oscillate. why?
pls explain to me where is wrong for me.
thanks.

Ihlblue, your "counter argument" regarding signal increase applies to each oscillator. But it isn`t really a counter argument. Why not?
Remember the oscillation condition: Loop gain=1 (0 dB).
In reality, this ideal condition can never be fulfilled.
Therefore, the following design approach is used:
* Design a loop gain at f=fo somewhat larger than unity. Thus, the oscillator will start safely (closed-loop poles in the RHP).
* As a result, the oscillation signal will continuosly increase (rising amplitudes) until limitation occurs at the supply rail (signal clipping)
* To avoid this clipping effects (and the associated distortions) an amplitude stabilizing circuitry is used (diodes, thermistor, FET control loop).

As a consequence, a signal increase (as mentioned in your post) will cause a corresponding gain REDUCTION - and the signal will decrease again.

This - or a similar - explanation how a linear oscillator works can be found in each textbook or internet-contribution dealing with oscillator principles.

i have some doubts also.
in my simulation, bpf f0=38K, gain@38k=5, then i add a non-inverting amplifier to form a loop, but the oscillating frequency is related to the loop gain. when loop gain=5, oscillating frequency is only 27K, until loop gain=1.05, oscillating frequency = 37.7K.
i want the oscillating frequency be equal to f0(38K) precisely, how to do it?
besides, the loop gain will change with process and temp, how to deal with this?
thanks.

- - - Updated - - -

how to make loop gain=1 precisely in all pvt (process, supply voltage, temp)?

lhlblue said:

i have some doubts also.
in my simulation, bpf f0=38K, gain@38k=5, then i add a non-inverting amplifier to form a loop, but the oscillating frequency is related to the loop gain. when loop gain=5, oscillating frequency is only 27K, until loop gain=1.05, oscillating frequency = 37.7K.
i want the oscillating frequency be equal to f0(38K) precisely, how to do it?
besides, the loop gain will change with process and temp, how to deal with this?
thanks.

- - - Updated - - -

how to make loop gain=1 precisely in all pvt (process, supply voltage, temp)?

Click to expand...

Your design value is 38kHz - and the actual frequency is 37.7 kHz?
And you are not satisfied? It is a deviation of 0.8 %.
Do you really expect more acuracy?
What are the tolerances of your parts (resistors, capacitors)???

how to make loop gain=1 precisely in all pvt (process, supply voltage, temp)?

If you could, you would win the Nobel price, perhaps.

maybe i forget to tell something, the bandpass filter gain@38K is 5, but i cannot get the formula of gain @ 38K (not a simple ratio of R2/R1), and the non-inverting amplifier gain is equal to 0.2 in my design, but for some case, bandpass filter gain @ f0 may change a lot. for example, when R has a violation of 30%, bandpass filter gain@ f0 may achieve 8 or 4 (and loop gain may be less than 1), so, the loop oscillation frequency will not be equal the f0 desired. how to deal with this?
f0=38k bpf gain @ 38k = 5
f0=50k bpf gain @ 50k = 8 (R has a negative violation 30%)
f0=29k bpf gain @ 29k = 4 (R has a positive voilation 30%)

- - - Updated - - -

besides, LvW, you are so humor! :lol:

- - - Updated - - -

and is there any other method to make the bpf oscillate at f0?

Ihlblue, I don`t understand your design strategy.
If the bandpas gain at fo is 5 you need another stage with a transfer ratio of 1/5=0.2.
However, I am surprised how you are able to realize a non-inverting amplifier gain of 0.2 (as mentioned by you).
Please post your circuit.

ok. my bpf is non-inverting, in order to obtail 0 loop phase, i add a non-inverting amplifier to close the loop, the amplifier is in the link(first circuit, and in chinese version):
**broken link removed**
gain=R2/2R1.
i want to detect the f0 with process violation, when without voilation, the oscillation frequency should be 38K (1% precision), when there is process voilation (R or C), the bpf f0 can change, so i want to observe the change of oscillation frequency by the loop to calibrate f0 to 38K. that is, i want the bpf f0 be 38K (1% precision) after calibration (by adjusting C value).
but when there is process voilation to make f0 have a change, the bpf gain @ f0(which has already changed) will change a lot.
for example:
f0=38k bpf gain @ 38k = 5
f0=50k bpf gain @ 50k = 8 (R has a negative violation 30%)
f0=29k bpf gain @ 29k = 4 (R has a positive voilation 30%)
now, loop gain may be larger or less than 1 (relative to case : bpf gain @ 38k = 5), then, the oscillation frequency will not be the desired value, or not oscillate at all. so, i cannot calibrate the changed f0 to 38K again.
what can i do?
is there any other way to make the bpf loop oscillate at f0?
thanks.

- - - Updated - - -

besides, if loop gain is larger than 1, why the oscillation frequency will change (not @ f0)? and the larger the loop gain, the larger the oscillation frequency voilation (relative to f0). what is the reason?
in my understanding, if loop gain is larger than 1, the oscillation frequency will be f0 also, just the amplitude will become very large, but can be limited by supply rail. where is wrong for me?

lhlblue said:

................
besides, if loop gain is larger than 1, why the oscillation frequency will change (not @ f0)? and the larger the loop gain, the larger the oscillation frequency voilation (relative to f0). what is the reason?
in my understanding, if loop gain is larger than 1, the oscillation frequency will be f0 also, just the amplitude will become very large, but can be limited by supply rail. where is wrong for me?

Click to expand...

* Sorry, but I cannot see any band pass filter in the given link. Thus, I cannot give any answer. All the circuits contain resistors only.
* In general - when the loop gain is >1 the poles move into the right-hand part of the s-plane and the frequency changes a bit (not very much).

As far as I have understood your task, you want to have an oscillator with a NOMINAL frequency of 38 kHz. Then, when some parts change their values you want to measure/indicate this change by observing the oscillation frequency. Is this correct? And your problem is that not only the frequency but also the loop gain is altered. Correct?
In this case, it is absolutely necessary to see the circuit because some elements have influence on the gain and some others not.

ok, thanks LvW for your reply, and your understanding of my purpose is right.
the link above is just the circuit of non-inverting amplifier.
for the bpf circuit, due to the encrypt consideration, i cannot upload the circuit, but it is similar to multi-feedback topology, and the transfer function is very complicated, from which, i cannot get a simple formula for gain expression (8 Resistor and 5 Cap in total).
so, how to suppress the alter of loop gain with process etc.?
and, besides the above mentioned method, is there any other way to make the bpf loop oscillate at f0?

- - - Updated - - -

besides, when resistor has a negative voilation to make f0 larger than 38K, Q can become larger too, and gain @ f0 (which has already get larger) becomes larger too.

Ihlblue, I am afraid that I cannot give you any help.
When - for a certain reason - you are required to use this uncommon oscillator ciruit (with FIVE capacitors), you are the only person who can find out how the gain and the frequency depends on parts values.
Without the circuit diagram, nobody else can say anything.

.....besides the above mentioned method, is there any other way to make the bpf loop oscillate at f0?

No, there is only one method: Loop gain=1. Or, what do you mean with "other method"?

ok, i know that.
if the bpf gain=R2*R3/R1/R4, and non-inverting amplifier gain=0.2, then, what should i do to suppress the loop gain alteration with process?
besides, is OP bandwidth or slew rate related to the loop gain alteration?

lhlblue said:

ok, i know that.
if the bpf gain=R2*R3/R1/R4, and non-inverting amplifier gain=0.2, then, what should i do to suppress the loop gain alteration with process?
besides, is OP bandwidth or slew rate related to the loop gain alteration?

Click to expand...

What do you man with "process"?

"process" just means process corner (tt, ss, ff, etc), thanks.
LvW, your advice makes me more clear about my problem, thanks.