Shorting opamp integrator output to rail (or target voltage) - allowed? - Page 1
Forum section: electronics / beginners

author: CircuitBreaker
author_rank: Contributor
timestamp: February 04, 2022, 09:32:21 am
content: Dear community,For a certain type of (audio) oscillator, I want to be able to reach either max or min output volume of an integrator opamp as quickly as possible.Up until now, I use a trigger to switch theinputof the integrator between the 'regular' slope value and min/max input voltage (+/- 5V) to let the integrator reach the corresponding max/min value that the integrator allows.My question at this point: is this theonlycorrect way to do this? Or is there a quicker way, like e.g. directly connecting the output of the integrator to either min/max voltage and then disconnect, to let the integrator start integrating from that new voltage?--> Note, that I would then like to use an integrator circuit like described in Figure 4 in Bob Pease's treatise:https://www.edn.com/bob-pease-on-bounding-and-clamping-techniques/, but then dual rail, around 0V

author: Kleinstein
author_rank: Super Contributor
timestamp: February 04, 2022, 09:57:02 am
content: To bring a integrator fast to a low value (near zero) one can have a switch in parallel to the capacitor - if needed this can be relatively low resistance.To reach a large output values it does not help to force the ouput directly to a value, as one would still need to charge the capacitor. So the method of choice would be a relatively beefy signal to the input. This can be a higher current than under normal use, e.g. with a separate path for the input current.So not just switching the input voltage before the integrator resistor, but send extra currrent from a seprate smaller resistor.To make it faster one could reduce the integrator capacitance and use thus less current in normal use.

author: CircuitBreaker
author_rank: Contributor
timestamp: February 04, 2022, 05:41:54 pm
content: Thanks for the helpful comment!Can I switch such an additional input with - say - a 4053 as well? I believe they are low-resistance enough to provide more current, correct? Or would you propose an alternative way to switch that higher current input?EDIT: O, wait, I see that those devices can support a max current of 10(s) of mA. That would perhaps be insufficient.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 04, 2022, 06:11:04 pm
content: The 4053 type (more like 74HC4053 or 74LV4053 for lower resistance) switches are Ok for switching the integrator input. The integrator input is at an essentially constant voltage and this no neede for a higher voltage swtich. For an OP integrator 10 mA are already a lot. The OPs output can not deliver much more current anyway. The integrtor speed also depends on the size of the integrator capacitor - to make it faster use a smaller capacitor and less current for the normal use. Going above 10 mA is hard. Going down to the nA range at the other end may be easier.What is considered fast in this context ?

author: CircuitBreaker
author_rank: Contributor
timestamp: February 04, 2022, 06:35:55 pm
content: It's for an audio oscillator, and it should serve as the "steep flank" for a pulse wave.So, quite arbitrary, but ideally it should go from (clamped) rail to rail in ca. 10-20 µs or so... (where I arbitrarily choose those to be +/- 5V or so)--> and no, I don't want to make pulsesonly, it should be a variable slope oscillator of which the steepness shall go "up to a pulse" waveform.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 04, 2022, 07:03:48 pm
content: For a slew rate of some 2 V/µs with a 2 mA current to the integrator the integrator capacitor of 1 nF. This a well reasonable size and with low leakage parts (e.g. FET OP) one can still hold a state reasonable well for up to 1 second or some.The alternative would otherwise be to keep the oscillator as a triangle and switch the waveform after the oscillator, a bit like old analog function genrators worked.

author: rstofer
author_rank: Super Contributor
timestamp: February 04, 2022, 07:36:43 pm
content: There are two components to the integrator timiing chain: the input resistor and the capacitor. It's too hard to change the capacitance but it's pretty easy to have some resistors in series and then short one or more of them as desired. This can be done with a mechanical switch.For analog computing, it is typical to use a 1 ufd capacitor and a 1 Mohm resistor and provide a separate 100k parallel resistor to multiply by 10. There are usually multiple inputs for each integrator and a summing junction keeps it all straight.Pick an RC combination that gives you the maximum desired slope and then add in extra series resistors to slow it down.

author: CircuitBreaker
author_rank: Contributor
timestamp: February 06, 2022, 02:21:58 pm
content: Thank you all for the replies. Have much to think about (and tinker with) the upcoming weeks!