I want to use a TL084 as an integrator (non inverting mode) as seen below. The op amp is supplied with +-9V, and will reach a chosen max ramp voltage before being reset digitally. What practical limitations determines a sensible maximum ramp voltage? The best reason that I have come up with is is that the Common Mode voltage is between +-4v, Therefore it would be best to have around 4V max ?

TL084 Datasheet

 

What common mode voltage?
I don't see any common mode signal in your circuit.

 

That is vcc+ -4V so if the opamp is upplied with +/-9V, then the recommended common mode voltage is -9+4=-5V to 9V-4V=+5V.

 

Thank you, so does the common mode not relate to my circuit? or is 5V a good limit to base it from ? I just want to know what a good output max would be and why

 

I want to use a TL084 as an integrator (non inverting mode) as seen below.

That's inverting mode, not non-inverting. A positive input voltage will cause a negative-going ramp, and vice versa.

The op amp is supplied with +-9V, and will reach a chosen max ramp voltage before being reset digitally. What practical limitations determines a sensible maximum ramp voltage?

The parameter that controls this is the Maximum Peak Output Voltage Swing:



On +/- 15V supplies, the output can go to +/- 12V under light load; i.e., it can go to within 3 volts of the supply rails. This difference is more or less constant with different supply voltages, so with +/- 9V supplies you can expect the output to go as far as +/- 6 volts. To be safe, plan on no more than a +/- 5V ramp.

The best reason that I have come up with is is that the Common Mode voltage is between +-4v, Therefore it would be best to have around 4V max ?

Input Common Mode Voltage has absolutely nothing to do with the maximum available output voltage range. And in fact, the input common mode voltage in your circuit is always zero, anyway so the inputs are always within the allowed range with +/- 9V supplies.

 

Capacitance loading of an op-amp output can compromise the output slew rate of the device.

The TL084 has a respectable output slew rate of 8V/us.

The voltage across a capacitor due to a constant current is given by:-

V = (I x t)/C

The specified maximum output current of the TL084 can be a low as 10mA.

Plugging this value into the above calculation gives a maximum capacitance loading of around 1nF before the device slew rate will begin to slow.

With a 100nF load, the slew rate will reduced to 0.1V/us and at 1uF be 0.01V/us.

 

Looking at two different datasheets Iol Current limit max is 27 mA, typical
output swing vs load curves show 20 mA in linear operation.

TI and Analog Devices datasheets.

Regards, Dana.

 

Capacitance loading of an op-amp output can compromise the output slew rate of the device.

The TL084 has a respectable output slew rate of 8V/us.

The voltage across a capacitor due to a constant current is given by:-

V = (I x t)/C

The specified maximum output current of the TL084 can be a low as 10mA.

Plugging this value into the above calculation gives a maximum capacitance loading of around 1nF before the device slew rate will begin to slow.

With a 100nF load, the slew rate will reduced to 0.1V/us and at 1uF be 0.01V/us.

Excellent, I didn't truly understand the meaning of slew rate. And I guess this is a good value since any lower C values will cause larger affects of bias currents due to larger resistor values, correct?

 

"Normally" slew rate refers to internal compensation capacitor derived slew
limitations.

https://www.edn.com/electronics-blogs/the-signal/4415482/Slew-Rate-the-op-amp-speed-limit

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=19&ved=2ahUKEwjLha_ipa7aAhUKSN8KHXbTCRMQFjASegUIARCyAQ&url=http://www.ti.com/lit/an/snoa852/snoa852.pdf&usg=AOvVaw3Zb1oBpx-HjvvrawsW-JLp


Regards, Dana.

 

apologies, im not sure why my comments are posting

I found the maximum peak output voltage vs load, but no current vs load that shows the 20mA?

 

Page 5 figure 9

https://www.google.com/url?sa=t&rct...L082_084.pdf&usg=AOvVaw38fG_1lxO1Q2c_RPLDsdDv


Regards, Dana.

 

Sorry, I have a TI Tl084, so im guessing it should be 10mA as hymie states. I can't find 10mA anywhere, and the output voltage vs load graph doesn't show it (to my understanding) either. its figure 5:

http://www.ti.com/lit/ds/symlink/tl082.pdf

they way that I am solving it is, look at my maximum voltage which is 5V, load at this point is 150ohm, giving about 30mA. much too high no?

 

I would say yes load a little stiff. What is max frequency and max Vout_pk-pk you
want ?

There are OpAmps that have greater drive capability, and slew rate and BW.

Regards, Dana.

 

I would say yes load a little stiff. What is max frequency and max Vout_pk-pk you
want ?

There are OpAmps that have greater drive capability, and slew rate and BW.

Regards, Dana.

Vp-p is 5V, max frequency is around 4000Hz, all driven by a 9V battery. how do I work out the affect of the bias current then ?

 

When an op amp is used as an integrator, it is not loaded with a capacitance in the normal sense and slew rate is not an issue unless the ramp rate is starting to approach the amp's slew limit. The output current of the amp can be a limiting factor if requirements are very unusual, for example where the same integrator must operate over many decades of frequency. If the capacitor charging requires significant current, it is usually best to buffer the amp output if the external load also needs some current.

The current through the integrator capacitor is exactly equal to the current through Rin in the circuit shown (ignoring amp input bias current, which is very low for the TL084) and is completely independent of the value of the capacitor.
Example:

Rin = 10k, C = 100 nF, Vin = 2 volts
- current through Rin = 2/10k = 200 µA - this is also the current through the cap and into the amp output, since it has nowhere else to go
- cap voltage changes according to δV/δt = i/C = 200e-6 A/100e-9 F = 2000 volts per second or 2 V/ms or 2 mV/µs
At 30 volts in (perfectly acceptable with inverting amp, as long as amp output hasn't hit the rail) the charge current would be 3 mA and the ramp rate would be 30 V/ms. If you wanted the ramp peak voltage (remember it is negative if Vin is positive) to be 6 volts, it would take 6/30 ms to reach the peak, assuming you started at zero, so that yields 5 kHz if the ramp reset time is zero.
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If you are dealing with the audio spectrum and want a reasonably high peak ramp amplitude, you would probably want to reduce the capacitance in the example to keep the current well under the limit that the amp can handle at its output. The input bias current, previously ignored, can become a significant error term, but because is so low with FET input amps you have a lot of room to maneuver (the lowest ramp rate of interest becomes the limiting factor since it requires the lowest current). Polypropylene film capacitors are a good choice for the integrating cap. C0G ceramic are usually pretty good.

A 6 volt peak ramp at 20 kHz is only 0.12 V/µs - far far below the amp slew rate spec, though slew rate may need to be considered for very fast ramp reset if you are trying to produce a sawtooth (as implied here by setting the entire period to the ramp time). Integrator reset for sawtooth generation is usually done by shorting the capacitor with an external device such as a FET, but the amp still needs to be able to keep up.