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adding extra large resistor for getting a good simulation result

 

Hello, I have built a PI controller as shown in the photos below.
It was taken from a manual.
They say that 1Tera ohm was needed to be added so the simulation whould run properly and give more accurate result.
How can I know in what case I need to add such large resistor ?
Thanks.

https://groups.io/g/LTspice/files/z_groups.io/Files-sorted-by-message-number/msg_ZZZZZZ/john23%20%232/1.zip
 

On 13/12/2024 08:42, john23 via groups.io wrote:
Hello, I have built a PI controller as shown in the photos below.
It was taken from a manual.
They say that 1Tera ohm was needed to be added so the simulation whould run properly and give more accurate result.
How can I know in what case I need to add such large resistor ?
In this case, it's because your circuit is an integrator. Even with no input voltage, a real circuit will most likely eventually saturate at one or other of the rails due to input offset, which in reality is never zero. However, it might not as capacitors aren't perfect either, and have a leakage resistance, which depends on the dielectric used. In simulation, sometimes the input offset really is zero and the capacitor is perfect - without leakage, so it doesn't happen. This will depend on the opamp model. Some models do have offsets built in.

--
Regards,
Tony
 
Змінено

On Fri, Dec 13, 2024 at 02:42 AM, john23 wrote:
How can I know in what case I need to add such large resistor ?
Try the simulation and see what happens.  In an LTspice simulation, after doing your .AC analysis, hover the mouse pointer over the op-amp's output net and read the DC (operating point) voltage in the lower left corner.  If that voltage is close to 0, or within range of the op-amp to drive and remain linear, then it is OK.  That is a quick way to tell when you need to add the large resistor.
 
Ask yourself, what is the voltage gain of your integrator at frequency = 0 Hz?  For an ideal integrator, it is infinite, and even one extra electron would drive the op-amp to saturate at the rails.  The 1T resistor limits the voltage gain to less than infinity.  The op-amp also has finite open-loop gain, but a SPICE model could in principle have infinite open-loop gain at DC.
 
LTspice capacitors already have a default conductance (Gfarad) in parallel.  Its default is 1e-12 mhos, so LTspice already puts a 1T resistor Rpar across it.  That should work in most cases, making it unnecessary to add your own 1T if you use LTspice.  On the other hand, adding an extra 1T might help because then the effective Rpar = 0.5 T, making the closed-loop gain at frequency = 0 Hz half as much as it would have been without it.
 
Please only upload files to the group's "Temp" folder.  The photos you uploaded (to Files) do not show anything useful.
 
Andy
 
 

Hello Andy, yes i see 0V at the output.Integrator gain=Rf/R1
Rf is resistance parralel to the capacitor
so the gain is very large when Rf is very large.
so integrator gain is always very large in such case  if the feedback is only a capacitor.
Then why its normal to have 0V on the output?
Thanks.
 

On Sat, Dec 14, 2024 at 02:22 PM, john23 wrote:
Then why its normal to have 0V on the output?
In this particular case, with the LM1028 SPICE model, its input offset voltage is 0 and its open-loop voltage gain is finite, so that the output voltage at 0 Hz is 0 V.
 
Many (not all) op-amp SPICE models have no input offset voltage because zero is the average offset voltage.
 
Andy
 
 
 

John23
You could place a V source in series with one of the op-amp inputs.
Set it to a few uV to see the effect of input offset on your integrator.
Set the V source to the data sheet's maximum offset for your op-amp to view that result.
Then you may set the "Large R" to the correct value to keep a sane output under that condition.
 
All for now

 
 
Sent: Saturday, December 14, 2024 at 2:22 PM
From: "john23 via groups.io" <yafimvar@...>
To: LTspice@groups.io
Subject: Re: [LTspice] adding extra large resistor for getting a good simulation result
Hello Andy, yes i see 0V at the output.Integrator gain=Rf/R1
Rf is resistance parralel to the capacitor
so the gain is very large when Rf is very large.
so integrator gain is always very large in such case  if the feedback is only a capacitor.
Then why its normal to have 0V on the output?
Thanks.
 

john23,
 
You probably extracted this integrator from a larger circuit, and the larger circuit probably had overall feedback that prevented the integrator from  saturating.  Therefore, the complete circuit did not need the large feedback resistor.  By pulling the integrator out, you cut that feedback path from around the integrator, and now as a stand-alone circuit it is capable of saturating even with 0 input voltage, unless you add the large feedback resistor.
 
There should be SOME way of controlling the offset voltage of an integrator.  This is true in both real and simulated integrators - especially real ones.  If the integrator is sufficiently non-ideal so that the combination of input offset voltage, leakage currents, and gain does not send the integrator's output towards the rails, then you may be lucky and you don't need additional compensation.
 
The size of the added feedback resistor could be anywhere from megohms to gigaohms to teraohms.  Teraohms make sense only in simulations, since they are impractical in most real hardware.
 
The op-amp you used (LT1028) has bias current cancellation and its mean input offset voltage is zero, so both of those effects are effectively zero in the simulation.  If you built the circuit (without the 1T resistor) and hand-selected the LT1028, you could get the same result as the simulation.  But if you tried several LTt1028 samples, you would find that the output might saturate without the feedback resistor.
 
Andy
 

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