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Function Generator Output Source/Sink Current Max ?

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I agree with Audioguru that there's nothing against using an OP with better performance.


F := 5000 ;Vp:= 20
-----------------------------

SR:= (2*3.142*F*Vp)/1E6

SR=[628.4m] - 0.63V/µs required
Up to now, the specification was 20 Vpp or 10 Vp, I think. Refers to 0.31 V/µs.
 
In some parts around the world, I was told by friends it was not that easy to obtain anything out of the ordinary. I am used to getting components delivered in a few days, but many are not that lucky. That is why I always aim to present a design solution with commonly available parts if possible for the benefit to other members.
 
The crossover distortion can easily be controlled with light DC loading at the output of the 358. The 4.7k resistor is placed for this purpose. The value can be adjusted for lowest distortion.
https://www.youtube.com/watch?v=9a-F59UnEY8 Watch from the 17:00 mark on or at 27:00 for the actual fix
Another example:
https://www.youtube.com/watch?v=VgodYtiD_F0


Most good specimens of the LM358 from reputable manufacturers can still perform well at 5 kHz in this circuit.

Using a single supply for this circuit, it performs better with just the local feedback. If you can add some negative supply, the distortion can be reduced an order of magnitude with global feedback.

I have bread-boarded this circuit configuration before using a 358 and will post some results later to illustrate.
----------------------------
With the supply slightly higher at 25 V, it gives a good 20 Vpp swing on the load.

With the DC load resistor (4.7k) reduced to 2.2k, the THD dropped down to 0.7%. Not bad at all, since the Agilent FG source produce about 0.05% of this.

At 6.5 kHz, the THD was up to 1% and climbing fast with an increase in frequency. At 7.5 kHz, it is severely slew-limited.

The GWInstek FG mentioned in the first post can't do better on distortion anyway.

Spec: (–55 dBc DC ~ 200kHz, Ampl > 0.1Vpp)

Hi,

I implemented your circuit in LT spice and created the library for the LM358 opamp.

The output is not good. I have attached the screen shot.

Please see my circuit what mistake I have done. Or since LM358 is from TI the LT spice will not work good for this part ?

Can I just go straight into building the prototype if the circuit you provide was 100% ok ?

Can I also please know why is the GND connection below LM358 is not connected any where ?

LM358.jpg
 

The nearest recommended LT equivalent of the LM358 is the LT1413.
On the full 10 Vpp input it clips bad on the bottom. In the simulation, I reduced the input slightly, but still shows clipping.

If you just add the -1V negative supply, it clears up nicely.

I had to reduce the DC loading resistor down to 1k with this part to get good results.

Can I also please know why is the GND connection below LM358 is not connected any where ?

It was just left after adding the negative supply to the pin. I could have deleted it.

- - - Updated - - -

If you really want to use a single supply, then it is better just to run with local feedback if you can tolerate the 0.7% distortion using a LM358 in your actual circuit.

In my practical circuit, I found it better to make R7 a 12k and R8 a 8.2k and use the FG level control to adjust the exact output required.
 

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  • LT1413_lfb.png
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Vikash, you must look at the datasheet when you use an IC. Your simulation shows you trying to feed 5kHz at 10V peak or 20V peak to peak into its pin 1 output. Pins 3, 4 and 5 are connected to something off the screen and the pin 8 positive supply pin and the pin 4 ground pin are not shown and are not connected. Since your input signal is AC that swings up to +10V and down to -10V then if you power the opamp with +24VDC and ground and feed its input with the AC signal, its input will be destroyed since its maximum allowed negative input voltage is -0.3V.
 
You must make sure your source setup is correct for the simulation or actual circuit to work.
 

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Vikash, you must look at the datasheet when you use an IC. Your simulation shows you trying to feed 5kHz at 10V peak or 20V peak to peak into its pin 1 output. Pins 3, 4 and 5 are connected to something off the screen and the pin 8 positive supply pin and the pin 4 ground pin are not shown and are not connected. Since your input signal is AC that swings up to +10V and down to -10V then if you power the opamp with +24VDC and ground and feed its input with the AC signal, its input will be destroyed since its maximum allowed negative input voltage is -0.3V.

Hi,

Actually I downloaded the pspice model for LM358 from TI

https://www.ti.com/product/LM358/toolssoftware

Below is the code

Code:
* LM358 OPERATIONAL AMPLIFIER "MACROMODEL" SUBCIRCUIT
* CREATED USING PARTS RELEASE 4.01 ON 09/08/89 AT 10:54
* (REV N/A)      SUPPLY VOLTAGE: +/-5V
* CONNECTIONS:   NON-INVERTING INPUT
*                | INVERTING INPUT
*                | | POSITIVE POWER SUPPLY
*                | | | NEGATIVE POWER SUPPLY
*                | | | | OUTPUT
*                | | | | |
.SUBCKT LM358    1 2 3 4 5
*
  C1   11 12 5.544E-12
  C2    6  7 20.00E-12
  DC    5 53 DX
  DE   54  5 DX
  DLP  90 91 DX
  DLN  92 90 DX
  DP    4  3 DX
  EGND 99  0 POLY(2) (3,0) (4,0) 0 .5 .5
  FB    7 99 POLY(5) VB VC VE VLP VLN 0 15.91E6 -20E6 20E6 20E6 -20E6
  GA    6  0 11 12 125.7E-6
  GCM   0  6 10 99 7.067E-9
  IEE   3 10 DC 10.04E-6
  HLIM 90  0 VLIM 1K
  Q1   11  2 13 QX
  Q2   12  1 14 QX
  R2    6  9 100.0E3
  RC1   4 11 7.957E3
  RC2   4 12 7.957E3
  RE1  13 10 2.773E3
  RE2  14 10 2.773E3
  REE  10 99 19.92E6
  RO1   8  5 50
  RO2   7 99 50
  RP    3  4 30.31E3
  VB    9  0 DC 0
  VC 3 53 DC 2.100
  VE   54  4 DC .6
  VLIM  7  8 DC 0
  VLP  91  0 DC 40
  VLN   0 92 DC 40
.MODEL DX D(IS=800.0E-18)
.MODEL QX PNP(IS=800.0E-18 BF=250)
.ENDS




The pin out can be seen from the above code. 

*                | INVERTING INPUT
*                | | POSITIVE POWER SUPPLY
*                | | | NEGATIVE POWER SUPPLY
*                | | | | OUTPUT
*                | | | | |
.SUBCKT LM358    1 2 3 4 5

I edited the above code and LT spice has an option of auto generation of the symbols using the code.
 

I do not know what is missing in the code for LTspice to show a box instead of an opamp and why its pin numbers are wrong.
 
Can I just go straight into building the prototype if the circuit you provide was 100% ok ?

Yes, just do that already. I don't think you are gaining any more information from doing simulations. Get the real circuit up and working.
 
Just remember that your FG has an output impedance of 50 Ω, and when connected to the high input impedance of the op-amp, the signal levels will be double of what is displayed on the screen or display. So you will need to use 50% of the level settings shown in the simulation source setup.
 
Hello all,

Many thanks for your help with the circuit design.

I have attached the picture showing the waveform in the oscilloscope for the different positions of the my resolver.

There is a small glitch on the reference signal but I was able to achieve what I was looking for. I will add the negative voltage in my actual PCB design.

So my next task is to find the degree of angle of the resolver position. One of the users guide of the resolver suggest that "to determine the angle with a resolver, we just need to take the arctan of the ratio of the two voltages supplied by the output windings of the resolver."

Using the oscilloscope I measured the sine and cos voltages for different angle of rotation of the resolver.

I have attached the image.

Upto 90 degree of my resolver rotation, my calculate value is approximately equivalent to the theoretical value.

After 90 degree my calculated value is of negative degree and I am not sure how to make it to positive to match my theoretical value ?

For example:

My resolver position is 135 degree and my calculated value shows -51.93 degree. Do I need to subtract 51.93 from 180 degree to get 128 degree approximately equal to 135 degree ?

So how do I calculate for other negative degree to make it positive ?

Please help me with this.



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20180103_112833_Richtone(HDR).jpg
 

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