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Calculating maximum allowable gain of the OPAMP

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For the OP484 opamp, the minimum open-loop (no negative feedback) gain shown on the datasheet is 150,000 for DC and very low frequencies.
For the OPA381 it is a minimum of 106dB which is 200,000 but it also depends on the source impedance that you do not know.
 

    V

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Can I know what will be the difference in using them in noninverting mode to get a gain of 100.
 

The datasheet for the OPA381 transimpedance amplifier does not show non-inverting.
For your project, other circuits use two opamps each with a gain of 100 for a total gain of 10,000.
 
Hello Thanks,
Can I know what can be the maximum gain that can be used by the sallenkey lowpass or highpass filter.
As here it **broken link removed**
says the gain should be kept less than 3 and also in this **broken link removed** design of widebandpass filter it says the filter has unity gain.
 

Sallen-Key filters can be implemented with any gain.
 

I have never fiddled with the gain and the parts ratios in a Sallen-Key filter. Maybe if the ratio of the resistors and/or capacitors is different than being equal or two-to-one then the gain can be more than 1 or 1.6.

Maybe you can make two opamps in series with gain of 100 each and each one with a capacitor parallel to the feedback resistor for a very simple lowpass filter like in the design I posted. Then use an extremely simple highpass filter that is just a coupling capacitor that does not affect the frequencies you want. Try it.
 

Yes,I used the two stage filtering as in the embeded lab blog with OP484 with a gain of 101 each and resulted in the pulses as shown in the two images

**broken link removed**

I analysed that
2)May be the original output of the phototransistor might be around 2.5mV becuase the 1st stage having a gain of 101 resulted in 250mV(form the image)
2)and the second stage of gain resulted in 2.5V (means the 101*250mV of first stage filtering output).
But the Dc component has not been removed after the first stage even there was a capacitor in the high pass of the 1st stage filtering.
The DC component of the signal is aorund 520mV I guess and for this purpose I added 2 cursors 1 at the 0(DC) level and other at the 1st stage filtering output.
Can I know why the DC component has not been removed by the capacitor of highpass stage in 1st stage of filtering and how to remove the DC component after the 1st stage itself. it.
 

Your attachment does not work.
You have too many threads about this project so that the schematic is not in this thread, instead it is in another thread.

- - - Updated - - -

I found the circuit in my recycling bin. Its 68k input resistor loads down the collector of the photo-transistor, reducing its output. Increase the 68k input resistor to 220k.

Your offset voltage might be caused by the DC input offset voltage of the opamp being amplified 101 times because a capacitor to ground at each 6.8k resistor is missing. The offset voltage might also be caused by leakage in electrolytic 1uF coupling capacitors. Use film capacitors instead.
 

sorry, about the attachement and here it is pulse.JPG.
I implemented the one in this https://embedded-lab.com/blog/?p=5508
But not this https://embedded-lab.com/blog/?p=1671
 

Hooray! You finally got the very simple circuit to work.
 

    V

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But still the DC component has not been removed after the first stage even there was a capacitor in the high pass of the 1st stage filtering.
The DC component of the signal is aorund 520mV I guess and for this purpose I added 2 cursors 1 at the 0(DC) level and other at the 1st stage filtering output.
Can I know why the DC component has not been removed by the capacitor of highpass stage in 1st stage of filtering and how to remove the DC component after the 1st stage itself. it.
 

    V

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There have been so many different, including wrong or incomplete circuits posted in this thread. Does anybody know which exact circuit we are talking about just now?

Genereally speaking, a single supply AC amplifiier needs a DC bias. It has to be removed somewhere in the signal path if you want a DC-free output signal. I presume you neither need a DC-free nor a biased AC signal but want to extract some information from the waveform. Which information are you looking for and what's your particular problem with the "DC component".
 
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    LvW

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    V

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This is the one I am using right now https://embedded-lab.com/blog/?p=5508
I wondering why the capacitor in the HP filter of the 1st stage filtering is not removing the DC.
particular problem with the "DC component".I am planning to use the gain of 200 in the 1st stage itself by using a 3.4Kohm resistor in place of 6.8K(1+680/3.4) so that it gives 201.
For this I want to know that 1)Is it ok to use the gain of 201 in the 1st stage(I am not sure if it is correct to get 201 gain with a single opamp) 2)Need the output signal to be at based at 0V instead of 520mV(so that the 2nd stage won't be needed.)
 

    V

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The circuit doesn't makes much sense, because it's missing a regular bias. Apparently the input AC is partly rectified, creating the said 500 mV output offset. But much of the input signal gets lost this way.

It would be better to apply a regular input bias and AC couple the output voltage.
 

    V

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Sorry,I just checked the circuit behaviour in the simulation tool by taking the phototransistor output(from analog board and by using the values in the simulation tool) and verified that the capacitor is perfectly working as it removing the DC component but it is the problem with the OPAMPs in the LPfilter stage
1)using OP484(opamp) in the LPfilter stage
**broken link removed**
it is again adding some DC value as in the image it is around 1.2V.
The image has 3 signals 1)green signal:eek:utput of phototransistor is at 2.9V2)red signal:Output of HPfilter(removed the 2.9V) 3)bluesignal:eek:utput of LPfilter(again added a offset voltage of 1.2V.
2)using MCP6001(opamp) in the LPfilter stage
**broken link removed**
The image has 3 signals 1)green signal:eek:utput of phototransistor is at 2.9V2)red signal:Output of HPfilter(removed the 2.9V) 3)bluesignal:eek:utput of LPfilter(but it is not perfectly filtered.)
The Output of the LPfilter is not filtered perfectly when compared to using OP484
Can some explain how to overcome this problems and is there anyother OPAMP which can does the jobs of the both OPAMP's perfectly
 
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    V

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Can anyone explain about the selection of OPAMP that can remove the offset voltage and can filter well as explained above.
 

    V

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The opamp has an "input offset voltage" that is amplified by the circuit because the simple circuit has a gain of 101 at DC. If you simply add a capacitor to ground from the 6.8k resistor that now goes to ground then the gain at DC will be only 1 and the DC output offset voltage will be extremely low. Of course you must properly calculate the value of the capacitor so it passes the low frequencies that need gain.
 

You mean adding a new capacitor to ground from 6.8K removes DC offset.
gain at DC will be only 1
Click to expand...
 will it show any effect on the amplification of the Pulse signal(which should be amplifed by afactor 100).
Can I know what will be the relation between 6.8K resistor and new capacitor for removing the offset and allowing the required frequencies to pass through and amplified so that I can calculate the required capacitor.
 

The standard formula for a coupling capacitor that forms a highpass filter where Fc is the cutoff frequency is Fc= 1 divided by (2 x pi x R x C) where R is in ohms and C is in Farads. The signal is at -3dB (0.707 times the level of much higher frequencies).

If you want the cutoff frequency to be 0.3Hz with R being 6.8k then C= 80uF. Use 100uF.
All coupling capacitors increase the cutoff frequency so they all must be considered.
 

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