Electronics > Beginners

Bogus parts from Reputable Dealer?

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wraper:

--- Quote from: Cerebus on January 20, 2020, 11:37:45 pm ---
--- Quote from: wraper on January 20, 2020, 10:48:03 pm ---... and BTW, this particular circuit with meter in series with input is not suitable for CMOS op amps like TLV2371 due to very high input impedance. You are basically connecting voltmeter in series with tiny capacitor.

--- End quote ---

Erm, that's a voltage source symbol, symbolising the offset voltage. Why would you even believe that was a meter? There would be no point of adding gain to then just go and read the unamplified signal at the input.

--- End quote ---
OK, I had a brainfuck.

wraper:

--- Quote from: Jwillis on January 21, 2020, 12:19:29 am ---
--- Quote from: wraper on January 20, 2020, 07:01:16 pm ---Please measure input offset voltage properly. First of all, it's as ebaster said. Secondly, "good opamps" most likely simply have input offset voltage of opposite polarity.

--- End quote ---

Before I get started on this experiment . Do I need a split supply , single supply or no supply .? I'm guessing VDD must be 9V  but I want to get this right . I want to be sure as to probe orientation . The  voltage at pin 3 must be exactly 4.5V ? I'll try to find exact or as close as I can for the resistors.
I would like to do this right so I give you the right measurements.

--- End quote ---
With circuit I posted you don't need split power. 2 resistor divider create a reference point at half of the power source voltage, which you can also consider as virtual ground. With split power, you don't need those resistors and can simply tie non-inverting input to ground.

rstofer:

--- Quote from: Jwillis on January 21, 2020, 12:19:29 am ---Before I get started on this experiment . Do I need a split supply , single supply or no supply .? I'm guessing VDD must be 9V  but I want to get this right . I want to be sure as to probe orientation . The  voltage at pin 3 must be exactly 4.5V ? I'll try to find exact or as close as I can for the resistors.
I would like to do this right so I give you the right measurements.

--- End quote ---

As Wraper's circuit is drawn, you need +9V at the Vdd pin of the op amp and 0V (the other side of the power supply) at Vee but we will call that "Ground".

Then you need to get a voltage APPROXIMATELY equal to Vdd / 2 (ie 4.5V) with 2 resistors.  Close is good enough.  Remember, I am advocating for a potentiometer with end resistors.  How well matched will that be when you twiddle the knob from end to end?

We're trying to get Vos but, really, we're looking for op amp functionality.  With the follower circuit, the output of the op amp should match the non-inverting input, more or less, due to offset voltage.  Again, close is good enough.  We know that Vos should be < 4.5 mV when the non-inverting input is almost exactly Vdd/2 but it shouldn't be wildly different anywhere else.

Jwillis:
I noticed that you made a mark from pin 3 to pin 6 so I took 2 measurements  with opposite polarities.
Vdd 9.06V    Resistors are 4.707K and 4.706K  Voltage at divider 9.06 V  4.536V Pin #3
Chip # 1 and # 2 are the ones that work in the circuit
Measurements in Volts

   Chip#         Output to grnd           Negative of probe at  pin#6  across to pin # 3          Negative of probe at Pin#3  across to pin # 6

     1                   4.535                                           0.033                                                                          -0.046
   
     2                   4.535                                           0.033                                                                          -0.046

     3                   4.535                                           0.041                                                                          -0.049

     4                   4.535                                           0.039                                                                          -0.048

     5                   4.535                                           0.040                                                                          -0.047

     6                   4.536                                           0.038                                                                          -0.048

     7                   4.535                                           0.036                                                                          -0.049

     8                   4.535                                           0.040                                                                          -0.046

     9                   4.536                                           0.038                                                                          -0.049

    10                  4.535                                           0.036                                                                          -0.048

Not sure what to make of this but chip 1 and 2 have identical measurements .

wraper:

--- Quote from: Jwillis on January 21, 2020, 02:26:35 am ---I noticed that you made a mark from pin 3 to pin 6 so I took 2 measurements  with opposite polarities.

--- End quote ---
What are you using for measurements? It's really strange that you are getting different reading by just switching the leads. Not to say that simply measuring output voltage shows only 1mV variation but measuring between non-inverting input and output 10+mV, or did you miss one zero? Also why it's in a volts range. Doesn't your meter have millivolt range?

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