# why put parallel resistor and cap to ground at opamp output?

Discussion in 'General Electronics Chat' started by tpny, May 6, 2012.

1. ### tpny Thread Starter Member

May 6, 2012
216
0
Hi, I want to use this opamp to amplify 0~1V (from temperature sensor output) to 0~5V (to adc input). Quesiton: Is this a good candidate? (Chose it because it supports single supply and rail-to-rail output range.) Question 2: Why does the sample circuit in datasheet, page 14 have an RL resistor and 100pF in parallel at the output of the opamp? Thank you so much!

2. ### crutschow Expert

Mar 14, 2008
22,206
6,468
That looks like a good amp for your requirements.

The RL and 100pF in parallel are just loads for test purposes to simulate typical actual loads. If you notice, they use different values of RL for the various test graphs on that page. You do not deliberately add them to your circuit.

3. ### tpny Thread Starter Member

May 6, 2012
216
0
thank you so much! That really cleared the confusion.

Another thought: since I'm using this opamp as single-ended input, (V- input tied to ground), is there such a concept as common-mode voltage (or input) here? (My understanding of common-mode voltage/input is only in the scope of differential inputs to opamp.) What does it mean in the scope of single-ended input, is it just ground? Thank you so much!

4. ### #12 Expert

Nov 30, 2010
18,076
9,683
The datasheet shows that it is permissable to input anything from -.2 volts to +4 volts (with a 5 volt supply). This can be said as, "Vcc minus one volt" on the positive end.
This is good on the negative end because many opamps fail when the input equals the negative supply voltage. This one doesn't.

I see more of a problem on the output. This chip can't output 5.0 volts with a 5.0 volt supply. Figure 21 shows that you can keep the error down to 10 or 20 millivolts if you only draw 100 microamps from the output. If that is good enough for you, you're done. If it isn't, you can fix it by giving the chip a positive supply of 5.05 volts to get 1 ma of output or 5.13 volts to get 10 ma out of it. Picky, isn't it!

I'm thinking, "Just give it 5.3 volts positive".

Next question?

5. ### tpny Thread Starter Member

May 6, 2012
216
0
Ha, I'm using 12V on opamp's +Vs so I would hope to expect 0~5V out on 0~1V in (from sensor) after 5x amplification. (12V just because it's a common power supply output.)

On the bottom of page 3, input characteristics:

Q1) It says common-mode voltage range is -0.2 ~ 4V. (I understand that means the supported input range but why call it common-mode?)

Q2) Underneath it says CMRR is 0~2V. What does CMRR mean in the scope of single-ended input (as opposed to differential)?

Q3) Underneath it says Input Impedance. What does 10^13ohm||2.8pF mean? What's the symbol '||'? I know input impedance to be an ohm concept but what does pF have to do with it? Thank you so much!!

Larry

6. ### #12 Expert

Nov 30, 2010
18,076
9,683
Q1) Opamps are usually used in a closed loop configuration, so both inputs are at the same voltage when it's working correctly. Thus, the voltages on each input are "common" to each other. That makes the inputs it's working with, "common mode" and they have to be within the common mode range of the chip.

Q2) That 0-2V is merely defining the range of input voltages for which the following specs are valid. Keep in mind that the 0-2V is listed for a 5 volt positive supply. The range is larger for dual 5's and dual 15's as supply voltages.

Q3) I can't find the "pipe" symbol on my keyboard so I'l just say that "//" means, "in parallel with". Apparently there are a couple of pf's inside the chip and they are in parallel with the input resistance (input impedance?).

7. ### crutschow Expert

Mar 14, 2008
22,206
6,468
To elaborate slightly:

Q2) CMRR really doesn't come into play in the inverting mode, but it does for a non-inverting amp since the common-mode voltage changes with input voltage in that configuration. For a non-inverting amp the amount of the CMRR could cause a slight non-linearity in input versus voltage.

Q3) The impedance value can be given in "ohms" but for a complex impedance this ohms value varies with frequency. For that reason the impedance is given as its combination of real and reactive components, which does not vary significantly with frequency. For the op amp this is a very high resistance in parallel with a small parasitic capacitance between the two input terminals (R||C).

(Note to #12: The vertical brace | is found on my keyboard as the shifted symbol above the backslash \.)

8. ### #12 Expert

Nov 30, 2010
18,076
9,683
OK. The symbol on my keyboard looks like a :
Probably has a scratch on it so I couldn't see it as a "vertical brace".
||

Got it.

9. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
Why are you tying the -input to GND? You will not get linear gain in this configuration unless there is an inverting amplifier in the feedback loop.
You need to post your schematic.

10. ### tpny Thread Starter Member

May 6, 2012
216
0
Hi Ron, I meant the sensor's output(-) tied to ground and only feeding sensor's output(+) to op amp's V+ (single-input non-inverting op amp config as opposed to differential-input in-amp config). And op amp's V- tied to op amp's Vout through 2 resistors to form 5X gain. (Sorry company network doesn't allow for uploading attachments.)

Could you please elaborate on this? It's very interesting!

I see. But why state -0.2~4V as common-mode input range and also 0~2V as CMRR. I thought CMRR was the op amp's ability to reject common-mode *noise*? But here common-mode is signal? How does it know the difference?? Thank you so much as always!

11. ### #12 Expert

Nov 30, 2010
18,076
9,683
I think you need to read slower or get some sleep.
0-2V can not possibly be a common mode rejection ratio because CMRRs are always stated as decibels or (some large number to 1)

As in the quote you presented right above your question, 0-2v is merely defining: the range of input voltages for which the following common mode rejection ratio statements are true.

Check again in the morning. It will probably read better when you are rested.

12. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
I think his point is why is the common mode range higher that the guaranteed CMRR range. If you don't know the CMRR over the entire input range, then you can't predict the performance over that range.

13. ### #12 Expert

Nov 30, 2010
18,076
9,683
Well...I got some sleep! Now..where are we at? Why is the common mode rejection ratio defined for 0 to 2 volts and the common mode input range is defined from -.2 to +4 volts.

I don't know why Analog Devices Inc. would do that.

14. ### crutschow Expert

Mar 14, 2008
22,206
6,468
In the non-inverting mode the voltage for both inputs is equal to the input voltage, unlike the inverting mode, where the inputs remain at (virtual) ground. Thus as the two inputs move in response to the input voltage, they experience a common-mode voltage change. This can affect the output as determined by the CMRR value. The effect is normally very small and is almost always neglected as a source of error (few even realize it is there).

15. ### tpny Thread Starter Member

May 6, 2012
216
0
So is there CMRR taking place in a standard inverting or non-inverting config where the input to the op amp is 1 wire (single-end) not 2 wires (differential). If so, how? How does it know Vcm between the two op amp inputs if only one is connected to signal source.

16. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
Vcm is not between the two inputs. Common mode means both inputs are the same voltage. They both vary relative to ground.
Think of a voltage follower. The output follows (approximately equals) the noninverting input. Since the -input is connected to the output, both inputs vary by the same amount when the input changes. This is common mode.
In fact, when an op amp is operating in the linear region with negative feedback, the inputs are always equal to the voltage that is on the noninverting input. In an inverting amp, the +input is at ground, or some fixed voltage relative to ground, The -input will be at that same voltage, so there will be no common mode voltage as a result of the input signal.

17. ### tpny Thread Starter Member

May 6, 2012
216
0
OK, I see the concept that in inverting config Vcm is held fixed while in noninverting config Vcm ivaries with input. But how does this relate to common mode rejection? What does it reject?

18. ### #12 Expert

Nov 30, 2010
18,076
9,683
It rejects any voltage that is exactly the same on both inputs. Very handy when the wiring picks up noise or hum.

19. ### tpny Thread Starter Member

May 6, 2012
216
0
So if you have this configuration below, how will noise originating from Vin arriving at V- input get rejected if V+ doesn't share the same noise since it's tied to ground?

Code ( (Unknown Language)):
1.
2.
3. [FONT=Lucida Console]Vin-------------- V-|\[/FONT]
4. [FONT=Lucida Console]                    | >---feedback to V- [/FONT]
5. [FONT=Lucida Console]              --- V+|/ [/FONT]
6. [FONT=Lucida Console]              | [/FONT]
7. [FONT=Lucida Console]             GND[/FONT]
8.
9.

So do I need to connect my sensor to op amp as below with two wires? But wil noise on sensor's output(-) line short to ground before reaching op amp's V+ input?

Code ( (Unknown Language)):
1.
2.
3. [FONT=Lucida Console]Sensor: output(+)-------------- V-|\[/FONT]
4. [FONT=Lucida Console]                                  | >---feedback to V-[/FONT]
5. [FONT=Lucida Console]        output(-)-------------- V+|/[/FONT]
6. [FONT=Lucida Console]                            |[/FONT]
7. [FONT=Lucida Console]                            |[/FONT]
8. [FONT=Lucida Console]                           GND[/FONT]
9.
10.

Or do I have to use an in-amp? (Note my sensor output range is 0~1V.) Thank you so much!!

Last edited: May 11, 2012
20. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
what kind of sensor are you using? How far is it from the op amp? Are they connected by a shielded cable?