# op-amp converters and other circuits

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#### Winterstone

##### Banned
Hi PG1995,

Q1: I think, you are right.
Q2: Yes, agreed
Q3: Again, you are right. I also think that there is no "switching". As soon as the capacitor charge approaches the input voltage the output voltage continuously goes to zero (not abruptly) and charging stops.

#### PG1995

##### Active Member
Thank you very much, Winterstone.

Best wishes
PG

#### PG1995

##### Active Member
operational transconductance amplifier (OTA)

Hi

Could you please also help me with the queries enclosed in the attachment?

Please note that there is a minor error in the attachment at one place, it should read: "This formula is referenced in Q5". Thank you.

Regards
PG

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#### Winterstone

##### Banned
Hi PG1995,

Q1: Real amplifiers have real output impedances (not zero and not infinite). An opamp should have a very small Rout (to act as a voltage source) and an OTA should have large Rout (current source). In reality, this cannot be achieved - however, as long as the Rout values approach the ideal case the device can be treated as a voltage resp. current source.

Q2: I really dont know. One reason might be that the input resistance for a particular application is not large enough (some OTAs have only some 10th of kohm as Rin). Then, because of symmetry reasons the second input gets the same resistor.

Q3: There is a buffer that converts the output curent (large Rout) into an output voltage (small Rout). Thus, the user can select the output node he wants.

Q4: I suppose it is Vrms (result of ac calculations).

Q5: Because the externally supplied current Ibias constitutes the "tail current", which flows through the common path of a differential amplifier.

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#### crutschow

##### Well-Known Member
Could you please help me with the queries in the attachment? It would be really kind of you. For high resolution copy of the attachment, please use this link: http://img851.imageshack.us/img851/7852/floydconverters8.jpg
RE Q3. When the input voltage goes below the voltage on the cap, the diode prevents the op amp output from following the voltage. This causes the loop to open and the op amp output will suddenly go to it's most negative value. That is what is meant by switching.

As this point the voltage on the cap stays at the peak value, affected only by any leakage currents.

#### Winterstone

##### Banned
RE Q3. When the input voltage goes below the voltage on the cap, the diode prevents the op amp output from following the voltage. This causes the loop to open and the op amp output will suddenly go to it's most negative value. That is what is meant by switching.
As this point the voltage on the cap stays at the peak value, affected only by any leakage currents.
Hello crutschow, can you please explain WHY "the op amp output will suddenly go to it's most negative value".
I think, it is an interesting question that deserves some explanations.

Theoretically, this happens only if the voltage at the inv. input terminal (across the capacitor) exceeds the dc input at the non-inv. terminal. However, some time before this rising voltage reaches a value slightly below the dc voltage (depending on the dc open loop gain) causing ZERO output voltage. Thus, the charging of the capacitor stops at this point and there will be NO "switching effect" to the most negative value.
This can be demonstrated via simulation using an ideal opamp model.
However, I think that in reality it is possible that such a "switch" to the negative supply voltage occurs. But this depends mostly on the time delay (storage capabilities) of the opamp itself.

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#### Winterstone

##### Banned
Hello crutschow, can you please explain WHY "the op amp output will suddenly go to it's most negative value".
I think, it is an interesting question that deserves some explanations.
..........
Its a pitty that Chrutschow up to now didnt find the time to respond to my question.
(Or didnt he even recognize it?) . Anyway....

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#### MrAl

##### Well-Known Member
Hi,

Isnt that when the diode does not conduct?
In some circuits when the diode does not conduct the feedback goes away, so the op amp moves into saturation in the other direction. It doesnt happen instantaneously though, it takes the time it takes the op amp to slew to that most negative level. For example, if the op amp has a slew rate of 1v/us and it is outputting 5 volts and then suddenly goes open loop, it has to slew all the way down to the negative power supply rail or close to it and with a supply of -10 volts that means it has to slew down 15 volts which will take 15us to accomplish. Likewise when it has to again operate in closed loop (diode conducts) it will take another 15us to slew back up to +5v. These slewing times have to be taken into account with designs like this because they greatly limit the operating frequency. If the signal did not have to slew that whole 15 volts (like maybe only 1v) it would only take 1us down and 1us back up, which is 28us faster than 15us down and 15us back up. That is very significant. Obviously if the signal itself recovers before the 30us it will work faster but that depends on the application. In the mean time some of that signal is lost so the average comes out lower than expected which may or may not be significant to the application.

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#### Winterstone

##### Banned
Dear MrAl,
I completely agree to your explanation. In principle, this was the meaning of my last remark (posting #7):
"this depends on the time delay (storage capabilities) of the opamp itself."
Thank you.

#### PG1995

##### Active Member
Hi

Could you please help me with the queries included in the attachment? Thanks a lot for the help.

Regards
PG

#### ericgibbs

##### Well-Known Member
hi PG,
Q1, look at this datasheet

Q2. .3Meg = 300K worst case [ lowest input resistance]

Q3. Is the Open Loop Gain , no feed back 50,000 [ typical 100K]

Q4 Ta = temperature Ambient, Vs = Vsupply , Rs Source impedance

E.

#### Attachments

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#### MrAl

##### Well-Known Member
Hello there,

To add a little to what Eric posted...

There are different versions of the same IC chip, some with better specifications
than others. The A version has better specs than the C version. It probably
costs more too.

As Eric pointed out, 0.3Mohms is the same as 300Kohms. To find out what value to
use, you would first note what IC version you are using, A or C, etc. Then under
that set of columns (and only that set) you would look for the spec.
You also see different rows even under one version of the IC. You also see major
rows and minor rows. The major rows are separated by a thicker horizontal line.
The major rows at the top of the blue section are labeled on the left starting with
either TA=25 deg C or TAMin<=TA<=TAMax. Thus the top major row is made up of three
rows, and the next major row down is made up of five rows. The top three blue rows
are for Ta=25 deg C and RL>=2k ohms. The bottom 5 rows are for TAMin<=TA<=TAMax and
also RL>=2k ohms.
Of the top three blue rows, the second and third refer to two different supply voltages
and their corresponding output voltages. The gain is 50v/mv which works out to 50000
for a source voltage set of plus and minus 20v and output voltage swing plus and minus
15 volts. For a source voltage of 15 volts (and swing plus and minus 10v) we see
a different value for two of the versions, where one is 50v/mv and the other is 200v/mv
which is 50000 and 200000 respectively, and the other is 20 and 200 which is for the
other version. The reason for the two numbers is because they are giving a typical
value and a minimum value. If you want to make sure the design is going to function
properly you have to use the worst case value which here is the min value.
The second major blue row is the five rows that start with TAmin<=TA<=TAmax. They
are giving the values of the gain again for all three versions, and for three different
values of source (plus and minus) voltages.
So understanding this data sheet means breaking up the rows and columns and reading the
values found there.

TA means "Ambient Temperature"
Vs means "Source Voltage(s)"
Rs in this context would mean the equivalent resistance in series with the source voltage(s).
To calculate the gain from the units of V/mV just divide the volts given by 0.001, so a
V/mV of 20 means a gain of 20000. This just means simply "Volts per Millivolt" which would
refer to volts output per millivolt of input.

#### PG1995

##### Active Member
Thank you very much, Eric, MrAl.

I have updated the question set in the attachment. Could you please help me with them? It would be really nice of you. Thank you.

Regards
PG

#### MrAl

##### Well-Known Member
Hi again,

You might be better off looking for a better data sheet, i think Eric posted one here.

If the space is blank then that means that they are not spec'ing the device for that set of parameters.

The designer is free to choose any version which he/she decides will work in the product. If he tries the LM741 for example and finds that he needs a better input offset spec, he might go to the 741A which has better specs for that. Obviously he would not go to a version that had worse specs as that would not help his design at all but even make it worse.

The part number chosen by the manufacturer is up to them really. This device probably got it's first part number as 741, then later there came other versions. the 741 might have fit between a better version and a worse version but they didnt want to have to change the name to 741B because that part number was already well known and fit many applications. Changing it to a B version might have thrown their customers a curve that they didnt want them to have to deal with.

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