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Continous Current Op-Amp

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I'm using this - Op-amp. LMC7101. V+ is 5V and V- is 0V.

I want to know whether the Op-Amp can provide continuous current output(100mA), while the input voltage is coming from DAC ranging 0V to 2V with 10mA current. we are using as a voltage feedback function along with maximum current output

In the Table LM7101A 5.0V Electrical Characteristics table, there it is mentioned as Output Short Supply Current as min of 120mA/80mA? So, its safe right? Or am I misunderstanding it? What does or when does Output Short Supply current occurs?

Also, in the same output short supply current , the values are given for Sourcing (Vout=0V) and Sinking (Vout=5V)? Shouldn't it be vice versa? Souring when Vout=5V and Sinking when Vout=0V?

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Hi,

the datsheet tells the "short circuit" current is guaranteed to be 120mA (200mA typ).
This says you are on the very limit.
But "short circuit" is an "error condition", not a normal operation condition.
At short circuit, you can not expect any useful output signal. The voltage gain is zero, the internal resistance rather high.

You get more useful values form the datasheet on normal operation when you look at the voltage drop.
But for this you should ask yourself about your load condition. I best case your load resistance is 5V/100mA = 50 Ohms. In worst case it it´s zero. (We can not know this. It depends on your very circuit)

The output voltage swing is given with 2kOhms and 600Ohms load connected to 2.5V. Thus the maximum ideal output current here is 2.5V / 600 Ohms = about 4mA.
This should give you a "feel" of normal operation condition. (It´s quite expectable that the OPAMP can drive much higher current, but with reduced performanc, like stability, distortion, timing)
(If the OPAMP is meant to easily drive 100mA under normal conditions, they surely mentioned this)

This means your 100mA is 25 times the value where the OPAMP output is specified.

So you have two values: 120mA under "erroneous" condition and 4mA for relaxed operational condition.

With your 100mA you are very far from the specified operational condition.

Conclusion:
* if you need 100mA (V_Out close to 2.5V) for a short time (because of power dissipation = heating) and you are satisfied with high voltage drop and high distortion, then it could be used.
* but if you expect it to drive the output close to the supply rails or you expect low distortion, then this OPAMP is not suitable.

So for many 100mA applications this OPAMP is not suitable. Better use a dedicated "high output current" OPAMP or consider to use an additional driver stage. (Which also depends on your requirements)

Klaus

Hi,

the datsheet tells the "short circuit" current is guaranteed to be 120mA (200mA typ).
This says you are on the very limit.
But "short circuit" is an "error condition", not a normal operation condition.
At short circuit, you can not expect any useful output signal. The voltage gain is zero, the internal resistance rather high.

You get more useful values form the datasheet on normal operation when you look at the voltage drop.
But for this you should ask yourself about your load condition. I best case your load resistance is 5V/100mA = 50 Ohms. In worst case it it´s zero. (We can not know this. It depends on your very circuit)

The output voltage swing is given with 2kOhms and 600Ohms load connected to 2.5V. Thus the maximum ideal output current here is 2.5V / 600 Ohms = about 4mA.
This should give you a "feel" of normal operation condition. (It´s quite expectable that the OPAMP can drive much higher current, but with reduced performanc, like stability, distortion, timing)
(If the OPAMP is meant to easily drive 100mA under normal conditions, they surely mentioned this)

This means your 100mA is 25 times the value where the OPAMP output is specified.

So you have two values: 120mA under "erroneous" condition and 4mA for relaxed operational condition.

With your 100mA you are very far from the specified operational condition.

Conclusion:
* if you need 100mA (V_Out close to 2.5V) for a short time (because of power dissipation = heating) and you are satisfied with high voltage drop and high distortion, then it could be used.
* but if you expect it to drive the output close to the supply rails or you expect low distortion, then this OPAMP is not suitable.

So for many 100mA applications this OPAMP is not suitable. Better use a dedicated "high output current" OPAMP or consider to use an additional driver stage. (Which also depends on your requirements)

Klaus
Thank you very much for your detailed answer! Really understood good points.

I just have a couple of questions:

In the Output swing parameter in the Electrical Characteristics table:

1. Why are the conditions only mentioned for load impedance of 2k and 600ohm? And what they give one normal value condition and one value with a condition that includes the junction temperature between -40 and +85? Why is this?

2. Also, doesn't OUTPUT HIGH mean +5V (according to my IC) ? Why did you consider as 2.5V? Is it because on the top of the table, they mentioned Vout = V+/2? But in my voltage follower configuration, shouldn't the output follow my input, which is 5V?

Thanks a lot in advance. @KLAUS.

Hi,

1) they (datasheet) tell you where the OPAMP works most perfect. It´s eaysy to drive no load (hogh ohmic), it´s more difficult to drive 2k, and it´s even more difficult to drive 600 Ohms.
I guess if the OPAMP was able to drive 100 OHms with good performance thy would "advertise" this in the datasheet.
Again I guess, that the OPAMP drops performans when driving loads less than 600 Ohms.

Temperture: They give you the information of what you can expect at room temperature and what you can get at extreme temperatures. It´s just worst case information. Good information.

In your case: driving high load current will also mean high power dissipation, is high temperature rise.
You can not expect to drive 100mA whil the case still remains at room temperature.

2) OUTPUT HIGH, means the output of the OPAMP is clamped or saturated (close) at the upper supply rail.
In this condition the OPAMP is not able to regulate or to compensate for output voltage errors anymore. The feedback does not work. It´s not the "linear" operation mode an OPAMP is designed for. It´s rather a "logic" output of a comparator.

Why 2.5V: Yes, because the datasheet says the load is expected to be connected to 2.5V.
Follow the input: In high current .. the OPAMP can not follw the input anymore. As already said: Gain drops, distortion increases, it can not regulate anymore it can not compensate voltage errors anymore, the feedback is useless.

--> I can only repeat my conclusion of post#2:
If you expect low distriortion, then this OPAMP is not the way to go.

Klaus

Hi,

1) they (datasheet) tell you where the OPAMP works most perfect. It´s eaysy to drive no load (hogh ohmic), it´s more difficult to drive 2k, and it´s even more difficult to drive 600 Ohms.
I guess if the OPAMP was able to drive 100 OHms with good performance thy would "advertise" this in the datasheet.
Again I guess, that the OPAMP drops performans when driving loads less than 600 Ohms.

Temperture: They give you the information of what you can expect at room temperature and what you can get at extreme temperatures. It´s just worst case information. Good information.

In your case: driving high load current will also mean high power dissipation, is high temperature rise.
You can not expect to drive 100mA whil the case still remains at room temperature.

2) OUTPUT HIGH, means the output of the OPAMP is clamped or saturated (close) at the upper supply rail.
In this condition the OPAMP is not able to regulate or to compensate for output voltage errors anymore. The feedback does not work. It´s not the "linear" operation mode an OPAMP is designed for. It´s rather a "logic" output of a comparator.

Why 2.5V: Yes, because the datasheet says the load is expected to be connected to 2.5V.
Follow the input: In high current .. the OPAMP can not follw the input anymore. As already said: Gain drops, distortion increases, it can not regulate anymore it can not compensate voltage errors anymore, the feedback is useless.

--> I can only repeat my conclusion of post#2:
If you expect low distriortion, then this OPAMP is not the way to go.

Klaus

Just a small clarification, my input to the non-inverting terminal of the op-amp is between 0-2V. So, since it's a voltage follower, the output of the op-amp will also swing between 0-2V right?

Now, why do we need to think about the output swing values in the table that has values around 5V?

Also, just a clarification, how to calculate the maximum normal current that the op-amp can deliver when the output (actually following the input) is between 0-2V? Can you tell how to make an approximate safe calculations for normal conditions?

Why don't you start with a load definition? Provided the OP would be able to source 100 mA at 0V, you need to short the output with a zero ohms load to receive current at 0V. So what's the real load case.

Available output current can be limited by two factors
1. internal current limit, which has been already discussed
2. power dissipation

100 mA at 0 V means 0.5 W power dissipation and would overheat the small OP package, you should definitely use an external current booster or an OP designed for larger power dissipation.

Hi,

Just a small clarification, my input to the non-inverting terminal of the op-amp is between 0-2V. So, since it's a voltage follower, the output of the op-amp will also swing between 0-2V right?

In post#1 one can not estimate/calculate your expected output voltage
In post#2 you talk about a voltage follower (now we know you want gain =1) and an inputt voltage of 5V (not 0-2V).
I´m confused.

But to answer your question: Yes, on a voltage follower with input of 0..2V the output ideally also is 0..2V

Now, why do we need to think about the output swing values in the table that has values around 5V?
Because in post#3 you wrote so.

Also, just a clarification, how to calculate the maximum normal current that the op-amp can deliver when the output (actually following the input) is between 0-2V? Can you tell how to make an approximate safe calculations for normal conditions?
Lets´s try again: It depends on your load and how it is connected.
All use ideal OPAMP in voltage follower configureation, all with max current of 100mA:
1) load connected to 0V: Load resistance is 20 Ohms. Max power dissipation on OPAMP: 3V x 100mA = 300mW (easiest condition for driving voltage 0..2V)
2) load connected to 1V: load resistance 10 Ohms. Max power dissipation 3V x 100mA = 300mW
3) load connected to 2V: load resistance 20 Ohms. Max power dissipation: 2V x 100mA = 200mW
4) load connected to 5V: load resistance 50 Ohms. Max power dissipation: 2V x 60mA = 120mW

To get a more precise answer, you need to give use the "load conditions" first.

BTW: why don´t you use a simulation tool? Then you can play around, measure every value you like....
They are free and easy to use...

Klaus

One option suggested is to use a buffer on the output.
This could be an NPN transistor (e.g. 2N3904) as an emitter follower (inside the loop).
That will readily source 2V @ 100mA with a 5V supply, and dissipate the resulting power.

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FvM

Hi,

To split hairs about a term used in the thread about 0 to 2V, in case it matters in the design - '0' V is very relative with outputs, could be 0.0001V, 0.001V, 0.010V, 0.100V for any/every OA with its Vos and even the tiny bit of noise.

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