Output current op amp-LTspirce simulation

Hi All,

I trying to do Howland current pump circuit, to get an output current depending o the input voltage to the non-inverting op amp input. The Howland current pump circuit hasn't been an issue. The problem is that I was planning to feed the Howland circuit with an op amp follower circuit to isolate the input of the Howland circuit. It didn't work quite well first try as the voltage at the output of the follower circuit was dropping unexpectedly. The reason is that the op amp chosen can't not draw enough current , so the voltage drops. The simulation is working now fine with a universal op amp from LT spice which doesn't have physical limitations.

I'm finding difficult to find an op amp that can draw enough out put current for my application. I would need something that can draw at least 25mA, when the op amp is connected from 0V to 15V.My real problem I that I don't know what parameters to look at to choose an op amp suitable for what I need. Anyone can advice please?

Below is the LTspice simulation:

Hi,

why do you use that low ohmic resistors?
--> try to use 10 times your resistor values for R1 ... R4.

Klaus

Use the "improved" Howland circuit as shown in this paper.
It allows a higher current with higher input resistor values.

But the output opamp will still have to supply that current.
You may need a higher current op amp or a buffer driver at the op amp output.

And use a real op amp model instead of the ideal one.

Isn't it wonderful that a severely overloaded circuit works perfectly in a simulation?

Hi All,

The current in the load is 22.4mA. I have chosen this op amp that I think it will do the job.

LM258ADT

https://www.farnell.com/datasheets/...MI_Yucxcno6AIVWuR3Ch37LgYKEAAYASAAEgJJL_D_BwE

The LM258A will have an output saturation voltage loss of about 3V when it is a "typical" one as shown on a graph on its datasheet and used in a simulation. In real life then ones with less than typical spec's will have more saturation voltage loss and the function will not work. The text in the datasheet guarantees a source output of 20mA only when the output is only 2V. Then the IC heats and the load gets only 2V x 20mA= 40mW. If your load of 22.4mA needs a voltage of 12V then it needs 12V x 22.4mA= 269mW and the IC stays cool.
Where can you buy a typical or better one? Only in a simulation.

Audioguru said:

Isn't it wonderful that a severely overloaded circuit works perfectly in a simulation?

Click to expand...

Not unexpected when you use ideal op amps.

- - - Updated - - -

Do you need bipolar voltage output?
If not, than you could add an emitter follower output to increase the current drive.

Otherwise you could use a higher current op amp such as an OPAx197 (x - blank, 2, or 4).

Hi Crutschow,

I have changed he circuit and I just need 1.87V at the op amp output to generated a current of 22.4mA. I am going now to use this op amp (TLV4171) which is a bit cheaper as I don't need now to dissipate that much wattage.

The TLV4171 also has a marginal output current rating for your application.

How much wattage do you need to dissipate?
What are the supply voltages?

The supply voltage for the op amp will be +/- 15V and the voltage that I need to output to achieve 22.4mA is 1.87V.

The TLV4171 is a low power opamp rated to drive a load as low as 10k ohms so it cannot do what you want.
An NE5532 dual opamp is rated to drive a load as low as 600 ohms and its will just do it (23.5mA typically). With your high supply voltage then the IC will get very warm but not too hot.

I am sorry but I am a bit confused. Is it possible to burn an op amp by outputting too much current or voltage?. If we short the output of an op amp what would happen?, would it run current until it burns out?, or it would just stop sourcing current at its Isc?. In other words, is it possible to burn an op amp by running it too hard?

Hi,

Is it possible to burn an op amp by outputting too much current or voltage?

Click to expand...

For sure.

Not current or voltage
--> But: current * voltage = power
where
* "voltage" is the voltage drop across the current path inside the OpAmp (usually from supply to output)
* "current" is the output current.
For details read the OPAMP´s datasheet.

and there is additional power dissipation by the internal circuitry

*************
This P = V x I is generally true, for each ohmic device. LED, incandescent lamp, bjt, MOSFET, diode, zener, resistor ....

*************

If we short the output of an op amp what would happen?, would it run current until it burns out?, or it would just stop sourcing current at its Isc?. In other words, is it possible to burn an op amp by running it too hard?

Click to expand...

Depends on the used OPAMP. --> read it´s datasheet.
* Some have current limitation, some don´t
* some have thermal shutdown, some don´t
* some have power dissipation limitation, some don´t

Klaus

The datasheet for many opamps show that the maximum output current is limited and say for the "Absolute Maximum duration of an output short circuit" then "Do Not Exceed Package Dissipation (heating) limit".
Of course all opamps have a maximum allowed supply voltage. The maximum allowed input voltage and current are also important.

Hi All,

I have got these Op amp as an example: TLV271 type D ( SOIC package).

https://www.ti.com/lit/ds/symlink/tlv272.pdf

The maximum Tj is 150°C and we can suppose an ambient temperature of 25°c.

According to the absolute maximum rating for current, it can output 100mA ( without specifying what supply voltage and what output voltage).

The junction to ambient thermal resistance for this device is 127.2 °C/W. The output current when VDD is 10V and V0 is 0.5V is 13mA (pag 9). The wattage dissipated by this device is (10V-0.5V) * 13mA = 123.5mW. If the RθJA is 127.2 °C/W * 123.5mW = 15.71 °C + ambient = 40.71 °C. This device is at 41 °c when driving at its maximum rate?. I know according to the absolute maximum rating for current it can drive 100mA, but the difference between typical figures and absolute maximum is massive.

The temperature of the device with the same conditions but outputting 100mA ( absolute maximum rating) is 120.84 °C + ambient = 145.84 °C. This approximates much more to the maximum junction temperature.

I don't know but it seems like the data for typical values( at least for current ) is not very helpful. Am I correct or I 'm misunderstanding something?, thanks.

A graph for the TLV271 datasheet shows that with a 5V supply the typical output current into a dead short is 32mA and 49mA.
With a 10V supply the typical output current into a dead short is 90mA and 99mA.
It is not a power amp with a metal fin to bolt to a heatsink so simply don't overload it.

When they say the output voltage is 0.5V from the rail then the graph shows that when the output is conducting 12mA or 13mA in a load then the load has of 9.5V across it. The current and heating are massive (almost 100mA) when it is driving a dead short because then the output transistor has the entire 10V across it.

I think I was missing that the voltage in the graphs is always voltage referenced to the positive or negative rail. It makes sense now, thanks.