0
\$\begingroup\$

As a laboratory task at my practical engineering school we were asked to make the simple circuit shown below, prove that it acts as a constant current source regardless of its load, and answer some questions.

One of these questions is: what limits the amplifier? Why would its output stay at a maximum value (clamp?) above a certain load resistance?
Another one is: Why is it necessary to supply the amplifier with a dual-supply? Wouldn't single-supply voltage work just as well?

Theoretically it should be able to supply constant current on R, which is the same current on the load, therefore the voltage should not be limited. I couldn't find an answer other than the fact that the amplifier has a power limit it cannot go beyond, but I am unsure as to whether that is the answer they are looking for.

Vcc=15v, -Vcc=-15v, R1=R1=5KΩ, R=1kΩ, RL=1kΩ
LM324 datasheet: http://www.ti.com/lit/gpn/lm324

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ Hint: a real op-amp is not the same as an ideal op-amp. Aside from the inputs and outpus, the other terminals on your drawing are related to the limiting behavior. \$\endgroup\$ – The Photon Dec 8 '14 at 22:48
  • \$\begingroup\$ I told my teacher that it cannot output anything beyond Vcc (when R=RL), but he said that wasn't the only answer. \$\endgroup\$ – idanp Dec 8 '14 at 22:57
  • \$\begingroup\$ Could the opamp output, say, 10 Amps? \$\endgroup\$ – George Herold Dec 8 '14 at 22:59
  • \$\begingroup\$ It probably couldn't. But the case here is a low constant current that is flowing through loads with different resistances - the current isn't supposed to go to such values, only the out-voltage is. \$\endgroup\$ – idanp Dec 8 '14 at 23:07
  • \$\begingroup\$ Have you tried to interpret the datasheet ? You might want to link the one you are using in your question. The answers to both your questions require a datasheet. \$\endgroup\$ – Spehro Pefhany Dec 9 '14 at 2:06
1
\$\begingroup\$

Have a look at the data sheet for the amplifier. on page 4 in the Electrical Characteristics section the max current (Io) is given as 60mA (the sign should just indicate the direction)

Opamps that can output large currents exist, just not this one. The output has a current limiting circuit built in which is shown in the schematic for the device sourcing current (consisting of the resistor above the output pin & the transistor to the left of that.) This works notionally as follows: As current flows out of the device, it passes through the resistor, developing a voltage drop across the resistor. The resistor is connected across the base - emitter junction of the transistor & as the voltage difference here exceeds the forward voltage of the base-emitter junction (nominally 0.7V - see other posts on that topic on this site), the transistor being properly biased, will start to conduct between collector & emitter. This current now flowing through that transistor used to flow (and be amplified by) the darlington pair at the top right of the schematic to produce the output current, but since that current is now diverted, it is unavailable to drive the output transistors.

The voltage output of the amplifier is limited by the supply voltage. Again look at the Electrical Characteristics, this time (VOH). It is given as 28V at VCC = MAX. Note 1 below the table says that 'MAX VCC for testing purposes is 26 V for LM2902 and 30 V for the others.' So for the LM324 MAX VCC is +30V, & the data sheet thus tells you that you will get at most 28V under this condition. (The Absolute Maximum Ratings section indicates that VCC MAX might be +32V, but lets not fight about that.)

This circuit should work perfectly well with only a single positive supply, particularly if both inputs are > 0V. This would usually be the case or there would be no load current.

Hope the above helps. It might have helped if the schematic transistors were labelled....

Just remember to study the datasheet for any component you use. The devil is in the details as they say.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.