Opamp linear  current regulator must come quickly out of positive saturation - Page 1
Forum section: electronics / projects

author: ocset
author_rank: Super Contributor
timestamp: September 17, 2017, 12:13:52 am
content: Hello,We have a linear LED current regulator involving an opamp which drives into the base of a FET and regulates the current in a sense resistor in the source of the FET (thers a reference voltage into the noninv input of the opamp). At first the opamp is powered but there is no voltage to power the LEDs….therefore the opamp current regulator is “hard-on” with its outputs banged up to its positive rail (in positive saturatrion). Then suddenly the LED driving voltage comes on and so the opamp has to slam very quickly out of positive saturation and quickly bring the LED current into regulation without LED current overshoot...In order that the opamp can more quickly come out of positive saturation, and start regulating the LED current, should we add circuitry to prevent the opamp’s output going any closer than about 2V away from its positive rail whilst it is in positive saturation?

author: Zero999
author_rank: Super Contributor
timestamp: September 17, 2017, 12:21:58 am
content: Preventing the op-amp's output from saturating sounds like a good idea but this does sound like bad design.Why can't the op-amp or at least the voltage reference be powered from the same source as the LEDs?

author: Ian.M
author_rank: Super Contributor
timestamp: September 17, 2017, 02:27:07 am
content: Clamping the feedback to prevent positive saturation wont help sufficiently. It will still let through a current spike due to the slew rate limiting the OPAMP + MOSFET's response time. The circuit needs to idle in negative saturation with the MOSFET off. If you can drive the reference from the switched LED +V supply rail, and that rail actually goes to 0V, you only need to add a pullup resistor to keep a permanent small bias current in the sense resistor so the OPAMP -In is above its +In when there is no supply to the LED. If you cant access that rail, or it doesn't properly go to 0V you need to add a low current source follower off the main MOSFET drain to feed the reference. LTspice sim attached.N.B. you also need a small capacitor across the LED (of the order of 1nF) to bypass the charging current of the main MOSFET's gate-drain capacitance so it doesn't cause the LED's If to exceed the desired maximum.

author: BrianHG
author_rank: Super Contributor
timestamp: September 17, 2017, 03:21:43 am
content: Quote from: treez on September 17, 2017, 12:13:52 amIf this is a linear configuration, you can try 2 things:1) add a cap from the FET output to it's gate slowing down it's turn on speed.2) add a cap from the output of the opamp to the negative feedback slowing down the opamp.3) add a resistor across the FET drain and source to create an initial minimum on current at powerup, sharing some of the load with the FET.4) any combination of 1),2),&,3).In a switching design, changing the location of your inductor in the output might help, like before the current sense resistor. But, you will need to adjust your feedback values and add a RC filter on the feedback.Finally, feeding your voltage reference to the + input on your op-amp, just feed that voltage through a series resistor like 1-10k, and add a 0.1 to 1uf cap on the +input to GND creating a soft slow power-up.

author: David Hess
author_rank: Super Contributor
timestamp: September 17, 2017, 05:41:53 am
content: Add a resistor between the current sense and the inverting input which may already be there for frequency compensation. Add a JFET or depletion mode MOSFET between the output of the operational amplifier and inverting input. When the LED supply is missing, the JFET or depletion mode MOSFET closes the loop preventing saturation.If the current levels are low and a continuous output is acceptable, then an easier way is to use current steering with a pair of diodes at the drain/collector of the output transistor to prevent it from saturating.

author: floobydust
author_rank: Super Contributor
timestamp: September 17, 2017, 06:41:52 am
content: It really helps to have the schematic portion or op-amp's part number etc.It's not like we're gonna steal the design and build our own LED fixture.Op-amps overload and "saturate" a few different ways. It can be the input stage diff amp, internals, or the output stage.You are assuming it's the output stage. And careful your op-amp does not phase-invert when overloaded.I would try add a Schottky diode to limit the differential input voltage to 0.3V max. and see if it recovers quicker.Otherwise the op-amp slew-rate is too low.

author: ocset
author_rank: Super Contributor
timestamp: September 17, 2017, 11:17:50 am
content: Thanks , We really need absolute minimum LED current overshoot....the supply goes on and off regularly....we really need to limit or preferably get rid of any LED current overshoot.The schem and offending led current overshoot waveform is as attached.There is also an LTspice simulation attached.This shows the exact situation that we have, but for copyright reasons, i unfortunately cannot tell or show the actual application.QuoteThanks, but sorry it cant.QuoteThanks, but sorry we cant do that…..we need it to come on pretty quick but with abs min overshoot.Do you think choosing a “rail-to-rail output” opamp would help?

author: Ian.M
author_rank: Super Contributor
timestamp: September 17, 2017, 01:30:12 pm
content: Try this.

author: Zero999
author_rank: Super Contributor
timestamp: September 17, 2017, 01:58:38 pm
content: I don't see why 200mA of overshoot, from 900mA to 1.1A is a big deal. The LEDs should be able to handle that with no problem. Why PWM the 50V? Why not the voltage reference?In any case, the added RC circuit in Ian.M's circuit is probably required to stop it from oscillating.

author: Damianos
author_rank: Frequent Contributor
timestamp: September 17, 2017, 08:02:42 pm
content: Quote from: treez on September 17, 2017, 12:13:52 amThe easiest way to test the LEDs is by the use of a properly selected SMU. The problem is that this may be a little expensive solution.So, to avoid the above, you are trying to do something else. But,to be really helped, some clarifications are needed. From the discussion it seems that there is a little confuse about the situation (there are proposals that are useful for msec, as they are simulated).I will try to identify the situation from your description:- there is a DC power supply for the load that is operated independently of everything else. It may be inactive, active for a period or continuously. Can it also operate in a pulsing mode?- there is a current controller (sink) that is always active and waiting to be hit from the above power supply!- the point, for the controller, is to guess when the hit will take place to react earlier!This goal, I think, is not possible with the topology that is presented, because the action of the independent power supply is involved inside the control loop. In other words, the result (current) must be there, before the controller starts to adjust the current.My proposal is a signal to be used, that will activate the current controller, after the output of the power supply. When the current is stabilized you can take the measurements.This signal can one that may be provided from the power supply or can be created from the output of it.... ... ... ...[I see now that you insist to have a controller that operates perfectly without knowing what it has to control!]

author: Alex Nikitin
author_rank: Super Contributor
timestamp: September 17, 2017, 08:58:30 pm
content: Here is one possible solution.CheersAlex

author: ocset
author_rank: Super Contributor
timestamp: September 17, 2017, 09:15:12 pm
content: Alex Nikitin and Ian.M ...both of your solutions are pretty well superb. Thankyou very much.

author: Ian.M
author_rank: Super Contributor
timestamp: September 17, 2017, 10:24:34 pm
content: Yes. Alex has a *very* neat solution if one has access to the high side of the LEDs. I like the P-JFET gating the feedback loop and he's done his 'homework' and is using it within its reverse gate breakdown voltage spec.

author: Zero999
author_rank: Super Contributor
timestamp: September 17, 2017, 10:39:34 pm
content: And here's my suggestion implemented: switching the reference.

author: ocset
author_rank: Super Contributor
timestamp: September 18, 2017, 05:15:55 am
content: Thanks Hero999, thats another great solution.

author: Damianos
author_rank: Frequent Contributor
timestamp: September 18, 2017, 11:39:58 am
content: Quote from: treez on September 17, 2017, 11:17:50 amQuote from: treez on September 17, 2017, 09:15:12 pmQuote from: treez on September 18, 2017, 05:15:55 amI am totally confused!Are all of these in the same discussion? The proposed solutions are in contrary with what you are stated, but you find them perfect, even if initially you refused them!The information given is incomplete:- has the power supply a fixed voltage or it can change? Maximum, minimum?- when it is deactivated is its' output actively to zero volts or it is disconnected by a switch/relay/transistor?- how short can be a "pause" before the next activation?- is the reference (that controls the current) fixed?...These affect more or less the solutions proposed so far.

author: ocset
author_rank: Super Contributor
timestamp: September 19, 2017, 04:54:25 am
content: Thanks, there has been some great info and i am grateful....unfortunately i am not able to give the full details of the app as my employer would not be happy

author: bson
author_rank: Unknown
timestamp: September 20, 2017, 07:58:51 pm
content: You can also turn off the reference voltage, or derive it from the LED supply assuming precision isn't important to drive LEDs. When off, it can be pulled to ground. A simpler resistor divider will do this, perhaps with a zener across the bottom half to make it less dependent on the LED supply if this is highly variable. Also, make sure the feedback loop is always closed, even with something like a 1M resistor, so the op amp never operates in open loop.