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LM675 in AC current source application

Gordon Su
Intellectual 285 points
Other Parts Discussed in Thread: LM675, TINA-TI

Hi,

I have a couple of questions regarding using LM675 as improved Howland current source to drive an inductive load. The circuit is shown below. R7, L1 and C1 represent the equivalent model of the inductor when measured with a LCR meter. The problem I am seeing is that the circuit is not stable. Without any signal input the output would oscillate between positive and negative rails at around 850 Hz. my questions are as below,

1. In LM675 datasheet Figure 15 the RC snubber is connected directly at the op amp output while in my case it is in parallel with the load. which location is a better one? http://www.ti.com/lit/ds/snosbp3e/snosbp3e.pdf

2. In TINA I couldn't find the spice model for LM675. Can someone provide a model for this IC? alternatively, can someone advise which IC might have similar performance and can be used instead of LM675?

3. I tried to measure the frequency response with output as the voltage across R9 (VR9 = Iout/2). I see only amplitude peaks at 2 times at around 1KHz. the phase shift is much less than 180 deg. the circuit should not oscillate according to the classic negative feedback theory. however, the below circuit has both positive and negative feedbacks. I am not very sure how to analyse it.

4. I am going to try increasing C2 to stabilize the circuit. any comments on this?

5. the inductor assembly in the application might vary from batch to batch. If parameters vary too much the circuit might become not stable again. If I want to set a LCR specification on this inductor how should I spec it? To spec the resonant frequency to be within certain range?

Thanks in advance

BR

Gordon

  • 0
    TI__Expert 6275 points

    I will provide assistance for you to stabilize this Improved Howland Currrent Source.  I will need to create some AC Macromodels for stability analysis and show you how to stabilize this.  I will work on this tomorrow and get you some details out in a powerpoint with TINA Schematics late tomorrow MST (DST).  From this analysis we can then detemine if LM675 can be used or if you will be better off with a different power op amp.  To stabilize will require loss of bandwith which is always true.  What is maximum frequency and peak current you are trying to get through the inductor? 

  • 0 in reply to Tim Green1
    Intellectual 285 points

    Hi Tim,

    thanks for your help and quick response.

    I have some updates and correction on my first post. First, the RLC value in the schematic was measured at 1KHz frequency. I realized that the RLC value changes with respect to frequency. thus, the frequency response with voltage output at test point VF1 is more indicative. With this I could see the amplitude peaking significantly at the oscillation frequency. The next step would be to find out why the latest batch of boards show more oscillation cases.

    the peak current is 2A. there is no spec on the frequency. the inductive load is part of voice coil motor. The motor is used to open and close a clamp. thus most of time the current passing through it is DC. Only during opening and closing certain current profile needs to be followed to minimize bouncing.

  • 0 in reply to Gordon Su
    TI__Expert 6275 points

    See two attached Powerpoint presentations (one in this post and a second in the following post)  One is an overview of the Improved Howland current pump and stability issues when driving inductive loads.  The other is a detailed analysis of your application with embedded TINA-TI Schematics you can use to analyze your ciruit.  The improved Howland Current pump is one of the most difficult circuits to stabilize.  It is extremely important that the load is accurately modeled, as you will see in attached, since it is directly in the feedback path of the op amp.  Review attached and feel free to let me know if you need further assistance or clarification of any points therein. 

    Howland Stability_TG.ppt
  • 0 in reply to Tim Green1
    TI__Expert 6275 points

    Second powerpoint relative to my previous post.

    LM675 Howland.ppt
  • 0 in reply to Tim Green1
    TI__Expert 6275 points

    Attached is a powerpoint which details the derivation of the proof that for dual feedback Beta = (Beta-) - (Beta+). It was not intuitive to me at first look.  Since I know this technique works such that Beta = (Beta-) - (Beta+) for the Improved Howland Current pump I thought you might find it useful to see the proof of the formula.  For your final design you must define accurately the load.  If the 500mohm resistor you put in the circuit to measure current is not in the final application it must be removed and the compensation re-done. In past applicatons I have analyzed the linear motors or valves have been predominantly resistive and inductive only.  Again, not sure that you may not have a unique load as you show quite a bit of capacitance in addition to the inductance and resistance.  If you research and find your load to be different let me know and I can help recommend best compensation.   Check with the vendor of your valve or motor and as a rule of thumb I have seen resistance tolerances of +/-15% and inductance tolerances of +/-30%. Once you get worst case tolerances you can check with TINA-TI SPICE that your circuit will be stable for all load variations.  

    Dual Feedback Beta Derivation.ppt
  • 0 in reply to Tim Green1
    Intellectual 285 points

    Hi Tim,

    thanks for your detailed explanation.

    Give me some time to digest the materials you've prepared. I will come back to you if I have more questions. If I have the choice of re-design the circuit I will definitely follow your suggestion. however, without re-layout the board I guess I could increase the capacitor value (10uF) for the snubber circuit. As the result the bandwidth suffers a lot. but this might be acceptable as short term workaround.

    BR

    Gordon

  • 0 in reply to Gordon Su
    TI__Expert 6275 points

    We can look at compensating correctly the load where you have the snubber located.  The key is precise characterization and modeling of the load.  If you can confirm the final characteristics of the load (with or without the extra 500mohm you show in series with the inductor-capacitor arrangement) I can analyze and recommend values that should be able to give good phase margin by analysis. 

  • 0 in reply to Tim Green1
    TI__Expert 6275 points

    If you need some information on Stability Analysis for Amp ciruits the link below is a good reference. Parts 1-4 wiol give you all of the tools you need for op amp stability analysis.  The remaining parts are mainly focused on ways to stabilize capacitive loads.  

    http://www.en-genius.net/site/zones/acquisitionZONE/technical_notes/acqt_050712

  • 0 in reply to Tim Green1
    Intellectual 285 points

    Hi Tim,

    Thanks for your great support.

    I  found the root cause of the problem. It is due the 1% resistor on both feedback paths. On the two boards that are not working the overall feedback becomes positive. After changing the resistor to 0.1% the problem is solved. I should not jump directly to suspecting the frequency response. In my defence the compensation method is really suspicious. On the next re-spin of the board i will put in the suggestion from you to make it right.

    below is the impedance vs frequency response of the inductive load. there are two tabs covering different frequency spectrum. Amplitude and phase were measured with Dynamic Signal Analyzer. inductance is calculated assuming only Rs and Ls. it seems parasitic capacitance should be included as well. The free version of Tina does not allow to use impedance curve, does it?

    7242.forcer axis electrical model 20130725.xlsx

    Thanks

    BR

    Gordon