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LM675: Case Number: CS0720693

Part Number: LM675
Other Parts Discussed in Thread: UA741

LM675 Pinout and Schematic.docxLM675 Inverting OPAMP.xlsx

  • Hi Don,

    Do you intend to mount the LM675 TO-220 package on a heatsink? If so, we need the heatsink spec's to derive the lowest resistive load it can drive without overheating.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • Thomas,

    I do not have the specs for the heat sink as it is out of the “junk” box. Have attached a photo of the re-created circuit that am now working with. Have also observed higher frequency sine wave outputs from an o’scope on the output of the LM675 while it overheats.

     

    Have previously tested this circuit in an earlier version and it worked fine accept for getting very warm. Can send a photo of that setup if necessary.

     

    Thanks,

    Don McCormick

  • Hi Don,

    Since you don't have the info about the heatsink an image of it might provide us something to work from. Since we won't have the exact thermal specifications about all that can be done is to try and find something similar and apply its specifications to determine the LM675 dissipation limits.

    You ask the question -

    Unity Gain LM675 Inverting Amplifier Schematic, Vin=60Hz, R2 =Rin=Rf =10K,  R1=1K,  C=0.22uF – Why are capacitors rated at 500kHz rather than at the operating frequency?

    The 500 kHz really doesn't have anything to do with the capacitor rated for a frequency of 500 kHz, but rather it is a factor in the R1C = 1 /(2 pi 500 kHz) equation. The equation assures that the LM675 which is normally required to be operated with a closed loop gain of 10 V/V or greater, can be stabilized when operated in this inverting G = -1 V/V configuration. The R and C values are positioned in the circuit and set to compensate it. Keep in mind the unity gain bandwidth for the LM675 operating at a G >= 10 V/V has a unity gain bandwidth (GBW) of 5 MHz. However, when the inverting gain of -1 V/V circuit is applied the GBW drops to 50 kHz. The bandwidth is much reduced as a result of the compensation.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • The heat sink is an Aavid Boyd. Not as tall but similar to the Aavid Boyd 58100802500G. It is on the right side of the attached photo.

    Thanks for your assistance.

    Don

  • Hi Don,

    I've been searching the Aavid Boyd website and the 58100802500G heatsink number you provided does not show up in a search. I think the correct number is something along the lines of 581008B02500G, and unfortunately that exact number doesn't up either. The closest number to that which they do provide information is 581101B02500G and it is a 4-fin extruded heatsink so it may be in the family and similar to what you are using.

    https://info.boydcorp.com/hubfs/Thermal/Air-Cooling/Boyd-Board-Level-Cooling-Extruded-5810.pdf

    If we use the 58100X heatsink and the thermal graph from its datasheet we can see that the heatsink's thermal resistance is a function of both the air velocity and the power being dissipated. The worst case heatsink to air thermal resistance occurs when the air velocity is zero. Since we don't have complete information about your heatsink lets use 10 °C/W for the thermal resistance from heatsink to ambient, ϴHS-A.

    Then, based on Figure 8 in the LM675 datasheet we can arrive at the maximum power dissipation. If the ambient temperature is 25°C, the maximum power dissipation is 10 Watts.

    The worst-case, maximum op amp power AC dissipation can be found from the following relationship:

    Popa(MAX_AC) = (2 ∙ V+2) / (π2 ∙ RL)

    Transposing the equation and solving for RL:

    RL = (2 ∙ V+2) / (π2 ∙ Popa(MAX_AC))

    For V+ = 15 V, and  Popa(MAX_AC) = 10 Watts 

    RL = (2 ∙ 15 V2) / (π2 ∙ 10 W) = 4.6 Ω

    That should get you pretty close to the minimum load resistance that can be used for the conditions stated and a heatsink having a ϴHS-A  of about 10 °C/W.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • PSpice Output File from schematic above. Note that amplitude is 9.1Vpk at f = 60 Hz.Actual O’scope output at R21 is approximately 2Vpk (reduced amplitude) at approximately 8 MHz (increased frequency). What is the problem here?   (Bypass Caps from LM675 power at pins 3 and 5 to ground (pin1).)

  • Hi Don,

    An unexpected 2 Vpk, 8 MHz sine wave, or distorted sine wave observed at the LM675 output would indicate the stage is oscillating. The R and C values you used for the LM675 stage look to be correct. However, the circuit is missing is the series 1 Ohm + 220 nF RC network from the LM675 output to ground. It is part of the overall compensation being applied and must be included.

    Do note that simulation models are simplifications of the real product. They produce close behaviors to the real product but it is difficult to design a simulation model to behave exactly like the real silicon amplifier. Therefore, especially with things like an oscillation you may find that the real product doesn't oscillate when the model version does oscillate and visa versa. I suggest you that once the output network is added you not only simulate the circuit, but actually test it on the bench and see what it does.

    Regards, Thomas

    Precision Amplifiers Applications Engineering  

  • Hi Thomas,

    The solution at this end was adding a 200pF cap between Vee pin 3 and grounded pin1 as well as another 200pF cap between VCC pin 5 and grounded pin1. I had failed to place these in the circuit previously as noted (My Error) on the above schematic. This stopped the oscillation of the LM675 and the circuit now provides the expected clean 60 Hz output signal at approximately 9 Vpk.  Thanks for your patience with and advice to me. At least for this particular circuit, the problem appears to be solved even without the 1 Ohm + 220 nF RC network from the LM675 output to ground. If you recommend, I will add those to the circuit for overall compensation. Many advise, however, "If it ain't broke, don't fix it."

    Best regards,

    Don

  • Hi Thomas,

    The solution at this end was adding a 200pF cap between Vee pin 3 and grounded pin1 as well as another 200pF cap between VCC pin 5 and grounded pin1. I had failed to place these in the circuit previously as noted (My Error) on the above schematic. This stopped the oscillation of the LM675 and the circuit now provides the expected clean 60 Hz output signal at approximately 9 Vpk.  Thanks for your patience with and advice to me.

    At least for this particular circuit, the problem appears to be solved even without the 1 Ohm + 220 nF RC network from the LM675 output to ground. If you recommend, I will add those to the circuit for overall compensation. Many advise, however, "If it ain't broke, don't fix it."

    What do you recommend, adding the compensation pair to the existing circuit or building another circuit and leaving this one as is?  

    Best regards,

    Don

  • Hi Don,

    I suspect the original series output RC network was developed as a solution when the LM675 is driving an appreciable output current. The safest bet is to include the series RC output network in case your next load is different than what you are currently using.

    You will note that all the circuits in the LM675 datasheet include the network regardless of the gain or configuration. My concern with just adding the 200 pF capacitors alone as you describe may work well for the LM675 devices you have in hand, but may not do the job with product you receive of in the future. Semiconductor product electrical parameters do have variance across production runs and for that reason a design needs to be designed for maximum robustness.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • Thomas,

    Will install the 1 Ohm, 200 pF compensation components as recommended and test again.

    Future loads will change. I used the 1K Ohm load on this circuit just to be safe. Hope to parallel LM675's to increase current and feed a 6.3 -120 transformer to build a small inverter.

    Appreciate your advice.

    Please remove me from receiving responses to your other inquiries.

    Best regards,

    Don McCormick

  • Thomas,

    When the LM675 is putting out 6.3 Vrms into a 1 ohm resistor we need almost 40 Watts across that resistor which overheats both the 5W resistor and the LM675 supplying its voltage.

     

    For noninverting UA741 Op Amp

    P RMS

    P RMS

    P RMS

    With Ri =10K, Calculating Rf Required to

    O'Scope

    Fluke

    Pspice

    Gain

    675 Pout

    675 Pout

    675 Pout

    produce 120Vrms out

    Pk -Pk Max

    RMS

    Peak

    (Rf/Ri)

    =E^2/R

    =E^2/R

    =E^2/R

    ICL 8038 Voltage Out

    7.40

    2.62

    3.70

    1

    R=1

    R=8

    R=16

    19/1 UA741 Vout 14.08K =Rf (Need)

    17.82

    6.30

    8.91

    2.41

    39.69

    4.96125

    2.48063

    R Watts

     

     

    In order to decrease the power required through that resistor without reducing its effectiveness, can the resistance value be increased to 8 or 16 ohms?

    Thanks,

    Don

  • Hi Don,

    It is evident from your calculations that the power decreases significantly as the load resistance is increased from 1 Ohm to 16 Ohms. Providing the LM675 produces the output voltage and current required for the application there is no reason to run it and the load at any higher power than is necessary. Reduced power dissipation and heating benefits the longevity of the components.

    Regards, Thomas

    Precision Amplifiers Applications Engineering