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ALM2402 power dissipatation

Yusuf Korkmaz
Prodigy 50 points
Other Parts Discussed in Thread: ALM2402-Q1, ALM2402F-Q1, ALM2403-Q1

Hello dear forum,

we are using in our application the ALM2402 to drive an load with approx. 100 Ohm impedance (1.5mHenry by 10Ohms DC resistance). The output of the amp is for both channels a sine wave with approx. 11.5 high peak and approx. 2V low peak, DC biased with approx. 6.7V - channel 2 same values just shifted at 180° degree. Both outputs are connected differentially to the load.

When I had a look to the datasheet (p.22), there is a formula to calculate power dissipation. I think the given formulas are dealing only the resistive load. Can you give me a hint how to calculate when a inductance is connected to the outputs ? 

My plan is to calculate with the approach:

PAMP = PINSupply + (POUTSupply - PLOAD)

for PINSupply: VBAT * ICC of the IC

for POUTSupply: VBAT * ILOAD  --> here i am not sure if the formula is correct, if I have to use the RMS or peak current. 

for PLOAD: VOUTAMP * ILOAD * Cosi(Phi) --> Cos(Phi): cos(RDC/ZLoad) --> here i am not sure if the formula is correct, if I have to use single ended voltage or the differential voltage for VOUTAMP, and also not sure if i have to use the peak or RMS value, and for the current not sure if to use the peak or the RMS

After finishing the calculation should I calculate the results times 2, since there are two channels in the IC ? There is no hint/mention in the datasheet, regarding this question, I think this is calculated only for one channel.

This calculation, I assume deliver the average power loss. Can you give me an useful hint how to calculate the peak power instead of the average power ?  

I found a DS of ALM2402 (maybe an old one): where the cos(phi) on the load is not considered, they used for the total calculation only the results from the formula: ALM2402-Q1 Dual Op-amp with High Current Output datasheet (Rev. D) (ti.com)

It would be gread if you can help.

BR and thanks in advance

  • 0
    TI__Guru 57241 points

    Hi Yusuf,

    Based on the description, you want to calculate the power dissipation for a resolver. TI has three resolver driver, namely ALM2402-Q1, ALM2402F-Q1 and ALM2403-Q1. I think that you are likely interested ALM2402F-Q1, which is suited for higher BW and amplitude resolver driving applications. ALM2402-Q1 works as well, but its slew rate is low and is limited for lower BW and small amplitude resolver due to limited full power bandwidth. 

    Yes, here is the easier way to calculate the average power dissipation in ALM2402F-Q1 or extract other power information from the simulation. 

    BTW, we created power calculation icon as shown below, which it made easier to extract the power. You may do the similar power calculations by multiply Vout*Iout at the excitation frequency from the simulation plots. You may place a current probe at Vcc power rail and voltage probe at the each power amplifier's output, and get the product of the two via post processing.

    Per your output voltage swing design requirements, I simulated the following scenario. It is close to what you defined, but I did not fine tune the output voltage swings yet, since I do not have your input information. 

    ALM2402F-Q1 Simplified Resolver Driver 01312024.TSC

    Yusuf Korkmaz said:
    an you give me an useful hint how to calculate the peak power instead of the average power ?

    In Tina simulation, while the plot display window is active, you may left-click on Process --> click on Average --> which it will show the average power dissipation of the curve. 

    If you use cursor a and/or b, you can find out the peak power instead, which the plot is shown at approx. 411mW or 0.41Wpk. 

    Please keep in mind that both ALM2402-Q1 and ALM2402F-Q1 have the input Vcm requirements, while the Vcm in ALM2403-Q1 is close to rail to rail.

    (FYI, ALM2402-Q1 and ALM2402F-Q1 are basically the similar design, except ALM2402F-Q1 has higher slew rate and other improvements). 

    If you have other questions, please let me know. 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    Prodigy 50 points

    Hi Raymond,

    thanks for the quick response - I appreciate it very much.

    Yes, it is a resolver application. I tried it also with the simulation and got similar results.

    The main question which I need support is:

    1.) for estimating the junction temperature i need the overall power dissipation of the IC. Therefore the approx. 400mW from the simulation, is then for one channel or for two channels? Since the IC has two amplifier stages in the package, I need the overall loss of both amplifiers U1 and U2 for calculating the TJ of the chip. Can I use approx, 400mW or need I mulitply it with 2?

    2.) I need also a confirmation/support about my calculation approach, since my goal is the get the information about the overall loss in mathematical way, a simulation is helpful to compare the simulation results with my calculation.

    3.) Should I use the overall average loss for the thermal calculation or the peak losses - any hints?

    BR

    Yusuf

  • 0 in reply to Yusuf Korkmaz
    TI__Guru 57241 points

    Hi Yusuf, 

    Yusuf Korkmaz said:
    Can I use approx, 400mW or need I mulitply it with 2?

    There are two power amplifiers per ALM2402F-Q1 package. So I would use 0.5W as simulated from above, because each power amplifier is dissipated approx. 0.2W at 25C. Please do not ignore the quiescent current which it contributes approx. Vcc*6mA of power. 

    In terms of Tj calculation, you may use the following equation Tj = Tamb + 27.6C/W*0.5W, and this should estimate Tj figure at the worst Tamb operating conditions. ALM2402F-Q1 has a thermal lug on the bottom of the package, and 90% of heat will be dissipated through the lug via PCB.   

    Yusuf Korkmaz said:
    2.) I need also a confirmation/support about my calculation approach, since my goal is the get the information about the overall loss in mathematical way, a simulation is helpful to compare the simulation results with my calculation.

    You may also take a look at our video training series about the derived calculation. 

    https://www.ti.com/video/series/precision-labs/ti-precision-labs-op-amps.html?keyMatch=PRECISION%20OP%20AMP%20VIDEO%20SERIES

    Yusuf Korkmaz said:
    3.) Should I use the overall average loss for the thermal calculation or the peak losses - any hints?

    For Tj and power dissipation calculation, the key variable are average power at the excitation frequency for resolver. All other power parameters are derived from the average power. For academic discussions, engineers will calculate and express the power in other terms. As you may know, there is only one power in the practical sense in the power calculation, which is the average power for AC drive.  Power company does not charge user by the peak power...

    I am going to close the inquiry for now. If you have other questions, please continue to post it and I will reply it.   

    Best,

    Raymond