AD8145 and AD8146 Load Current Calculation

Hi,

   Dissipation on the power due to the load drive depends on the particular application and generally for each output, the power due to load drive is calculated by multiplying the load current by the associated voltage drop across the device. The power dissipated due to all of the loads is equal to the sum of the power dissipation due to each individual load.

   From this FAQ you can find load current calculation at MAX9959: FAQ Calculate Sense Resistor Value Based on Maximum Load Current - Documents - Automated Test Equipment (ATE) - EngineerZone (analog.com) & Load current of amplifiers - Documents - Amplifiers - EngineerZone (analog.com) Since it is an other part.

Thanks,

Poornima

As per my understanding, i have calculated the current consumption of IC, could you verify it once. Dual supply voltage is used. Since feedback current is less. I have neglected it. Calclation are as follows:

Total current Consumption - 161.66mA

Then for Total Power Dissipation, I think i can consider either 363mW or 338mW, (either he will drive low or High). I s this correct?

As per my understanding, i have calculated the current consumption of IC, could you verify it once. Dual supply voltage is used. Since feedback current is less. I have neglected it. Calclation are as follows:

Total current Consumption - 161.66mA

Then for Total Power Dissipation, I think i can consider either 363mW or 338mW, (either he will drive low or High). I s this correct?

Hi,

 As per specification "AD8145/AD8146 is specified to operate on a single +5V supply (i.e +5V single-supply operation, that VS+ is connected to +5V, VS- is connected to 0V, and the GND pins are connected to 0V ). So Please take the positive supply voltage power dissipation value 363 mW.

 Also Please note that, Using only negative supply to power the AD8145/AD8146 can be done as long as it is within the supply ranges but some of the internal circuitry of these ICs will not work properly because these internal circuits were designed to operate in either single positive supply or dual supplies. Also, using negative supply will not provide proper output signal.

It is better to follow the recommended supply voltage ranges indicated in the datasheets in order to get your desired output.

Please refer below Expert suggestion regarding "Power supply requirements are based on system implementation".

  The power dissipated in the package (PD) is the sum of the quiescent power dissipation and the power dissipated in the package due to the load drive for all outputs.

  The data sheets contain all the current information needed to calculate complete system power requirements.  Regarding the AD8145, the data sheet shows the quiescent power requirements however you must also include the output load requirements.

As a simple example: Assume the ADV8145 is being supplied by +-5V, the load is 100 Ohms and the input is 1Vrms sinewave.

So P5Vquiescent = 5V * 70mA = 350mW

P-5Vquiescent = -5V * -65mA = 325 mW.

P5Vload = P-5VLoad = 100 * 1/QRT(2) / 2 = 35mW

Pad8145_5V = P5Vquiescent + P5Vload = 350mW + 35mW = 385mW

Padad8145-5V = P-5Vquiescent + P-5Vload = 325mW + 35mW + 360mW

The same basic approach applies to all the other devices you've listed.  I would pencil sketch a system, then build a simple excel spread sheet that can quickly calculate you power needs.

Is there any necessity to use negative voltage or dual supply and risks involved if only positive voltage is used ?

   Split power rails are useful if the source singles are close to ground which often is the case.  Single rails can be done but more care has to be taken with AC couplings and signal processing.

Solution about voltage supplies to enhance the overall performance of these components ?

  A solution is dependent on what your system looks like.   To save power you can disable parts when the signal path is not used.

Thanks,

Poornima

PoornimaSubramani said:

8

if my supply voltage is 5 V, Load = 100 Ohm,  1Vrms-sinewave

Then , Power Consumption due to Load = ((Vrms)^2)/R) =(1/100)= 10mW

Could you please explain how you got P5Vload = P-5VLoad = 100 * 1/QRT(2) / 2 = 35mW ?

Hi,

  This calculation we have taken it from expert comment at Power consumption - Q&A - Video - EngineerZone (analog.com)

  Will check with expert and let you know about this.

Thanks,

Poornima

Hi Poornima,

Any update regarding this?