REF3325: how to calculate the power consumption of Voltage reference

Dong Ba

Part Number: REF3325
Other Parts Discussed in Thread: LM4050-N, REF3425

Hi Team,

Could you please help to figure out how to calculate the power consumption of Voltage reference? like DCDC or LDO or normal IC?

Please help to provide an example with 1mA load with Series voltage reference and Shunt voltage reference seperately, e.g. REF3425 and LM4050-N.

B.R.

Eric

over 2 years ago

0 Tony! over 2 years ago

TI__Intellectual 1735 points

Hi Eric,

I am looking into this - I will get back to you shortly.

-Tony

0 Tony! over 2 years ago in reply to Tony!

TI__Intellectual 1735 points

Hi Eric,

Here's how you calculate the power consumption on Series and Shunt Voltage references:

Pseries = ((Vs-Vref)/Rs)^2 Rs

Pshunt = Vs(I_L+I_Q)

-Tony

0 Dong Ba over 2 years ago in reply to Tony!

TI__Intellectual 825 points

Hi Tony,

Thanks for the quick response, but it seems something puzzled,

Pshunt = (Vs(I_L+I_Q), Correct.

Pseries = (Vs-Vref)^2 /Rs, what does Rs mean? Could you please help to provide an example with REF3425 ?

-Eric

0 Tony! over 2 years ago in reply to Tony!

TI__Intellectual 1735 points

Hi Eric,

My bad, I flipped the equations. It should be:

Pshunt = ((Vs-Vref)/Rs)^2 Rs

Pseries = Vs(I_L + I_Q)

-Tony

0 Dong Ba over 2 years ago in reply to Tony!

TI__Intellectual 825 points

Hi Tony,

Please help to provide an example seperately. Thanks.

1.Series, REF3425

2.Shunt. LM4050-N

B.R.

Eric

0 Tony! over 2 years ago in reply to Dong Ba

TI__Intellectual 1735 points

Hi Eric,

I'll take a look at it.

-Tony

+1 Tony! over 2 years ago in reply to Dong Ba

TI__Intellectual 1735 points

Hi Eric,

For Series:

Vs = input voltage

I_L = load current (which you initially provided as 1mA),

I_Q = Quiescent current ( which is 72uA for REF3425 in active mode)

Let's assume Vs = 5V and I_L = 1mA

Thus,

P = 5V (1mA + 72uA) = 5.36mW

For Shunt: Below is a schematic from LM4050-N

If you want to calculate the power dissipated through Rs, then use the equation I provided earlier Pshunt = ((Vs-Vref)/Rs)^2 Rs , where Vref = Vout (from the schematic) or reversed breakdown voltage VR as stated on the datasheet, and Rs is determined by the supply voltage (Vs), load and quiescent current (I_L & I_Q), and the reversed breakdown voltage VR. Rs = (Vs-VR)/(I_L+I_Q)

If you want to calculate the power consumption of the voltage reference as entire system, the equation is Pshunt = [(Vs-Vr)/Rs]*Vs (same variables as explained above.

Let Vs =5V, Vr = 2.5V, and Rs = 900 Ω

Thus, P = 13.9mW

0 Dong Ba over 2 years ago in reply to Tony!

TI__Intellectual 825 points

HI Tony,

Thanks for the detailed feedback.

If we only consider the reference part and remove the load.

For Series: Pseries =Vs(I_L + I_Q)-Vout*IL, the same with LDO right?

For Shunt: if we consider Rs and the voltage reference as a Group, Pshunt = [(Vs-Vr)/Rs]*Vs-Vr*IL

B.R.

Eric

0 Tony! over 2 years ago in reply to Dong Ba

TI__Intellectual 1735 points

Hi Eric,

The unit analysis doesn't match on your suggested equation.

[(Vs-Vr)/Rs]*Vs-Vr*IL

[Voltage/Ohm] [Voltage] [Amp] = [Amp][Voltage][Amp] = [Amp]^2 [Voltage] ≠ [Watt]

0 Dong Ba over 2 years ago in reply to Tony!

TI__Intellectual 825 points

HI Tony,

Thanks for the detailed feedback.

You misunderstand the equation.

[(Vs-Vr)/Rs]*Vs-Vr*IL=[(Vs-Vr)/Rs]*Vs-(Vr*IL)

So Am I right about description below?

-----

For Series: Pseries =Vs(I_L + I_Q)-Vout*IL, the same with LDO right?

For Shunt: if we consider Rs and the voltage reference as a Group, Pshunt = [(Vs-Vr)/Rs]*Vs-(Vr*IL)

-----

B.R.

Eric

0 Tony! over 2 years ago in reply to Dong Ba

TI__Intellectual 1735 points

Hi Eric,

Your equations seem reasonable if you are interested only in power being dissipated by the Vref.

-Tony

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REF3325: how to calculate the power consumption of Voltage reference