FS6500/FS4500 power efficency

[Jack_Zhao]

Hi Sir,

     We are design a automotive project and has some questions for the efficency of FS6500 and 4500 as below, please kindly give your comment, thanks a lot:

1. As to FS6500: Are the vcore DC/DC, VCCA, VAUX and VCAN all generated by 6.5V;
   As to FS4500: Are the vcore LDO, VCCA, VAUX and VCAN all generated by 6.5V

2. Is the LDO efficency is calculated as below:
LDO input Voltage= 6.5V
5.0V output: 5/6.5=76.9%
3.3V output: 3.3/6.5=50.8%

3. What is the efficency in different load?
Vpre and Vcore

Light loads:
Out current
1mA=?
10mA=?
100mA=?

Normal loads:
300mA=?
500mA=?
700mA=?

It will be great appreciated if you can kindly recommend your light load but high efficency parts(for automotive) to us?

[guoweisun]

1. As to FS6500: Are the vcore DC/DC, VCCA, VAUX and VCAN all generated by 6.5V;
   As to FS4500: Are the vcore LDO, VCCA, VAUX and VCAN all generated by 6.5V

[gw]All input from VPRE.

2. Is the LDO efficency is calculated as below:
LDO input Voltage= 6.5V
5.0V output: 5/6.5=76.9%
3.3V output: 3.3/6.5=50.8%

[gw]No,you should use the Efficiency= VOUT*IOUT/(VIN*IIN).

3. What is the efficency in different load?
Vpre and Vcore

Light loads:
Out current
1mA=?
10mA=?
100mA=?

Normal loads:
300mA=?
500mA=?
700mA=?

It will be great appreciated if you can kindly recommend your light load but high efficency parts(for automotive) to us?

[gw] you can use the EVB to test the light load efficiency not too much data here:

[Jack_Zhao]

Hi GW,

     Per your comment: 1. VPER=6.5V; 2. LDO Efficiency= VOUT*IOUT/(VIN*IIN) and following the LDO feature IOUT=IIN, so LDO efficiency=VOUT/VIN(6.5V).

     For example, Vout=5V, LDO efficiency =5V/6.5V=76.9%

    Please correct me if i misunderstand your meaning.

    Thanks a lot.

[guoweisun]

No,the input current is not equal with output current for efficiency calculate.

[Jack_Zhao]

Hi GW

Do you have some efficiency data for VCCA, VAUX and VCAN？

It seems Vcca&Vaus are not pure LDO as what we know.

We would like to know the  efficiency for those ouput with light loads especially.

NXP web site has a tool:FS6500-FS4500 Power Dissipation Calculator  https://www.nxp.com.cn/downloads/en/calculators/FS6500-FS4500-power-dissipation-calculator.xlsx Could we refer to the excel file for efficency evaluation？

We calculated with this tool and found the efficency(Vcca and Vaux) is very high in light load such as 1mA-10mA.(the result is as below picture)

They should be Linear voltage regulator.

We are not sure if it is correct or not？

Could you help to double check?

Thanks

From Foxconn RD Qinguang and Jack

[guoweisun]

Yes, you can use  Power Dissipation Calculator  to estimate data.

[Guang-Qin]

Hi GW

We have doubts about the high efficiency of these two circuits (Vcca and Vaux). Is this really correct?

The efficiency does not change with the load. It is unreasonable. Can you help to confirm?

Another question is, They are  linear regulator, how can it achieve such high efficiency? ==> More than 99%?? It is unreasonable also.

FXN  Qin

[guoweisun · Accepted solution]

For VAUX used the external FETs which mean the FETs dissipation no calculate inside of SBC,so the efficiency will be higher.