why there have so lagre Cap 1.5mF in the OPT8241_CDK_EVM?

Hi

The two huge caps are for energy buffering. During integration time, the high current required by the laser diodes are supplied from these caps.

Hi Nithin,

Thanks for you information.

Q1:The high current is Transient current,all right?

Q2: The energy stored in Cap: P=1/2*C*U^2, Asumed laser diode VF is forward voltage,IF is forward current , according to Conservation of energy, we can solve C：C=2*VF*IF/U^2, all right?

Thanks&Regards

None

Hi None_feiu,


High current meant the current the forward current current through laser diode during integration time. Because of the presence of the inductor , average voltage at the ouput of the current source = 4* VF*0.5

I will show an example case for determining the energy buffering required.
Now you have to decide , what is the maximum integration time you are allowing. If I assume 100uS of integration time,

Assuming
VF =2.4 V per laser at 2.5A
Energy taken up by the lasers = 2*2.4*2.5*100u = 1.2mJ
Now you have to account for the losses in linear source as well as in the switching mosfet.
losses in mosfet = 2.5A^2*27m Ohm * 50uS = 8.4 uJ
Assume that you are allowing 150 mV drop across the linear current source .
Losses in linear current source = 150mV*2.5A*100uS = 37.5 uJ
Hence take total losses as say 50mJ
Total energy required = 1.25mJ
Initial voltage in Capacitor V1 = 5V
Here say I am ready to allow a max drop of 150mV
Final voltage V2 =4.85 V

Now as you suggested
.5*C*(V1^2 - V2^2) =energy required = 1.25mJ
Hence C = 1.7mF
Assuming a derating factor of 0.8 , required capacitor become 2.12 mF.
Assuming a safety margin , two 1.5mF capacitors are connected in parallel.

Thanks,
Nithin

Hi Nithin,

for you explain,i am clear about this. but i am also have a question about this?

Q1:The Power is supply by the DC 5V and integration time is 100us or maybe this time is large than 100us ,it seem like this time is not fast,So I think the current through the laser is Completely can acquire by the DC Power supply.

what you say,in my understand,if integration time is so short,it makes the DC power supply can not supply this need current in this short time,so it need large Cap to store energy to response this short time?

Thanks&Regards

None

Hi None_feiyu

the capacitors act as a fall back in case the input current could not be 2.5A. For example, OPT8241_CDK can be used on USB only input. In that case ilium board gets 100mA input only. Then this current is supplied by the capacitors.

Hi Nithin,

Thanks for you information.

I check the opt8241_CDK_EVM sch file, when the usb supply only,i found there have TPS2552 limlit the current to the illum board.

Now,my quesetion is :

Q1: Can you give me a example to explain why limit the current to illum board only 100mA? why not 50mA,it also work well.

Thanks&Regards

None

Hi None
The lasers may work upto 4.5V. We have set 100mA current limit to illum board so that we can support higher integration duty cycles in general. As long as you are able to replenish the capacitor, you should be okay. With 50mA current limit, you will be limited to lower integration duty cycles.

Thanks & regards
Nithin

Hi Nithin,

I know what you mean, in my opinion,as long as can able to replenish the capacitor in the time t=(Quad time - integration time),so the 50mA current will be ok.

About the time t=(Quad time - integration time) what my understand is right or not?

Thanks&regards
None

Hi Nithin,

Thanks for you information.