How reduce power loss on current sense resistor in boost SMPS drived by UC3843?

[carpenter]

I'm going to the boost isolated converter from +12V to +400V (20W)
I use about UC3843 on frequency 130kHz.
I do not like the losses to current sense resistor 150m Ohm.
I would like to reduce it.
It occurred to me to use 5mOhm resistor and amplifier with gain 30.
The question is whether it is a good idea, what will be the effect on the stability of the control loop?
And above all, what amp to use?

 

[barry]

The current limit circuit is not part of the control loop, it simply turns the output off, so I don't think that's a worry.

But I should ask, you've got a 20Watt converter, that means your output current is 50mA. 50mA through a .15 ohm resistor is .000375 watts. Why are you concerned about that?

 

[schmitt trigger]

An Unitrode Semiconductor (Now Texas Instruments) app note indicated that for such instances, a current transformer is required. The app note gave tips on how to implement the proper circuit to interface it to the controller, including the all-important flux reset.

Let me see if I can find the note.

 

[carpenter]

The current limit circuit is not part of the control loop, it simply turns the output off, so I don't think that's a worry.

But I should ask, you've got a 20Watt converter, that means your output current is 50mA. 50mA through a .15 ohm resistor is .000375 watts. Why are you concerned about that?

How 50mA , primar is 12V, it is boost converter from 12V to 400V. Current about 1,7A RMS.
Current control is one from error amplifier in UL3843, i thing current control is part of control loop. In any case , deactivate current control in circuit powered from 100Ah battery is impossible .

 

[barry]

How 50mA , primar is 12V, it is boost converter from 12V to 400V. Current about 1,7A RMS.
Current control is one from error amplifier in UL3843, i thing current control is part of control loop. In any case , deactivate current control in circuit powered from 100Ah battery is impossible .

1.7 amps on the INPUT, 50 mA on the OUTPUT. It's the OUTPUT current that is flowing through the sense resistor.

And, again, the current limit circuit is NOT part of the control loop. Look at the data sheet.

- - - Updated - - -

How 50mA , primar is 12V, it is boost converter from 12V to 400V. Current about 1,7A RMS.
Current control is one from error amplifier in UL3843, i thing current control is part of control loop. In any case , deactivate current control in circuit powered from 100Ah battery is impossible .

1.7 amps on the INPUT, 50 mA on the OUTPUT. It's the OUTPUT current that is flowing through the sense resistor.

And, again, the current limit circuit is NOT part of the control loop. Look at the data sheet.

 

[FvM]

I presume you are talking about UC3843 current sense pin? An optional sense amplifier should be able to track the current waveform, then there won't be a problem. Several 100 kHz bandwidth is probably necessary. I think, the shunt power dissipation should be in a reasonable relation to overall switcher losses. It doesn't make much sense to improve efficiency by 0.1% or less, but 1 % might matter. A low ohmic shunt will ususally need 4-wire ("kelvin") connection and a respective amplifier circuit.

 

[carpenter]

I'm not sure if you fully understand.
I'm talking about the sensing resistor in the Source of switching transistor to Primary.

 

[schmitt trigger]

I repeat my advice, check the app note on using a current transformer.

 

[carpenter]

Yes I see App note and hut search for very small current transformers ,**broken link removed** seem too big, for this aplication.

- - - Updated - - -

I presume you are talking about UC3843 current sense pin? An optional sense amplifier should be able to track the current waveform, then there won't be a problem. Several 100 kHz bandwidth is probably necessary. I think, the shunt power dissipation should be in a reasonable relation to overall switcher losses. It doesn't make much sense to improve efficiency by 0.1% or less, but 1 % might matter. A low ohmic shunt will ususally need 4-wire ("kelvin") connection and a respective amplifier circuit.

It's not quite so little is
With an average current 1.6 A over sense resistor talking about power dissipation 380mW on 150Ohm
If you know of a similar circuit that enables the construction of boost power supply with greater efficiency I'll like

 

[zynixoriginal]

I did that by redusing the Vc to a lower voltage say 0.1V. This will reduce the Rs power dissipation to 1/10
To do that, I configure the error amplifier by connecting a 7k5 resistor from VFB to VREF and a 2k4 resistor
from VFB to COMP. When you have done that, the output of the error amplifier (measured at COMP pin) is
1.7V thus Vc = (1.7V-1.4V)x(R/3R) = 0.1V.

For regulating the output voltage, I use photo coupler to feed back to output voltage error signal to the VFB pin.
Have a good try. Hope this will help you.

 

[FvM]

I did that by redusing the Vc to a lower voltage say 0.1V. This will reduce the Rs power dissipation to 1/10
To do that, I configure the error amplifier by connecting a 7k5 resistor from VFB to VREF and a 2k4 resistor
from VFB to COMP. When you have done that, the output of the error amplifier (measured at COMP pin) is
1.7V thus Vc = (1.7V-1.4V)x(R/3R) = 0.1V.

Does this solution guarantee a current limit for the output transistor? Closed loop operation can use a lower current sense voltage range, but out of loop control (start-up, overload) the maximum output current will be still governed by the 1V voltage clamp at the inverting comparator input. You can reduce the current range by sinking current at the output compensation pin, but source current range has too large type variation to do this precisely.