SHUNT resistances value

[tnnelectro]

Hi,

I'm trying to create a Driver circuit for a laser diode stack (see attached diagram)

To adjust the power of the laser stack, on pin 3 of the operational IC3, using a voltage ranging from 0 to 4V, I adjust the maximum current that the driver can supply.

With the R1 trimmer I carry out the calibration.

+ Vo1 and + Vo2 = + 12Vdc
-Vo1 and -Vo2 = -12Vdc
On V + and V- there are 24Vdc
On LD + and LD- the laser diode stack is connected, it can have a maximum absorption of 60 amps for a few milliseconds e.g. 30ms.

If you kindly give me some advice on the value of the SHUNT resistances (R25, R24, R23, R22) to use ...

is 0.1 ohm 10Watt ok?

 

[Harald Kapp]

is 0.1 ohm 10Watt ok?

You have 4 resistors in parallel, so the effective resistance is 1/4 of a single resistor.
60 A × 0.025 Ω = 1.5 V. This is well within the adjustment range of 0 V to 4 V and therefore should be good.
Power dissipation is total: P = 1.5 V × 60 A = 90 W.
Each resistor has to dissipate 90 W / 4 = 22.5 W. That is peak power.

Whether 10 W resistors are suitable depends on the pulse rating of the specific resistor in question. You'll have to check the datasheet and find the pulse rating for 30 ms and the respective duty cycle allowed. Pulse power rating is usually quite higher than the permanent power rating.

For normal operation you do the same math but use nominal current instead of peak current.

 

[tnnelectro]

Thanks for your answer...
then what do you think of the following reasoning:
1. The voltage drop across the shunt resistor must be approximately 1% of the supply voltage, therefore:
Vs = 1% of V + = 1% of 24 V = 0.24 V (volts eliminated through the shunt resistor at maximum current)
2. shunt resistor value Rs = Vs / I = 0.24 V / 15A (60A max current divided by 4 because it is distributed over 4 MOSFETs) = 0.016 Ohm
3. Power dissipation required for Rs
P = I x I x Rs = 15Ax15Ax0.016Ohm = 3.6Watt
What do you think about it?

 

[Harald Kapp]

Reasonable.

 

[AnalogKid]

Looks good so far.

But ...

What if the current sharing among the four FETs is unequal? Worst case would be all 60 A through one FET. As I read the schematic, the circuit would not see this condition as an error, and the lone shunt resistor would be hit with 58 W - ish.

ak

 

[tnnelectro]

Looks good so far.

But ...

What if the current sharing among the four FETs is unequal? Worst case would be all 60 A through one FET. As I read the schematic, the circuit would not see this condition as an error, and the lone shunt resistor would be hit with 58 W - ish.

ak

right observation ...
how can you avoid this?

 

[AnalogKid]

First thought -

Add four Shottkey signal diodes, one each anode to the tops of R22-25. The junction of the four cathodes represents the peak value of the four currents minus one diode drop. This plus a comparator gives you a first-order error signal. For example:

If the diodes come on at 0.3 V, then the cathode node pulls up from GND when the voltage across *any* of the four shunt resistors is above 0.3 V. This happens at 18.75 A. If the comparator trip point is set to 0.1 V, then the trip voltage is 0.4 V, and the error signal goes true at 25 A. Note that all four FETs might still be conducting, but one way more than the others.

ak

 

[Harald Kapp]

Add a fuse in the source leg of each transistor.

 

[tnnelectro]

Add a fuse in the source leg of each transistor.

it could be a good idea...
Could delayed 16 Amp fuses work well?

like this:
https://www.tme.eu/it/details/0001.2516/fusibili-5x20mm-ritardanti/schurter/