Electronic – Mirroring constant current

constant-currentcurrentlm317

I'm trying to draw a constant 60mA from a battery but I get a draw that is too high with the following setup. Does anyone know what I'm doing wrong?

Regards

EDIT: Sorry I added the wrong measures. It should be correct now

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

First impression : the base-emitter junctions inside the BCV61 cannot be independent circuits since accurate current mirroring requires the same Vbe in each transistor.

So ... you need to connect the emitters together.

If you need isolation, you'll have to find another approach.

EDIT : as Arsenal points out, the test circuit (fig 14) appears to show isolated emitters, something I don't understand. His simulation appears to confirm my prejudices. [Wild speculation deleted since schematic corrected] As - substantially - does the revised question. With isolated emitters, there is very little current transfer.

However it may be worth experimenting with identical emitter resistors in each circuit; this commonly improves matching and may help here.

Another consideration is power : at 60mA,3.6V, 0.2W is a lot to dissipate in this tiny package. In free air, at 500K/W (Table 6), that suggests a 100K temperature rise.

It'll heat the RH transistor considerably, reducing its Vbe by 2mv/K and thus increasing its Ic. To some extent this is balanced by warming the reference transistor, but still...

NB the datasheet (p.4) rates Ie2 at 5mA (not 90 or 200!) for correct operation with Vce=5v. Clearly, thermal effects beyond these ratings affect accuracy.

I'm coming back to the emitter resistor recommendation.

  1. it'll improve balance.
  2. it takes power dissipation out of the package.
  3. I'd add a collector resistor to C2 to keep Vce on that transistor within reasonable limits (say 1V max) further decreasing power.