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?


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


simulate this circuit – Schematic created using CircuitLab


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.
  • \$\begingroup\$ So the test circuit in the datasheet is wrong? \$\endgroup\$
    – Arsenal
    Nov 12 '15 at 13:25
  • \$\begingroup\$ OK, let's just say I can't see how the test circuit can possibly work then! \$\endgroup\$ Nov 12 '15 at 13:28
  • \$\begingroup\$ I also couldn't but that figure threw me off. I tried simulating this just now, without connected emitters it just didn't work at all. Maybe they got a ground loop in their testsetup (current source and voltage source connected somewhere)... \$\endgroup\$
    – Arsenal
    Nov 12 '15 at 13:38
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    \$\begingroup\$ Also the 3.6V at 60mA already equals to 216mW power dissipation. The other side is 0.7V at 60mA which adds another 40mW. The BCV61 is rated at 250mW, so the circuit - as is - is already slightly above the total power dissipation. I'd expect the transistor to die pretty soon in this configuration. \$\endgroup\$ Nov 12 '15 at 13:39
  • \$\begingroup\$ @Nils : overlapping edits! \$\endgroup\$ Nov 12 '15 at 13:59

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