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LM395 does not turn off

Other Parts Discussed in Thread: LM395

Hi,

I am evaluating some LM395 TO-220 samples.  I am using the device as described in the datasheet (www.ti.com/.../lm195.pdf) Figure 26 in conjunction with a CMOS logic output.


The LM395 has emitter to ground, open collector switching a LED + resistor.  I am not able to get the device to fully turn on or off and need help.


I created a simple test breadboard where I used a pull down resistor (33k) from base to ground and switched the base to V+ (5V) to turn on.  With Vbe @ 5V, the device should be in saturation.  My Vce = 1.3V, which seems high.  With the base pulled to ground, my Vce is 2.5V.  I have tried this with V+ at 24V as well with similar results.


This circuit works fine with a 2N2222 NPN transistor.  I have tried various combinations of base resistors and even pulling the base hard to ground without success. 


Please advise what I'm missing!

  • The minimum Vce of 1.3V would seem to be in line with Typical Performance Characteristics (Figure 5). If you are looking to sink 10's of mA, rather than 100's of mA, the LM395 might not be the right choice.

  • Thanks- I can see how the on Vce would be ok.  I am looking to switch 100s of mA, up to 1A, so I am ok with this.


    I still do not understand why I cannot get the device to turn off.  I have not measured the current in the off state yet, but I'm getting Vce = 2.5V when my supply voltage is at 5.0V and the base is grounded through a pull down or directly shorted, so Vbe = 0.

  • "... but I'm getting Vce = 2.5V when my supply voltage is at 5.0V and the base is grounded ..."

    For your testing in the milli-amp load range, that could still be normal operation as the LM395 has a typical Quiescent Current (Iq) of about 2 milli-amp. The LM395 is an integrated circuit and requires some nominal bias current to keep the internal circuitry operational even when Vbe= 0.0V.