# got a too low VCE on a Infrared Phototransistor

I used an Infrared Light Emitting Diode QED123(VF=1.7V) and an Infrared Phototransistor QSD123 in pair. I learnt from QSD123's datasheet that its VCE(SAT) under IC=0.5mA is 0.4V max. But the VCE I measured with an multimeter is only approximately 0.05V... Here is my schematic:

I put the led and transistor head to head like this: So VCE is just voltage on OUT1... whether 0.05V is VCE(SAT) or just VCE, is it too low for a transistor? The IC I measured is approximately 1.01mA.

I also found an article about VCE(SAT):

Photo-transistor: Collector Saturation Voltage:VCE(sat)(V)

Collector-to-emitter voltage in relation to a prescribed collector current(IC) of the photo-transistor on the light-receiving side, when a prescribed forward current(IF) flows through the LED on the light-emitting side. Since this voltage changes greatly depending on the forward current(IF), collector current(IC), and also individual variations in the current transfer ratio (CTR) and its degradation, the forward current and collector current values must be designed taking these factors into careful consideration.

The article says that VCE(SAT) is greatly depending on IF... But I still got no sense on the 0.05V VCE. Is my measuring wrong or I have a bad understanding about VCE(SAT)...

Reply is desired, thanks:)

• So what exactly is the problem? The 0.4V is a maximum. Assuming you're actually measuring what you think you are, the part is well within spec. – Matt Young Dec 30 '13 at 6:59
• Is Vce rising to about 5V when D1 is off? If not you will have to shield T1 from the room light as this will turn T1 on fully or partially. – Russell McMahon Dec 30 '13 at 9:09
• .05 is what we call less than .4. -1 for stupid question. – Olin Lathrop Dec 30 '13 at 13:13

Uh, when they say $$\{V}_{CE}\$$ max they mean the worst-case (e.g. largest) collector-emitter voltage drop, not the smallest.