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PR1 is floating between 10V and 81.3mV. I am finding it difficult to explain why PR1 is not floating between 10V and 0V...

I tried to read about it, but it feels as if I'm getting voltage out of no where. Is there always tiny voltage associated with schmitt trigger..?

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    \$\begingroup\$ Your schematic doesn't show a Schmitt trigger circuit. \$\endgroup\$ – The Photon Nov 14 '16 at 16:52
  • \$\begingroup\$ Take the pull up off and see what happens. \$\endgroup\$ – Voltage Spike Nov 14 '16 at 17:59
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You don't mean 'floating', you mean switching. And it's just a comparator, not a Schmitt trigger, as @ThePhoton notes.

PR1 is the collector of a bipolar transistor with the emitter grounded (through a very low value - 4\$\Omega\$ - resistor. It has 4.7K to 10V, so about 2.1mA is flowing into the collector. Vce of a saturated transistor would typically be in that range (< 100mV) but it will not be exactly zero. From the datasheet:

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For example, take a 2N3904 under similar conditions with 100uA base current:

schematic

simulate this circuit – Schematic created using CircuitLab

If you run the above simulation you'll find the collector voltage is around 63mV.

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  • \$\begingroup\$ Could you explain without use of such terms as 'emitter', 'bipolar', 'collector', 'saturated'... I really don't know what you referring to when you speak about an emitter or bipolar or collector... Also, I tried to work out how u got 4.7mA from 4.7k and 10V by Ohm's law but it's not working out, how did you reach that conclusion of 4.7mA? I cant seem to get the second circle either... \$\endgroup\$ – Lukali Nov 14 '16 at 22:04
  • \$\begingroup\$ It's 2.1 mA actually, 5 was an error. \$\endgroup\$ – Spehro Pefhany Nov 14 '16 at 23:03
  • \$\begingroup\$ When the transistor is 'on' it and conducting it will have a bit of voltage drop. Anything less than about 0.6V means it's turned on pretty good. 0.083 is much less than 0.60. \$\endgroup\$ – Spehro Pefhany Nov 15 '16 at 15:54

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