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I have a doubt about using opto-couplers to isolate the high voltage side from low voltage side.

This might be a dumb question :D

enter image description here I'll ask based on above picture. SO when 5V is applied to the input of optocoupler , it will turn on and pass the 5V to the base of darlington transistor.

But inside the opto-coupler is a transistor , right? so in order to turn the transistor on , B-E should be forward biased, i.e., Base +ve and Emitter- GND. But in above circuit Emitter of Opto coupler is not connected to GND, so how applying 5V will turn it On and how 5V will reach base of darlington transistor?

and also is it okay to say "5V will reach base" or current flows from C to E of optocoupler to base of second transistor?

thanks

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First, it's not ok to say "5V will reach base"?
If you means the output-side "+5V" reach the darlington's base, you're right.
If you means the left "5V" reach the optocoupler's transistor base the answer is, obviously, not, because the optocoupler is used exactly to maintain disconnected input and output.

Second, Phototransistor in optocoupler is used in switch mode. When the light is "enough" the phototransistor saturate. Then the 5V in the output-side pass through the phototransistor, that act like switch, and reach the darlington Q1-Q4.

You not need to work with \${V_{BE}}\$ because in this case we use the photovoltaic effect (there's a little bit of phisics we need to know).
Also, you don't need top put emitter to the ground, like said in comments, because without current the Darlington pull down the phototransistor's emitter.

If you put a LED, as asked in comments, it work like normal BJT switch, like in this schematic (LTSpice file at bottom); you can see that the \$V_E\$ is zero while microcontroller (V1 in schematic) don't assert output. enter image description here

You can read more about phototransistor in the first page of this Fairchild Application Note, in some optocopuler's datasheet and in the productor's knowledge base, like:

  1. Toshiba
  2. Vishay

LTSpice file:

Version 4
SHEET 1 880 680
WIRE 128 -240 112 -240
WIRE 144 -240 128 -240
WIRE 128 -80 128 -240
WIRE -96 -32 -112 -32
WIRE 64 -32 -96 -32
WIRE 240 16 240 -16
WIRE 240 16 128 16
WIRE -112 64 -112 -32
WIRE 240 64 240 16
WIRE 240 176 240 144
WIRE -352 256 -352 160
WIRE -112 256 -112 144
WIRE -112 256 -352 256
WIRE 240 256 240 240
WIRE 240 256 -112 256
WIRE 240 272 240 256
FLAG 240 272 0
FLAG 128 -240 +5V
FLAG -352 80 +5V
FLAG 240 16 Ve
FLAG -96 -32 Vin
SYMBOL diode 224 176 R0
SYMATTR InstName D1
SYMBOL res 224 48 R0
SYMATTR InstName R1
SYMATTR Value 1k
SYMBOL npn 64 -80 R0
SYMATTR InstName Q1
SYMBOL voltage -112 48 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(0 5 0.5)
SYMBOL voltage -352 64 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value 5
TEXT -386 296 Left 2 !.tran 1
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  • \$\begingroup\$ In the Fairchild application note-common collector configuration, there is a resistor connected in collector to GND(I guess the purpose is pull down). so when transistor gets base current(when led turns ON), base is +ve and Emitter is -ve so FB and it turns the transistor ON. But here Emitter of opto-coupler not connected to GND. Shouldn't that be a problem,? \$\endgroup\$ – Athul Aug 8 '16 at 9:42
  • \$\begingroup\$ @Athul, no, (you means \${R_E}\$?) because of the principle behind the operation (photovoltaic). Se also the Vishay's Application Note 02 for examples. \$\endgroup\$ – Antonio Aug 8 '16 at 9:50
  • \$\begingroup\$ so, if i connect a LED at the emitter (replacing darlington transistor), it will turn on when opto-isolator turns on. and it doesn't require emiitter connected to gnd.? Also forget about opto-copuler. If it's a normal transistor , configured same as above, base connected to micro controller, collector to VCC and load to Emitter say and led. Will it turn on when apply a base voltage? If yes, why it doesn't require Emitter connected to GND to FB base emitter(I my knowledge transistor only turns on when EB is FB) \$\endgroup\$ – Athul Aug 8 '16 at 10:05
  • \$\begingroup\$ @Athul, You don't need to put emitter to GND, the led+resistor put the emitter to ground for you (while current is zero the voltage drop is zero). If you could simulate with LTSpice you could see that the \$V_{E}\$ is zero while microcontroller don't assert output. If you want I can add LTSpice picture at my reply to explain this. \$\endgroup\$ – Antonio Aug 8 '16 at 13:12
  • \$\begingroup\$ that would be helpful. Also which one is correct drive.google.com/open?id=0B5_IWI68ZqZTRXF0NHNMaTA5bVE \$\endgroup\$ – Athul Aug 8 '16 at 13:17
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When 5V is applied to the input of optocoupler , it will turn on and pass the 5V to the base of darlington transistor.

Correct. The transistor base current is generated by light from the LED. This injects a small base current which is amplified by the transistor. The effect is to reduce the collector-emitter resistance. Current flows in the direction of the arrow.

But inside the opto-coupler is a transistor, right? so in order to turn the transistor on , B-E should be forward biased, i.e., Base +ve and Emitter- GND. But in above circuit emitter of opto coupler is not connected to GND, so how applying 5V will turn it On and how 5V will reach base of darlington transistor?

There are other transistor configurations besides emitter to ground. In this case the opto-isolator is in common collector configuration.

... and also is it okay to say "5V will reach base" or current flows from C to E of optocoupler to base of second transistor?

Nearly, but you need to think about currents when dealing with transistors. A current will flow through the circuit based on (1) the on resistance of the opto-transistor, (2) R1 and (3) the forward voltage drop of the Darlington transistor which will be two diode drops - about 1.4 V.

If the component values are chosen correctly the Q4 base current will be enough to drive Q4 into saturation so that there is minimum voltage drop across it and maximum voltage across the load.


[OP's comment:] I think i'm having trouble understanding common collector configuration.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Common collector mode and (b) Common emitter mode.

  • In common collector mode we lose 0.7 V across Q1's base emitter junction. The emitter tries to "follow" the base voltage but with a "diode drop" of about 0.7 V. We are unable to get the base voltage high enough to drive the transistor into saturation.
  • In common emitter mode we drive the transistor into saturation and the collector-emitter voltage can drop to as little as 0.3 V. This makes it a more efficient switch with more of the supply voltage being dropped across the load (which is where we want it).

Note that (a) does not invert the signal whereas (b) does.

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  • \$\begingroup\$ But in commaon collector configuration Collector is connected to ground, right? But here it's connected to 5V \$\endgroup\$ – Athul Aug 8 '16 at 9:34
  • \$\begingroup\$ okay, so in common collector configuration collector can be tied to Ground or power level, en.wikipedia.org/wiki/Common_collector But there shouldn't be a resistor in emitter connecting it to GND \$\endgroup\$ – Athul Aug 8 '16 at 9:44
  • \$\begingroup\$ In this case the NPN collectors are tied to +5V. There is a path to ground through R1 and Q4 base-emitter. \$\endgroup\$ – Transistor Aug 8 '16 at 10:01
  • \$\begingroup\$ I think i'm having trouble understanding common collector configuration. Forget about opto-copuler. If it's a normal transistor , configured same as above, base connected to micro controller, collector to VCC and load to Emitter say and led. Will it turn on when apply a base voltage? If yes, why it doesn't require Emitter connected to GND to FB base emitter(I my knowledge transistor only turns on when EB is FB) \$\endgroup\$ – Athul Aug 8 '16 at 10:33
  • \$\begingroup\$ See if the update helps. \$\endgroup\$ – Transistor Aug 8 '16 at 11:00

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