2
\$\begingroup\$

I am analyzing following circuit, but I am not getting its flow.

In the following schematic SW1 is coming from controller and connected to opto coupler ( cathode of LED ) and J4B is connected to heater.

Please explain me its operation and when Mosfet will conduct?enter image description here

\$\endgroup\$
3
\$\begingroup\$

its simple, the controller generate a square wave in S1, (probably its at VCC level at the beginning) your circuit U9 (Optocoupler) will work when the diode inside it is conductive at low level, so the transistor Q1 is operational (ON) and you can enable the heater.
The LED D11 is conductive when the transistor Q1 is ON.

\$\endgroup\$
2
\$\begingroup\$

When SW1 connects the LED cathode to ground, the LED lights up (inside the optocoupler, in infra-red). The light causes the phototransistor to conduct. This raises the gate voltage of the MOSFET, and the MOSFET conducts. Current will flow through the MOSFET from the 24V supply, via the heater.

So, to turn the heater ON, connect SW1 to GROUND. Note that this must be the ground of the Vcc power supply, which may be separate from the ground of the 24V supply.

\$\endgroup\$
  • 1
    \$\begingroup\$ The ground of Vcc must be separated from the ground of +24V to avoid U9 short circuit. \$\endgroup\$ – R Djorane Nov 28 '14 at 15:23
  • 1
    \$\begingroup\$ @codo No, the grounds can be connected. U9 will not be short-circuited. \$\endgroup\$ – John Honniball Nov 28 '14 at 15:24
  • 1
    \$\begingroup\$ @Techknowlogic R34 is used to switch the MOSFET off by ensuring that the gate voltage goes to zero when the phototransistor is off. You can work out the gate voltage by considering R33 and R34 as a voltage divider. \$\endgroup\$ – John Honniball Nov 28 '14 at 15:26
  • 1
    \$\begingroup\$ @Techknowlogic its a pulldown MOSFET resistor. \$\endgroup\$ – R Djorane Nov 28 '14 at 15:27
  • 1
    \$\begingroup\$ @JohnHonniball At my knowledge, the opotcoupler ground input must be separated from the Optocoupler ground at the output! (galvanic isolation) \$\endgroup\$ – R Djorane Nov 28 '14 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.