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i have a circuit for which i have simulated its output. But i can't seem to understand how it works. Here is the circuit..

schematic

simulate this circuit – Schematic created using CircuitLab

For analysis purpose, PWM source = 50% duty cycle with period of 20ms, and pulsed value of 2V. Vin = 10 Vdc Load = 1k L1 = 1H C1 = 1uF

Here is a snippet of waveform i got through simulation.

enter image description here

In above circuit,. I need to know the behavior of circuit. Like When Q1 is on what voltage appears at load and when switch is off then what happens??

I am stuck at;

1 - When Q1 is on, Vin appears at L1, and D1 will be open (reverse biased), so only path of current would be to C1 and Load through L1. But why doesn't any signal appear at load in this time?? why doesn't current flow from Vin to C1 or Load through L1??

2 - When Q1 is off, why does some voltage appear across Load, when there is no Vin because it is cutoff by Q1 switch?? and how does the current flow in C1+Load part, because there is only 1 wire connecting this part to L1 since D1 is open??

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    \$\begingroup\$ Due to lack of time for complete answer: You should look up the principle of operation of a dc dc buck - boost converter. Wikipedia has an explaination describing the multiple states (on - off) of the switch and the other fundamentals like the inductor storing energy in a magnetic field to be used when the switch is "off" (this is why there is a voltage when there us no Vin). But try searching for "buck boost converter" basics and fundamentals. \$\endgroup\$ – Remco Vink Oct 1 '18 at 8:43
  • \$\begingroup\$ @RemcoVink Thank you, now that i know what this is, i ca study it more.. thank you for the hint.. \$\endgroup\$ – BetaEngineer Oct 1 '18 at 12:26
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1) When Q1 conducts, Vin is across L1 -> a constant voltage across an inductor leads to a linear current rise in that inductor. Vin is in parallel to L1 (neglecting the Transistor). The Circuit then looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

2) When Q1 turns off, the current from Vin to L1 abruptly cuts off but since the current in an Inductor is always steady (thats a fundamental law), it starts decresing but still flows through the inductor in the same direction. Since Q1 is off, the current flows via the C1 and D1 which now is vorward biased (C1 is charged by this current). Circuit looks like this:

schematic

simulate this circuit

3) Now if the current through L1 is close to 0 Q1 should conduct again. This the principle how a DC/DC Inverter or Buck/Boost converter would work. Remko Vink already gave you the right hint.

Concerning your question about singnals seen at the Load: I do not understand what you mean, maybe you should put it more precisely. But imagine C1 and Load are in parallel all the time. So the current through the Load (R) will always be: Voltage across C1 divided by R. In other words the current through the load is proportional to the Voltage C1 is charged up to.

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  • \$\begingroup\$ Yes, i understand..thank you very much @stowoda, i didn't even know it was something like boost converter so i had a hard time searching it on internet.. and yes by load i means the same as volts across C1, so thank you again..now i can study it in more detail.. \$\endgroup\$ – BetaEngineer Oct 1 '18 at 12:25

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