@hobby16

I need to implement a low side switching buck converter for one of my projects. I tried implementing the circuit shown by you. However I am getting an output = Vin for all duty cycles. Details are as under
Vin = 12V
L = 100uH
C = 10uF
Freq = 20kHz
Diode -> IN5819
Mosfet - IRFZ44 driven by BC847 from input rail, whose base is triggered by Arduino.

Please help for debugging.

Thanks


Mods Note:
Please don't hijack other member's thread, now you have your own.
This thread was split from -- Doable buck converter design ?

 

I am not sure what you mean by "low side switching buck converter". Can you give us a schematic showing what you have in mind?

 

I am not sure what you mean by "low side switching buck converter". Can you give us a schematic showing what you have in mind?

Probably he was talking about the P MOSFET is the high side and the N MOSFT is the low side, but sometimes the high side still using N MOSFET, so the input should be using an ic to drive two N MOSFETs.

 

For reference, the schematic mentioned in post #1 is below.

 

If you keep the mosfet off, then the output voltage across C1 surely must be zero, otherwise you don´t have it connected correctly.

 

By low side switching I mean that the FET is connected on the ground side of the circuit as opposed to the normal Vcc side.
Please find the full schematic that I am using.

 

Apart from the unconnected ground, I think the inductor should be connected at the junction of the fet and diode. The way you have it now makes it very hard to measure output voltage across the load.

 

Please check the blue frame.
The values of R2 increase to 8.2K, another pin of R3 move to Gnd and increase to 12K, the values of R8 increase to 4.3K or 4.7K.

 

kubeek had the problem diagnosed correctly.

I simulated the circuit with the inductor where you have it, and moving it down to the more conventional spot between the bottom of D1 and C1.

Both worked. But in your circuit, neither side of the output voltage is at ground, so you cannot measure it they way you are trying to. You need to measure both sides of the load resistor to get the voltage.

If you move the inductor, you are already measuring the voltage input, you the rest of you circuit works, you just have to subtract your two voltage measurements to get the output voltage.

\Bob