Can you tell me the purpose of two inductors in these circuits??

Hi ,

i was goin through Power Supply circuit.I think that the power supply circuit is based on buck boost inverting topology..But there exists two inductors in the circuit that confuses me..
Can somebody tell me what the additional one prior to the diode is being used for?

Regards

what you see as 2 inductors are in fact a single tapped inductor - done this way to help get the high o/p volts , and allow a sort of flyback action as the high current built up in one winding by the FET is transformed (at turn off) into a higher voltage (lower current) in the other part of the winding. This is known as a tapped inductor buck-boost.
Hope this helps, Regards, Orson Cart.

So is the secondary isolated from the primary in this case??
May be this configuration helps to avoid selecting MOSFET with large VDS ? is it so?

there is no isolation, there is a part winding, followed by another part winding - with 3 connections, Yes a lower voltage fet can be used.
Regards, Orson Cart.

Thank you Orson, for the replies..
One more question.. i want to tweak this design..
i was wondering ,how to calculate the required turns ratio for primary & secondary turns..So i think of a method as below please correct me if im wrong..

i feel like we know the output voltage, reduce it by suitable low voltage in terms of turns ratio..
then design the first stage for that much low voltage..using normal buck boost inverting configuration..Now add the secondary inductor..into the calculation and design suitable filter at the output..But in the one of the above design they use the secondary L & an output C as filter but it second case they are using C L C filter configuration..
Thanks & Regards,

It doesn't really work with 2 seperate inductors, the total design has a bit more to it than perhaps you envisage from the below comment, firstly what are the min/max i/p and o/p voltages required and max o/p current? Desired freq of operation?

In one of the examples they have more filtering on the o/p than the other - to reduce ripple volts at nominal full load.
Regards, Orson Cart.

4.5 to 5.5V input, current can be 500mA.
-100V ,2.0mA at output
Frequency of operation desired to be fixed i.e not being varied during operation.But no strict specific frequency(say 100KHz) requirement is there.
So can you tell me how could i proceed to the inductor design now?
Thanks & Regards

the design here goes like flyback.
Your primary winding is vertical one
and secondry winding is sum of those two

OK, to arrive at the coupled inductor needed, firstly lets assume the P-fet can allow its drain to go to -20V (plus spike), and the min input is 4.5VDC, thus the duty cycle will be 85% or so worst case. If the max o/p power required is 0.2watts (100V x 2mA) sufficient energy has to be stored in the choke core (or gap) every switching cycle. At 100kHz we need to store about 2uJ per cycle plus a bit to allow for losses (say 50% extra). If we let the current ramp to 200mA max, the inductance is then 150uH (100kHz, 4.5Vin, low on drop fet) and this gives us 0.3watt capability. So now we have 20 volt on the left hand winding and need 80V on the right hand part of the winding (a turns ratio of 1:4). The total inductance in circuit during the inductor discharge phase is now 1350uH and will take 2.7uS for the current to ramp to zero, which means the converter may be just operating in continuous inductor current at or near full power.
So a core and winding combination is needed that can handle 0.2amps peak at 150uH, with an overwind of 4 x the turns used to get the 150uH, close coupling of the windings is beneficial. A small gapped ferrite EE would do, or a small Kool-mu type toroid. [if we let the current go to 400mA peak, then the LHS L can be 36uH, still with the 4:1 overwind for the RHS part of the choke). Make sure you get the phasing correct (dot notation) when you connect the windings to each other and to the circuit. You will need an 150V diode for a 4:1 turns ratio and a max input volts of 5.5.
Regards, Orson Cart.