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enter image description here

Let's assume that Vin is a DC battery and Vout is a spacecraft main power bus.

It is given that this circuit is DC-DC converter with bidirectional power flow.

But to me this circuit exactly looks like boost converter except unidirectional switches are replaced by bidirectional switch. Even $$V_{\text{out}} = \frac {V_{\text{in}}} D$$

And hence similar to boost converter, power should always delivered by the battery (discharging) to bus.

But after read I found that in this circuit battery can deliver as well as absorb (charging) the power from bus.

Is this only because we replace the type of switch or due to some other reason?

Which factors determine whether battery is charging or discharging?

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  • \$\begingroup\$ Usually bidirectional current flow DC DC converters have 4 switches \$\endgroup\$
    – Voltage Spike
    Jul 27 at 18:49
  • \$\begingroup\$ @Voltage Spike♦ Isn't this Circuit also act as bidirectional power flow? \$\endgroup\$
    – user215805
    Jul 27 at 19:21
  • \$\begingroup\$ I think you need to show a more complete circuit. Specify values for the voltages at Vin, Vout, S1, and S2 for the two cases where "power" is flowing in two possible directions. \$\endgroup\$ Jul 27 at 19:24
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schematic

$$V_{\text{out}}=V_{\text{in}}\cdot D$$

Where \$D\$ is the duty cycle of the upper transistor (M1), the lower transistor switches synchronously - duty cycle is \$(1-D)\$.

schematic

simulate this circuit – Schematic created using CircuitLab

$$V_{\text{out}}=\dfrac{V_{\text{in}}}{1-D_{M_2}}=\dfrac{V_{\text{in}}}{D_{M_1}}$$ Where \$(1-D)\$ is the duty cycle of the lower transistor (M2), the upper transistor switches synchronously - duty cycle is \$D\$.

It can work as bidirectional: boost battery to bus /and/ buck from bus to battery, without changing any hardware components, just adapting the duty cycle time.

The current flows from higher potential to lower potential, so if you put two batteries on both ends, the current shall flow from one to another. It would make sense to add a dead band - if both voltages are more or less equal (with regard to the boost/buck equation) then you do noting. In the other way, the converter would just fill and empty the inductor with all corresponding losses.

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  • \$\begingroup\$ Hi ! Thanks for answer , I'll simulate but can you please tell intuitvily what condition determine whether battery is charging or discharging? \$\endgroup\$
    – user215805
    Jul 27 at 19:50
  • \$\begingroup\$ The timing of the FET drive signals. \$\endgroup\$ Jul 27 at 20:15
  • \$\begingroup\$ +1, adapting duty cycle , but how circuit automatically decide that now i have to adjust my duty cycle and start charging the battery? \$\endgroup\$
    – user215805
    Jul 27 at 20:31
  • \$\begingroup\$ @user_1818839 how circuit automatically decide that now i have to adjust my duty cycle and start charging the battery? \$\endgroup\$
    – user215805
    Jul 27 at 20:32
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    \$\begingroup\$ @user215805 By measuring the input/output voltage, there is no other option. \$\endgroup\$ Jul 27 at 20:36

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