battery charger

tyblu

Joined Nov 29, 2010
199
The schematic is not complete. There needs to be something connected to the gates of the transistors. D1 is a protection diode for Q1. D2 and L are part of a "Buck" circuit; it basically drops the input voltage (see http://en.wikipedia.org/wiki/Buck_converter ). I'm not sure what Q2 is doing there. Maybe provide a load for when the battery is disconnected.
 

Thread Starter

noobuser

Joined Oct 19, 2010
9
Sorry, I should have been a little more clear. The schematic is for a bidirectional dc/dc converter. The signals of Q1 and Q2 are complements of each other. When the circuit is charging the battery, Q1 and D2 are conducting, and when the battery is being discharged, Q2 and D1 are conducting.

What I don't understand is in the case of charging, why does D2 need to conduct? (instead of going from Vbus - Q1 - L - battery - Vbus) Similarly, when the circuit is discharging, why does D1 need to conduct instead of the current just starting from Vbatt go to L - Q2 and back to the battery in a loop?

Hope that clairify's my question.

Thanks.
 

#12

Joined Nov 30, 2010
18,224
When Q1 turns off, the inductor will force the current to keep flowing for just a moment as its magnetic field collapses and D2 will allow the current to flow instead of letting the inductor try to get current through the "off" transistor (Q1). Same thing the other way around. Q2 turns off, the inductor won't let the current stop instantly, the current flows through D1 instead of frying Q2.
 

Thread Starter

noobuser

Joined Oct 19, 2010
9
When Q1 turns off, the inductor will force the current to keep flowing for just a moment as its magnetic field collapses and D2 will allow the current to flow instead of letting the inductor try to get current through the "off" transistor (Q1). Same thing the other way around. Q2 turns off, the inductor won't let the current stop instantly, the current flows through D1 instead of frying Q2.
i see...so what you are saying is that the diodes are not always conducting within a cycle? and just during the time before the inductor reaches "steady state" again?

I just want to verify this point since the source i am reading says that the transistor and diode are "alternatly conducting" which i thought implied that it was not being used as a protection diode as you have mentioned.
 

tyblu

Joined Nov 29, 2010
199
Basically 4 stages:
1. Q1 on; Q2 off: current flows from "main power bus" through Q1 and L into the battery, setting up the inductor's magnetic field.
2. Q1 off; Q2 on: current continues to flow through L as the magnetic field collapses, because that is what inductors do, pulling current through D2. Vds<0, so no current flows through Q2 (negligible amount from source to drain).
3. Q1 off; Q2 on: the field in L has completely dissipated, and now current flow out of the battery, through L and Q2, setting up a new magnetic field in the inductor, but flowing the other direction.
4. Q1 on; Q2 off: current continues to flow out of the battery and through L, because that is what inductors do, pushing current through D1 back to the "main power bus". Vds<0, so no current flows through Q1 (negligible amount from source to drain).

If the timing is done correctly, the inductor will just barely saturate before its current "feeder" switches off, making the collapsing magnetic field create a voltage across L. Towards the right, this drops the voltage; towards the "main power bus", this boosts the voltage. Similar theory for bidirectional buck/boost circuits is used for electric car regenerative braking systems.
 
Hi Guys,

I am simulating a bi-directional buck-boost converter.I am able to boost voltage from 48 to 400 Volts. But i am not able to generate pulses during buck operation.Can any one help me please...

circuit1.png

circuit2.png

Thanks in advance
 
Top