LM5149-Q1: negative current problem with LM5149

Hi Manu,

I want to get clarification here on the implementation:

There are two switches SW1 and SW2. During Normal operation - SW1 is open and SW2 is closed. 

For current reversal - the customer wants to open SW2 and Close SW1. 

1. Is my understanding correct? 
2. What are the input and output voltages required? 
3. What is the load at the output? 
4. During and after switchover - what does the customer want the LM5149 to do? Do they just want the current to flow from the output to the input? 

If it is just current sinking - the LM5149 does allow that (which is required in FPWM mode) - however, I am not sure what exactly the customer wants the controller to do after switchover has happened? 

Thanks,

Best Regards,

Taru

Hi Taru,

Pls check the schematic like below. When SW1 is on and SW2 is off, the load current is from Vin through TEC to Vout, and this is sink current to the LM5149.

When SW2 is on and SW1 is off, it is like a normal load and current went from Vout to load. 

Input is 12Vin and output is 6Vout, load is about 3-8A. The load is a TEC and it need the bidirectional current to control the temperature of the system.

The concerns and questions from customers is, when switching the load current direction, there is a sink current when turning SW1 on, what would this sink current cause? Can LM5149 handle this and why? Customers observed that after switching the inductor current might hit to 0A, what could be the possible reason?

BR,

Manu

Hi Manu,

I'll get back to you on this - this use case will require some analysis and internal discussion. Appreciate your patience here. 

Thanks,

Best Regards,

Taru

Hi Taru,

Any updates regarding this question?

Thanks

BR,

Manu

Hi Manu,

Apologies for the delay - I'll get back to you by tomorrow positively on this, 

Thanks,

Best Regards,

Taru

Hi Manu,

We had a discussion on this topic. Please see my comments:

1. If the requirement is a simple reversal of the potential difference across the terminals of the peltier - then a simpler implementation looks like the following:

SW1 and SW2 are SPDT switches. When SW1 and SW2 are in position 1 - current flows in the positive direction. 

During switchover - move SW1 to position 2 then SW2 to position 2 - the polarity across the PEC will reverse.

2. With the current implementation - there is no steady state. When you reverse polarity across the PEC - Current 'tries' to flow from the Vin to Vout. Let's say Vout is > 6V - the device will turn on Q2 and the voltage at output will fall (since the device wants that voltage to be regulated at 6V) - however, the moment it falls down below 6V - it will turn on Q1 to pull up the output voltage to 6V - trying to force current to the output. There are two opposing effects at play here - the peltier trying to force current into the buck - but the buck will try to force current in the peltier when output voltage falls below the set point. 

3. Can you also share the peltier part number? - the implementation that you have shared is using three terminals - however, I have typically seen peltiers with two terminals only. 

Hope this helps. 

Let me know your thoughts. 

Thanks,

Best Regards,

Taru