Turning LM25085 into an electronic load

[king.oslo]

Hello there,

I am still working on my first power supply. I decided to make a switcher with the capability to source and sink current.

Please see the attached schematic. Current will be sunk and sourced through the header called "Binding post". The relays will direct the current the way I want it.

Whilst sinking current I reckon I have to input an alternative signal for FB-pin and the differential amplifier: V_FB needs to fall when input voltage rise. Second, the output of the differential amplifier will become negative whilst sinking current, so I need to rectify this as well.

So my questions:

First, what about the RT and VIN pin? I reckon V_RT needs to fall when V_IN rises? Otherwise the One-shot-module will not function properly? But what about the VINpin?

Second, any additional constructive criticism is welcome.

Thank you for your time.

Kind regards,
Marius

[tom66]

You cannot readily sink current with a switching converter; unless you have somewhere to store the energy in.

[king.oslo]

Current is stored in the coil before it is passed to GND. Did I miss something?

Thanks.M

[tom66]

Current is stored in the coil before it is passed to GND. Did I miss something?

Thanks.M

How are you dissipating the load power? It has to go somewhere, as heat usually.

[king.oslo]

It is sunk:

Binding post -> Rsense -> caps -> inductor -> PMOS -> GND.

How does that sound?M

[tom66]

It is sunk:

Binding post -> Rsense -> caps -> inductor -> PMOS -> GND.

How does that sound?M

So, where does all the heat go? Into PMOS? Otherwise, it is just a dead short.

[king.oslo]

The pmos will switch at 300khz. When it is on, current stored in the inductor will go to gnd. When it is off, the inductor will charge up.M

[tom66]

The pmos will switch at 300khz. When it is on, current stored in the inductor will go to gnd. When it is off, the inductor will charge up.M

You aren't dumping the heat anywhere. Current will flow through the inductor, but then it turns off. Inductors don't like sudden current interruptions (=HV/breakdown), so you need a return path. This return loop must be where heat is dissipated; either that, or on the input through some resistors or similar.

My suggestion: Forget about using the switching regulator to make a current sink. It won't work reliably as the regulator is compensated and designed for a specific feedback configuration. Once you start overdriving the feedback with an op-amp you introduce additional phase shifts which shouldn't be there and they'll cause a whole lot of trouble for you.

I would just add a powerful MOSFET array, and a typical current sink control (you could share the high-side measurement.)

[king.oslo]

Thanks for your suggestion I will keep it in mind. So for the sake of learning, when the MOSFET switches off, the magnetic field of the coil collapses, and current flows through the diode, and all is sweet? Or not?

What exactly is the problem with this? Are you saying high voltage will appear at the inductor/PMOS node which will destroy the MOSFET?

Could high speed opamps cure problem with phase shifts?

Thanks.M

[tom66]

Thanks for your suggestion I will keep it in mind. So for the sake of learning, when the MOSFET switches off, the magnetic field of the coil collapses, and current flows through the diode, and all is sweet? Or not?

In a typical (non-synchronous buck converter):
 - high side switch turned on
 - current flows through high side switch (MOSFET)
 - current flows through inductor
 - current flows through output capacitor (charging it)
 - high side switch turned off
 - current flows through inductor
 - current flows through output capacitor (discharging it)
 - current flows through buck diode

What exactly is the problem with this? Are you saying high voltage will appear at the inductor/PMOS node which will destroy the MOSFET?

No, it probably won't destroy the MOSFET, but it won't work as an electronic load. It will be a dead short to your device under test.

Could high speed opamps cure problem with phase shifts?

Possibly, but I would think high speed op amps could introduce even more problems (noise sensitivity and higher frequency oscillations being some.) The chip is compensated (stabilised) for a few particular configurations - operating it outside of this range can lead to anything happening.

[jahonen]

If you push current into the output of a synchronous step-down regulator, it causes regulator input voltage to rise. It turns into a step-up regulator, working from output to input. Unless there is an overvoltage lockout, you'll just destroy the regulator due to overvoltage.

Regards,
Janne