Regulator High current draw and in rush current problem

[danny davis]

I talked with my manager today about this regulator issue

He said that the filter capacitors on the input to the regulator gets discharged and lowers the voltage down when there is a in rush current or high current draw on the output of the regulator.

Also there is a voltage drop across the regulator it self which has a resistance from input to output of the regulator

So it's like a voltage divided with the LOAD

They have been trying different regulators that are 12 volts at 1 amp , but each one behaves differently when there is a HIGH current draw on the output

He was saying that when there is a HIGH current draw on the output that the LOAD is not constant, which i don't understand , I would think the load would be constant , he is saying the load is not constant

I'm also not sure why a linear regulator discharges the filter capacitor

When using a Switching Regulator , it doesn't discharge the filter capacitor on the input of the regulator, why is that? is it because a switching regulator is isolated?

He was saying that a switching regulator using a PWM oscillator which has a "switching Mode LOSS" only in nanoseconds, that the leading and falling edges of the PWM signal will create a voltage drop on the output of the regulator

A Switching Regulator doesn't have a voltage drop across it or a resistance from input and output , so there is no voltage loss , is this true?

 

[Darktrax]

the filter capacitors on the input to the regulator gets discharged and lowers the voltage down when there is a in rush current or high current draw on the output of the regulator.

There may well be filter components at the input, but the primary job of the capacitor at the regulator input is energy storage. It only gets refreshed from rectified peaks at low frequency. The voltage ripple will be more as the load is increased. Using a larger capacitor here rated to take the ripple current would help.

Having the unregulated voltage high enough that the ripple waveform is always higher than the output voltage plus the regulator drop ensures smooth output, but can mean the power dissipated in the regulator gets high. It is always better to try and have the input voltage high enough, but no higher than necessary. If the ripple frequency is low, as in rectified AC supply via transformer, most linear regulators can work fast enough to regulate out the ripple.

The voltage drop across a linear regulator is how they work. The difference between the capacitor voltage and the output voltage is dropped across the regulator, and it self-adjusts to keep it so. LDO(low drop-out) linear regulators are able to remain operating with lower voltage across when the input gets low, but there has to be some difference to operate the regulator's own internals.

Be aware that some LDO linear regulators can "latch up" in a high current shorted state if started up with the load connected, or if starved into drop-out.

I would think the load would be constant , he is saying the load is not constant

This is so. Regulators can be used where the loads are constant, but the input supply can vary greatly. They can also be used where the input supply is OK, but the load varies. If the need is to maintain the voltage, in spite of load changes, then the regulator self-sdjusts to deliver the constant voltage. Of course, regulators are expected to do both, simultaneously.

I'm also not sure why a linear regulator discharges the filter capacitor

Any current taken at all, as it goes through the regulator to the load, will discharge the capacitor. If the capacitor is being charged from rectified AC, then between the charge pulses, the capacitor voltage drops. This is what makes the ripple waveform. If the load applied is so high that even the main supply to the capacitor collapses, then the supply is not adequate to to the job. If it was a transformer, then it may be too small, and is becoming overloaded.

When using a Switching Regulator , it doesn't discharge the filter capacitor on the input of the regulator, why is that? is it because a switching regulator is isolated?

A switch-mode regulator, depending on its type, can boost a voltage as well as reduce it. Switching regulators work at high frequency, and store energy in inductance as well as capacitance, and by their principle, do not have to dissipate heat in linear mode. Switching regulators can work even if the input voltage is lower than the output, BUT, be aware that in this state, they are taking higher current from the input.

This is OK if the input is capable of supplying it, but if the whole reason the input voltage is getting low is it is collapsing under load, then a switching regulator will make the situation quickly worse! It demands ever more current, and the voltage keeps dropping, until the input is totally overloaded. It cannot give you out more energy than is put in.

He was saying that a switching regulator using a PWM oscillator which has a "switching Mode LOSS" only in nanoseconds, that the leading and falling edges of the PWM signal will create a voltage drop on the output of the regulator

PWM (pulse width modulator) works using a very high frequency, and a duty cycle. The energy you get out is decided by the ON time compared to the OFF time. Unloaded, it will be making short duration ON pulses, with the rest of the cycle being OFF, These pulses, when filtered, will deliver the voltage at minimal load. When heavily loaded, the PWM control self-adjusts to make the ON time proportion longer, so that the cycle only has very short periods of OFF.

The transition from OFF to ON and also ON to OFF is done in nanoseconds. The switch element does not dissipate power when OFF, and almost none when it is fully ON in a low resistance state. The only power lost is during the brief time during the switchover. That is what makes switched mode energy-efficient.

A Switching Regulator doesn't have a voltage drop across it or a resistance from input and output , so there is no voltage loss , is this true?

The switching regulator as a whole must have a different voltage at its input as at its output. It might be a voltage drop, or with a "buck-boost", it might even be a voltage rise. Because of the time-proportioned way the switcher delivers energy, this is done without a continuous high dissipation drop across the switching element itself. It spends time ON, and OFF, but very little in between. The rest of the components store energy and smooth the output.

 

[danny davis]

Thanks for the help

The switch element does not dissipate power when OFF, and almost none when it is fully ON in a low resistance state. The only power lost is during the brief time during the switchover.

Why is a switching regulator has very low resistance? a linear regulator has resistance

When is the Switchover, do u mean when its transiting or in between ON time to off time? this is in nanoseconds too

Be aware that some LDO linear regulators can "latch up" in a high current shorted state if started up with the load connected, or if starved into drop-out

I think this is what was happening, because when you power on the unit there was an IN RUSH current and a constant HIGH current draw , which put the regulator into a current limiting mode

Can you talk more about these states?