Power Management circuit - resistance across voltage sources

Power Management circuit

hi,

i am currently reviewing the design of our company's power management circuit; for your background in our design - our power management is composed of different blocks for 5V to 3.3V then 3.3V to 2.25V then 2.25 to 1.8V these blocks are used to supply the different VCCs of our cores inside our product, and here is what I found:

--> there is a significant resistance across each voltage sources for example across the 3.3V (disconnected from the load) there is a resistance of 5ohms but this is subjected to supply as high as 5A...

the question is, is the 3.3V supply capable to handle this amount of load (5A)?

on my point of view, this is not valid because this voltage supply is limited only to a maximum of (3.3/5) 660mA current which is by considering its internal resistance of 5ohms...

need your opinions...

thanks....

regards,
Simeon

Power Management circuit

I believe there is a flaw in your reasoning. While you did not state it, I believe you are measuring across linear regulator ICs using a low power ohmmeter. The 5 ohms you measure is just the bleeder resistor across the pass transistor. The pass transistor, which you cannot measure is in parallel with this resistance.
The internal circuitry will control the pass transistor toallow sufficient current to pass to keep the output voltage at the 3.3V level.
If a linear regulator is rated at 5A it will be able to pass this much current providing that two conditions are met. First, the input voltage must have sufficient headroom above the output voltage to allow regulation to take place. Secondly, the heatsink on the regulator must be able to keep the IC's temperature at a reasonable level. Since poor heatsinking can destroy power regulator chips, most chips now sense overtemp and will turn themselves off if they get too hot. Many people blame the regulator as faulty when it is really in thermal shutdown due to a poor heatsink.

As an example, assume that your 3.3V regulator is being sourced from your 5V supply. The regulator must require a headroom of less than(5V - 3.3V) or 1.7V. In addition, at its max current rating, the headsink must dissipate (1.7V * 5A ) or 8.5 Watts. Of course, if the circuit only ever draws 1A from a 5A part, everything is still OK. In this case, the heatsink only needs to dissipate 1.7Watts.

Re: Power Management circuit

banjo said:

The 5 ohms you measure is just the bleeder resistor across the pass transistor. The pass transistor, which you cannot measure is in parallel with this resistance.

Click to expand...

Why is it that I didnt measure the resistance of the pass transistor if it is in parallel with the bleeder resistor?

How if I did not use a regulator IC instead I only use a discrete linear regulator using a comparator OP Amp and a PMOS to convert 5V to 3.3V? Will it drive a 5A load with an internal resistance of 5ohms?

Power Management circuit

There are a couple of reasons that you cannot measure the resistance of the pass transistor. First, the ohmmeter does not have enough output voltage to enable any of the silicon devices. This is by design so that you can measure in-circuit resistors more accurately without worrying about the conductance of the semiconductors.
Another reason is that the pass transistor is a three terminal device. It requires either an enabling current or enabling voltage before it will conduct. (Whether it requires current or voltage depends on whether it is a BJT or FET. ) Your ohmmeter can only drive two of the three terminals, therefore, the device remains off.

When fully on, power MOSFET transistors will have a resistance on the order of 0.025 ohms. Do the math, if the input voltage is 5V and the load is 3.3V at 5A, then the effective resistance of the regulator is 1.7/5 or 0.34 ohms. This is thirteen times higher than the lowest resistance of the MOSFET, therefore, it will operate in the linear region, not the saturated region.

Power Management circuit

When you measure and find the resistance that it is giving 5 ohms that time the unit is off.
when you live the unit with potential then its impedance will change.
the best way to find the load is with the application of rated voltage not by the ohmmeter.

Power Management circuit

Thanks for your inputs guys....

Btw, I am not that convinced that the internal resistance that I measured is the OFF-resistance of the MOSFET because if the MOSFET is in its OFF state the output voltage that I will measure is zero...