[SOLVED] Help with Voltage Regulator

[Leethal]

Hi, I have added R6 to M2 of this circuit in a bid to limit the current to 6A. It seems that the value has to be very precise to achieve this and would vary with temeperature. Would anyone any knowledge of current limiting a LDO regulator and is this a viable method or can someone show me a better method??

 

[crutschow]

In order to limit the current, you need to sense the current. R6 doesn't do that.

Below is your basic circuit modified with a PNP transistor and shunt resistor added to provide a ≈6A current limit. When the voltage across the shunt reaches about 0.66V it starts to turn on the transistor which increase the gate voltage to M1 (reducing Vgs), limiting the current.

This simple method is somewhat temperature sensitive (it has a negative coefficient of about 0.3%/°C).

Note: I didn't have the model for the MOSFET you used so I just used an appropriate one in my model list.

 

[Leethal]

Thank again crutshow, that works well although there is some power loss/heat associted with the shunt resistor which I would like to avoid as much as possible. Possibly adding a difference amp with some gain and a smaller shunt resistor will achieve this. While this circuit will perform the task required of it for the time being, do you know of any further reading where I might improve this circuit?

 

[crutschow]

Below is the circuit using a rail-rail input and output op amp for the current limit to lower the shunt resistor power dissipation

A 0.01 ohm shunt resistor is used to detect the current giving a 60mV threshold for 6A which is generated by R6 and R7.

I couldn't use the op amp you used since it isn't rail-rail input.

Note I had to increase the value of C1 to avoid an overshoot glitch upon power up.

Edit: Since the LT1498 is a dual op amp, I also tried it in the simulation for U1 and it seemed to work the same as the one you used. Thus you only need one package for both op amps.

 

[Leethal]

Hi,

Thank you again. Would you be able to describe how you calculated the values for the difference amp. I figure you have 60mV across the shunt resistor and you need 600mV to forward bias the diode. I don't quite understand how you set the gain to 10??

 

[crutschow]

There is no specific voltage gain as such. It's really a current to voltage loop gain The circuit you posted is not equivalent to the circuit in the voltage regulator. Your circuit has no feedback, giving it a gain equal to the op amp open loop gain. My circuit is a closed loop, with negative feedback from the voltage across the shunt resistor, due to the current, to the gate of the MOSFET which controls the current.

The diode is to prevent the op amp output from affecting the MOSFET control voltage when the current is below the limit and the op amp output is at zero volts. The diode forward drop is inside the feedback loop so its effective voltage is reduced by the open loop gain of the op amp, and becomes negligible.

So basically you have two control loops. The voltage feedback controls the loop below the current limit point, and the current feedback loop controls the loop when at the current limit.

 

[Leethal]

I see. The voltage divider 180R/100.18k * 30 gives approximately 60mV which is the voltage drop across 0.01R shunt resistor at 6A. The op amp keeps the voltage at the gate such that the voltages at its inputs are equal via the transconductance of the FET.

 

[crutschow]

I see. The voltage divider 180R/100.18k * 30 gives approximately 60mV which is the voltage drop across 0.01R shunt resistor at 6A. The op amp keeps the voltage at the gate such that the voltages at its inputs are equal via the transconductance of the FET.

I used 33V but otherwise, you got it. :-D

 

[Leethal]

Any ideas how I may protect the op amp. I seems to get hot and fail if I briefly short the output.