Op-amp voltage regulator - Page 1
Forum section: electronics / beginners

author: Unknown
author_rank: Unknown
timestamp: May 31, 2015, 10:33:22 pm
content: Hi,i found this voltage regulator on the schematic of the pickit2, it uses an op-amp and a p channel mosfet.I have some doubt about R34 and R7, can someone explain their function?Thanks for the help.

author: jeroen79
author_rank: Frequent Contributor
timestamp: June 01, 2015, 02:30:36 am
content: R34 will discharge C6 and C7.R7 will pull V_tgt to Gnd when Q1 is closed.

author: rs20
author_rank: Super Contributor
timestamp: June 01, 2015, 02:32:56 am
content: Quote from: jeroen79 on June 01, 2015, 02:30:36 amSeems legit. I totally agree on R34, what else could it possibly be doing, but doesn't 2.7k seem like an awfully small and specific value? Does a 5V cap even need a bleeder resistor? Wouldn't 10k, 100k work just fine? It seems odd.

author: Ian.M
author_rank: Super Contributor
timestamp: June 01, 2015, 03:39:31 am
content: R34 is needed to discharge C5 quickly enough for the PICkit 2's PIC18F2550 MCU to reset properly if the USB lead is unplugged then immediately replugged. Its needed because the firmware cannot have the BOR reset enabled in the CONFIG as therail voltage tends to dip when switching power to the target board.I suspect the value was determined by experiment. It should be noted that it violates the USB standard for suspend current.

author: dom0
author_rank: Super Contributor
timestamp: June 01, 2015, 10:04:27 am
content: This is the regulator for the target power? I'm surprised it is stable with low-ESR loads you'd expect from any kind of logic board...

author: Unknown
author_rank: Unknown
timestamp: June 02, 2015, 12:12:54 pm
content: Thanks for the help.Quote from: dom0 on June 01, 2015, 10:04:27 amCan you explain a bit more this problem? In a project i'm designing i need a low power controllable rail (max 20 ma) and was thinking of using this circuit.

author: dom0
author_rank: Super Contributor
timestamp: June 02, 2015, 12:30:53 pm
content: Well the capacitance at the output node contributes a pole to the regulators transfer function (why? because the common emitter output stage of a LDO has high dynamic output resistance, in contrast to an emitter follower - if the output resistance is very low the capacitance's phase shift is neglibile). Now here's the catch: the ESR plus the capacitance also do something, they give you a zero (which gives positive phase shift, increasing stability). If C and C_esr are controlled, the zero and the pole will about compensate, and the regulator is stable. (Also see National/TI AN-1148)So the issue is, that such a simple regulator is usually not stable with loads that have high capacitance or low ESR (e.g. ceramic caps).

author: Ian.M
author_rank: Super Contributor
timestamp: June 02, 2015, 03:42:40 pm
content: After this regulator, it has 10uF (16V) electrolytic || 0.1uF ceramic for decoupling then it goes through the P channel half of a FDC6420C dual MOSFET for target power switching and a ZHCS1000 Schottky diode before it reaches the target board. Also the target power capability is only supposed to be 25mA, so this regulator should never be powering a large logic board or one with large total decoupling capacitance.I suspect the MOSFET and diode add enough series resistance for stability with all normal loads.