LM2940 Voltage Regulator Question

Thread Starter

mikewashere05

Joined Oct 26, 2009
15
I have a very simple circuit, a 12v 500mA wall wart going into a LM2940 voltage regulator connected in series with a 130Ω resistor and an LED. When I specified the voltage regulator to order, I gave the part # for the 5v. Case markings indicate 5v (LM2490T-5.0). All signs point to this being a 5v regulator.

HOWEVER, in my circuit, when voltage is measured from Vout of the regulator to ground of the regulator approximately 10.7v is measured. Current draw should be 5v / 130 ~ 38mA. The datasheet says for 5mA < Io < 1A there is a typical output of 5v (being a 5v regulator and all).

Why am I seeing 10.7v? I tried this on two regulators, both are the same way. I'm afraid to go further in the project (attaching microcontroller, etc.) without making sure it will only see 5v.

EDIT: Without the LED I measure 9.6v...
 
Last edited:

k7elp60

Joined Nov 4, 2008
562
Do you have the recommend external capacitors connected to the regulator? The 0.47uF in the input and the 22uF on the output. The data sheet says the 22uF is required for stability. Generally the input capacitor(.47uF)is required if the regulator is greater than 6" away from the filter capacitor. The 6" generally refers to the length of the input + wire.
 

Thread Starter

mikewashere05

Joined Oct 26, 2009
15
Do you have the recommend external capacitors connected to the regulator? The 0.47uF in the input and the 22uF on the output. The data sheet says the 22uF is required for stability. Generally the input capacitor(.47uF)is required if the regulator is greater than 6" away from the filter capacitor. The 6" generally refers to the length of the input + wire.
Badabing.

So what exactly was I seeing without the capacitor? Does the DMM read Vp, Vpp, Vrms? Was it just very noisy and the capacitor helped smooth it out? Why is the capacitor necessary and why isn't it just integrated into the IC?

Just some questions so I know the why behind the how.
 

SgtWookie

Joined Jul 17, 2007
22,230
So what exactly was I seeing without the capacitor?
Hard to say, without looking at your output using an oscilloscope.

Does the DMM read Vp, Vpp, Vrms?
You tell us - what did you have the DMM set to read?

Was it just very noisy and the capacitor helped smooth it out?
Without input/output capacitors, many monolithic voltage regulators will oscillate. I have a few old Motorola 5v linear regulators in my parts collection; if there is no capacitance on the output, they oscillate at between 3MHz-8MHz; the output voltage varies wildly.

Why is the capacitor necessary...
To ensure stability that it doesn't oscillate. It lowers the resonant frequency to below where it would oscillate.
... and why isn't it just integrated into the IC?
It wouldn't fit!

It is simply not practical to include large values of capacitance on the substrate of an integrated circuit such as a regulator.
 
Last edited:

THE_RB

Joined Feb 11, 2008
5,438
LM2940 will give all sorts of problems if you dont have input and output caps both connected, and pretty close to the pins of the regulator. Sometimes they refuse to power up too depending on the nature of your load, or have source voltage issues.

I have a bag of LM2940 5v here that I just stopped using because I got sick of their fussy operation.
 

Audioguru

Joined Dec 20, 2007
11,248
The LM2940 is a low dropout regulator. Most low dropout regulators use a PNP transistor as its adjustable series resistance. This PNP has voltage gain and phase-shift unlike the NPN transistor that has no gain and a small phase-shift that is used in ordinary regulators.
So low dropout regulators must use a pretty big capacitor at their output to prevent oscillation.
 

bountyhunter

Joined Sep 7, 2009
2,512
Why is the capacitor necessary and why isn't it just integrated into the IC?
The LM2940 is an LDO (low dropout regulator) which has a PNP pass device. It is a two pole control loop because it drives the load from the collector of the PNP which models as a current source, unlike the NPN regs which drive off an emitter which acts like a pure voltage source. The output cap is required to reduce the bandwidth and the output cap's ESR is required to cancel out one of the loop's poles.

A cap that large can not possibly be fabricated into an IC.
 
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