[SI-LIST] Re: Regarding Output Voltage Rise time of linear Regulators.

Vivek,
the LDO's I am familiar with are collector output. As such the output Z,
above the frequency where the loop controls it, is high. The older non-LDO
regulators tended to have emitter outputs and consequently lower output Z at
high frequencies.

You can determine the transient response of an LDO by simply running a step
load test. An oscilloscope, a piece of wire, a resistor, and a
connecting/disconnecting the wire is all that is needed. From that you can
build a usable model.

    Dave Cuthbert

On Fri, Feb 27, 2009 at 1:56 AM, Vivekkumar M-TLS,Chennai <
Vivekkumarm@xxxxxx> wrote:

> Dear All,
>
> We need to find the output Voltage regulation time of linear regulators.
> Generaly when using LDO's(Linear Dropout Regulators) the Output Voltage
> regulation time is given by the manufacturer himself. But in the
> scenario when data is not avaliable in the spec, please can any one
> confim if the following method is acceptable:
>
> (Input Voltage - Output Voltage)/Load Current =  Series pass resistance
> Value.
>
> Let this resistance value be Rs.
>
> We will have adjustable resistors which provides a feedback of the
> output voltage. Let them be Ra,Rb.
>
> Hence using Thevenin's theorem,
>
> The effective resistance would be Rs + (Ra II Rb) = Reff.
>
> The Effective output capacitance will be the total capacitance added at
> the LDO output. = Cout.
>
> Hence the Regulation time ( Load changing from 10% to 90%) = t 5*Reff*Cout.
>
>
> Requesting everyone to give their comments on this.Your immediate
> response will be sincerely appreciated.
>
> Regards
> Vivek
>
>
>
>
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