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'[EE]: RC time constant (was: [PIC]: Using PWM to v'
2002\03\16@094220 by Edson Brusque

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flavicon
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Hello Spehro et all,

<SNIP>
> Maybe try  PIC---[1K]-----x----> contrast
>                           |
>                         [470R]
>                           |
>                          0V
>
> If you need a cap, the output Z of the above is around 320 ohns, so
> the value should be maybe > 3 /(320 * fpwm)
> Eg. 1kHz PWM, get 9.3uF, use a 10uF-22uF electrolytic..

   this remember me a problem I haven't solved. When you have this RC
filter circuit:

   IN----R----*-----OUT
              |
              C
              |
             GND

   The RC time constant is 2*PI*R*C, so for a 2K2 resistor and 100nF cap
its:
       t = 2 * PI * 2200 * 100e-9 = 1.4e-3 = 1.4uS
   And the cutoff frequency is:
       f = 1/t = 723.4Hz

   Now, how can I calculate the RC time constant for the circuits below?

   IN----R1----*-----*----OUT
               |     |
               C     R2
               |     |
              GND   GND

   IN----R1----*-----*----R3----OUT
               |     |
               C     R2
               |     |
              GND   GND

   Best regards,

   Brusque

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2002\03\16@103836 by Spehro Pefhany

picon face
At 10:30 AM 3/16/02 -0300, you wrote:

>     this remember me a problem I haven't solved. When you have this RC
>filter circuit:
>
>     IN----R----*-----OUT
>                |
>                C
>                |
>               GND
>
>     The RC time constant is 2*PI*R*C,

RC time constant is just R * C, your formula is correct for 1/f (Xc = R, so
3dB
down), or lose the 2*Pi and calculate f in radians/sec.

>  so for a 2K2 resistor and 100nF cap
>its:
>         t = 2 * PI * 2200 * 100e-9 = 1.4e-3 = 1.4uS
>     And the cutoff frequency is:
>         f = 1/t = 723.4Hz

Yes.

>     Now, how can I calculate the RC time constant for the circuits below?
>
>     IN----R1----*-----*----OUT
>                 |     |
>                 C     R2
>                 |     |
>                GND   GND

Just take the Thevenin equivalent- assuming no loading on OUT
we have an ideal voltage source Vin as below:

Vin = IN * R2 * (R1+R2) --------[Rx]-----x--- OUT
                                         |
                                         C
                                         |
                                         GND
Rx = R1 *R2/(R1+R2)

Then you can use your method above with the equivalent circuit.


If you're not familiar with this method, here's a web page
that 'splains it fairly well:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html



>     IN----R1----*-----*----R3----OUT
>                 |     |
>                 C     R2
>                 |     |
>                GND   GND

The same as above unless there is loading on the circuit.  ;-)
If you short the output to ground and look at the current, then the
3dB down frequency would be

f = 1/( 2 * Pi *C * R1 || R2 || R3)

where A || B || C = 1/(1/A + 1/B + 1/C)


Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
.....speffKILLspamspam@spam@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
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2002\03\16@103917 by Dave Dilatush

picon face
Edson wrote...

{Quote hidden}

Actually, what we refer to as the "RC time constant" is usually the
product T = RC, without multiplying by 2*pi.

This product is used when calculating the time-domain response of an RC
circuit such as above: when a voltage step is presented to the input,
going from zero volts to some voltage Vin, the output waveform is
described by the expression

 Vout = Vin * (1 - (e^(-t/T))),

where t is the time elapsed since the input voltage step and T = RC.

For AC circuit analysis, in which we're using the circuit as a low-pass
filter, we use the 2*pi factor along with T to get the corner frequency
of the filter, as:

 f(-3db) = 1/(2*pi*RC)

just like you had.

(I'm not disputing your frequency calculation, just clarifying the usage
of the term "time constant".)

>
>    Now, how can I calculate the RC time constant for the circuits below?
>
>    IN----R1----*-----*----OUT
>                |     |
>                C     R2
>                |     |
>               GND   GND

The way to do this is to consider R1 and R2 to be in parallel with one
another with regard to AC, since R1 is connected from C to a circuit
node (the input) which has zero impedance with respect to ground (that
is, the input is driven by a voltage source), just as is R2.  So in
terms of the resistance that the capacitor "sees", R1 and R2 are in
parallel.

So the cutoff frequency calculation for the above circuit becomes:

 f(-3db) = 1/(2*pi*C*(R1*R2/(R1 + R2)))


>    IN----R1----*-----*----R3----OUT
>                |     |
>                C     R2
>                |     |
>               GND   GND
>

Now, this one depends on what the output is connected to: if the output
is connected to a high-impedance load, such as a PIC A/D converter input
or a digital voltmeter, then R3 simply "disappears" and the calculation
is just like in the previous circuit.

However, if R3 is connected to a short-circuit load (such as the
inverting input of an opamp configured as a summer or an inverting
amplifier), then R3 has to be taken into account.  In this case, for
calculation purposes, R3 appears in parallel with R2 and R1, and the
cutoff frequency expression becomes:

 f(-3db) = 1/(2*pi*C*(1/((1/R1)+(1/R2)+(1/R3))))

And finally, if R3 is feeding into a load resistance Rload which is
neither infinite nor zero, then you do the calculation as above but with
(R3+Rload) replacing R3 in the formula.

Hope this helps a bit...

Dave

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2002\03\16@112835 by Edson Brusque

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flavicon
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Hello Spehro and Dave,

Spehro says:
> RC time constant is just R * C, your formula is correct for 1/f (Xc = R,
so
> 3dB down), or lose the 2*Pi and calculate f in radians/sec.
Dave says:
> Actually, what we refer to as the "RC time constant" is usually the
> product T = RC, without multiplying by 2*pi.

   oh, big confusion here. Now it clarified. Thanks.

Dave says:
>This product is used when calculating the time-domain response of an RC
>circuit such as above: when a voltage step is presented to the input,
>going from zero volts to some voltage Vin, the output waveform is
>described by the expression
>  Vout = Vin * (1 - (e^(-t/T))),
>where t is the time elapsed since the input voltage step and T = RC.

   Sorry for my math ignorance but what's that "e" in the formula?

>(I'm not disputing your frequency calculation, just clarifying the usage
>of the term "time constant".)

   thank you again for clarifying this.

Dave says:
{Quote hidden}

   Great explanation.

Spehro says:
<SNIP>
> If you're not familiar with this method, here's a web page
> that 'splains it fairly well:
> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html

   thank you again. Great site.

Spehro says:
{Quote hidden}

   Good.

Dave says:
>And finally, if R3 is feeding into a load resistance Rload which is
>neither infinite nor zero, then you do the calculation as above but with
>(R3+Rload) replacing R3 in the formula.

   Actually it's connected to the non-inverting input of a comparator. I
think this makes the calculation more complex as we have the input impedance
of the comparator, the positive-feedback resistance (schmitt trigger) and
the pull-up resistor at the comparator output.

   But your lesson give me clues of what to do and what to expect.

   Thank you very much again,

   Brusque

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2002\03\16@114900 by Dave Dilatush

picon face
Edson wrote...

>    Sorry for my math ignorance but what's that "e" in the formula?

That's the natural logarithm base, e = 2.7182818

>Dave says:
>>And finally, if R3 is feeding into a load resistance Rload which is
>>neither infinite nor zero, then you do the calculation as above but with
>>(R3+Rload) replacing R3 in the formula.
>
>    Actually it's connected to the non-inverting input of a comparator. I
>think this makes the calculation more complex as we have the input impedance
>of the comparator, the positive-feedback resistance (schmitt trigger) and
>the pull-up resistor at the comparator output.

Hmmm...

Unless you're generating a REALLY big amount of hysteresis in your
Schmitt trigger, the positive-feedback resistance is going to be rather
big compared to the rest of the resistances in the circuit.  And if
that's the case, R3 and everything beyond it will have little shunting
effect on the RC network, and you might be able to ignore it.  Remember,
capacitors usually have rather wide tolerances and it's often pointless
to make extremely precise calculations when they are involved.  :)

Dave

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