[SI-LIST] Re: Negative Voltage signal Return Path
- From: Rohit MISHRA <rohit.mishra@xxxxxx>
- To: "si-list@xxxxxxxxxxxxx" <si-list@xxxxxxxxxxxxx>, "ah.vinod@xxxxxxxxx" <ah.vinod@xxxxxxxxx>
- Date: Mon, 16 Apr 2012 21:16:32 +0800
Hello Vinod,
The coupling capacitor is basically working as a high pass filter and hence
decoupling dc common mode voltage of amplifier & receiver.
To understand the return path of negative swing, you should first try to
understand how series capacitor converts +ve swing to -ve swing.
To explain the concept I have drawn a figure where amplifier/driver is
connected to some resistor Rterm ( Rterm can be terminating resistor or input
resistance of receiver) through a coupling capacitor. Terminal A of capacitor
is connected to driver while terminal B to Rterm.
A B
Vp -------------||-----------------
___
|
DRIVER | | Rterm
| |
|
0V --------------------------------
Return Path
Case 1 :
Amplifier generates a pulse of voltage Vp,
The positive supply will charge the capacitor so
At T = 0, Voltage across Rterm = + Vp, Voltage across Capacitor = 0 Volt
At T = 5 X Time Constant, Rterm = 0 Volt, Voltage across Capacitor = +Vp (Node
A positive w.r.t B)
So after 5 X time constant, capacitor is fully charged and voltage across Rterm
is 0 V.
Case 2 :
Amplifier output goes from Vp to 0V, In this case amplifier is basically
shorting node A of capacitor with Return Path ( Assuming o/p resistance of
amplifier is zero)
At T = 0, Voltage across Rterm = Voltage across Capacitor = - Vp,
At T = 5 X Time Constant, Voltage across Rterm = Voltage across Capacitor = 0
Volt ( All energy stored in capacitor has dissipated through Rterm )
In case 1, when amplifier output was + Vp and positive swing across receiver.
Direction of current :
driver -> capacitor -> Rterm -> Gnd(Return Path)
While in case 2, when amplifier output was 0 V and negative swing across
receiver,
Direction of current :
capacitor -> driver -> Gnd -> Rterm
Though amplifier is generating a positive pulse but receiver is seeing +ve as
well as -ve swing and current direction in case 2 is opposite to case no. 1.
Now if positive potential at return path during negative swing is bothering you
then you should understand that potential difference is just a relative
concept. To understand this take two cells A & B of 1.5 volt and connect them
in series to get total 3 Volt. Now connect two resistors across A & B. Both
resistors will see same 1.5 V, If you assume that the point where cell A & B
meets i.e. + terminal of A & -ve terminal of B is the return path of resistor,
you can see though return path is connected to +ve as well as -ve potential of
cell but still it is at same potential. That's something happening with your
board during negative swing though whole board is at same potential.
Does it clear your doubt ??
Rgds,
Rohit Mishra
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of steve weir
Sent: Monday, April 16, 2012 12:37 PM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Negative Voltage signal Return Path
I = C*dv/dt.
Steve
On 4/15/2012 11:52 PM, vinod ah wrote:
> Hi All,
> I have one very basic question on how does negative signal return i.e.
> return path of a negative voltage signal. I am working of HD component
> video signal which swings from -0.3V to +1V at the connector. The signal
> comes from an amplifier (which is a single supply op-amp) and ac coupled
> before going to the connector. The amplifier is powered using a positive
> supply, so the design makes sure that output and input of op-amp is having
> positive swing signal. But since we do AC-coupling, we get negative signal
> at the connector. So i am just thinking, whole board is working on
> positive supply and 0 Volts ground, but only at connectors i see a negative
> signal i.e. ground at higher potential than the signal.
>
> So i am feeling that ground is at different potential at connectors when
> compared to other places in board? But this is not possible!!!. So where am
> i missing in understanding the return path of negative swing signal?
>
> Regards
> Vinod A H
>
>
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