## which is more reactive?

Expand Messages
• is a floating bicore more or less reactive than a grounded bicore? if so, why? Adam
Message 1 of 5 , Apr 3, 2005
• 0 Attachment
is a floating bicore more or less reactive than a grounded bicore? if
so, why?

• The floating or suspended resistor bicore is more reactive than a grounded resistor bicore if by reactive you mean sensitive to electrical noise. The why is
Message 2 of 5 , Apr 3, 2005
• 0 Attachment
The floating or suspended resistor bicore is more reactive than a grounded resistor bicore if by "reactive" you mean sensitive to electrical noise.

The why is of this is not a simple matter.  Just after changing state both are equally reactive as the voltage level at each inverter input is at Vcc or at Gnd.  These voltage levels start to rapidly decay and it  is when the voltage at the inverter input approaches the switching threshold that the sensitivity to noise becomes noticable.

In both cases the voltage at the inputs start to change towards the voltage reference that the resistor is connected to.

For a grounded resistor bicore , the voltage at the active input is initially at Vcc and  drops exponentially towards Gnd. For a suspended bicore, the voltage waveforms at both inputs change from Vcc and Gnd respectively towards Vcc/2

The sensitivity to noise of the bicores depends on the rate of change of this voltage as it crosses the switching threshold.

It is not easy to translate this into everyday experience but here goes:

Think of the waveform as a skislope which is almost vertical at the start and then gradually changes to horizontal at the end of the slope.

Just using gravity, a skier would have maximum speed at the top and still have considerable speed when he crosses the  midpoint of the slope. If the midpoint is the finish line (74HCxxx with Vth = Vcc/2) , then whether or not he cheats and reaches out with his pole (noise) to cross early will make little difference in the race time.   If the finish line was near the horizontal end of the slope (74HCTxxx with Vth=Vcc/3), his speed by then will have slowed to a crawl (and he might not even make) reaching out with the pole will make  a big difference in the time it takes especially if the pole was of significant length ( ie 10% of the slope 8^)

Now the case of the suspended bicore is a little more difficult to portray since  gravity doe not have polarity.  But lets assume that we have two slopes that face each other and meet on in the middle when both slopes are horizontal. Now two skiers starting at the top of each slope.  Which ever crosses the fnish line first ends the race. If the finish line is right in the middle (74HCxxx)  they have equal chance of getting there and equal chance of cheating by usingthe pole to cross the finish line first.   If the finish line is higher up on one or the other slope (74HCTxxx) ,  that side will always win.

I hope this helps to paint a picture of what it means when we say,

"The sensitivity to noise of the bicores depends on the rate of change of the exponential timing voltage waveform as it crosses the switching threshold of the inverters"

wilf
• thanks wilf, i think i got understand it. so it only actually matters right before the threshold of the input, if there is more current in the motor at that
Message 3 of 5 , Apr 3, 2005
• 0 Attachment
thanks wilf, i think i got understand it. so it only actually matters
right before the threshold of the input, if there is more current in
the motor at that time, it will trigger the inverter sooner than if
the motor wasn't stalled or what-not. do i got that right, or is that
backwards somehow?

--- In beam@yahoogroups.com, "Wilf Rigter" <wrigter@d...> wrote:
> The floating or suspended resistor bicore is more reactive than a
grounded resistor bicore if by "reactive" you mean sensitive to
electrical noise.
>
> The why is of this is not a simple matter. Just after changing
state both are equally reactive as the voltage level at each inverter
input is at Vcc or at Gnd. These voltage levels start to rapidly
decay and it is when the voltage at the inverter input approaches the
switching threshold that the sensitivity to noise becomes noticable.
>
>
> In both cases the voltage at the inputs start to change towards the
voltage reference that the resistor is connected to.
>
> For a grounded resistor bicore , the voltage at the active input is
initially at Vcc and drops exponentially towards Gnd. For a suspended
bicore, the voltage waveforms at both inputs change from Vcc and Gnd
respectively towards Vcc/2
>
> The sensitivity to noise of the bicores depends on the rate of
change of this voltage as it crosses the switching threshold.
>
> It is not easy to translate this into everyday experience but here
goes:
>
> Think of the waveform as a skislope which is almost vertical at the
start and then gradually changes to horizontal at the end of the slope.
>
> Just using gravity, a skier would have maximum speed at the top and
still have considerable speed when he crosses the midpoint of the
slope. If the midpoint is the finish line (74HCxxx with Vth = Vcc/2) ,
then whether or not he cheats and reaches out with his pole (noise) to
cross early will make little difference in the race time. If the
finish line was near the horizontal end of the slope (74HCTxxx with
Vth=Vcc/3), his speed by then will have slowed to a crawl (and he
might not even make) reaching out with the pole will make a big
difference in the time it takes especially if the pole was of
significant length ( ie 10% of the slope 8^)
>
> Now the case of the suspended bicore is a little more difficult to
portray since gravity doe not have polarity. But lets assume that we
have two slopes that face each other and meet on in the middle when
both slopes are horizontal. Now two skiers starting at the top of each
slope. Which ever crosses the fnish line first ends the race. If the
finish line is right in the middle (74HCxxx) they have equal chance
of getting there and equal chance of cheating by usingthe pole to
cross the finish line first. If the finish line is higher up on one
or the other slope (74HCTxxx) , that side will always win.
>
> I hope this helps to paint a picture of what it means when we say,
>
> "The sensitivity to noise of the bicores depends on the rate of
change of the exponential timing voltage waveform as it crosses the
switching threshold of the inverters"
>
> wilf
• Very close Adam. This sensitivity to noise causing early triggering only matters right before the waveform reaches the threshold. The increased motor load can
Message 4 of 5 , Apr 3, 2005
• 0 Attachment

This sensitivity to noise causing early triggering only matters right before the waveform reaches the threshold.  The increased motor load can cause more noise in the system caused by supply voltage drop, motor brush noise or (if connected directly to the motor) when the bicore output voltage drop rises from increased output current.

The latter is the most direct path as the motor generated noise is coupled directly from the motor terminals back to the bicore inputs.

This feedback effect can be obtained in circuits with motor drivers (which normally isolate the bicore from the motor) by connecting the bicore capacitors, not to the 240 outputs, but to the motor driver outputs.

wilf

wilf
----- Original Message -----
Sent: Sunday, April 03, 2005 8:44 PM
Subject: [beam] Re: which is more reactive?

thanks wilf, i think i got understand it.  so it only actually matters
right before the threshold of the input, if there is more current in
the motor at that time, it will trigger the inverter sooner than if
the motor wasn't stalled or what-not.  do i got that right, or is that
backwards somehow?

--- In beam@yahoogroups.com, "Wilf Rigter" <wrigter@d...> wrote:
> The floating or suspended resistor bicore is more reactive than a
grounded resistor bicore if by "reactive" you mean sensitive to
electrical noise.
>
> The why is of this is not a simple matter.  Just after changing
state both are equally reactive as the voltage level at each inverter
input is at Vcc or at Gnd.  These voltage levels start to rapidly
decay and it  is when the voltage at the inverter input approaches the
switching threshold that the sensitivity to noise becomes noticable.
>
>
> In both cases the voltage at the inputs start to change towards the
voltage reference that the resistor is connected to.
>
> For a grounded resistor bicore , the voltage at the active input is
initially at Vcc and  drops exponentially towards Gnd. For a suspended
bicore, the voltage waveforms at both inputs change from Vcc and Gnd
respectively towards Vcc/2
>
>  The sensitivity to noise of the bicores depends on the rate of
change of this voltage as it crosses the switching threshold.
>
> It is not easy to translate this into everyday experience but here
goes:
>
> Think of the waveform as a skislope which is almost vertical at the
start and then gradually changes to horizontal at the end of the slope.
>
> Just using gravity, a skier would have maximum speed at the top and
still have considerable speed when he crosses the  midpoint of the
slope. If the midpoint is the finish line (74HCxxx with Vth = Vcc/2) ,
then whether or not he cheats and reaches out with his pole (noise) to
cross early will make little difference in the race time.   If the
finish line was near the horizontal end of the slope (74HCTxxx with
Vth=Vcc/3), his speed by then will have slowed to a crawl (and he
might not even make) reaching out with the pole will make  a big
difference in the time it takes especially if the pole was of
significant length ( ie 10% of the slope 8^)
>
> Now the case of the suspended bicore is a little more difficult to
portray since  gravity doe not have polarity.  But lets assume that we
have two slopes that face each other and meet on in the middle when
both slopes are horizontal. Now two skiers starting at the top of each
slope.  Which ever crosses the fnish line first ends the race. If the
finish line is right in the middle (74HCxxx)  they have equal chance
of getting there and equal chance of cheating by usingthe pole to
cross the finish line first.   If the finish line is higher up on one
or the other slope (74HCTxxx) ,  that side will always win.
>
> I hope this helps to paint a picture of what it means when we say,

> "The sensitivity to noise of the bicores depends on the rate of
change of the exponential timing voltage waveform as it crosses the
switching threshold of the inverters"
>
>  wilf

• This is interesting. I can see (I think) how this works with an AC245 as a motor driver, for instance, but can it be done using an H-bridge? Wilf Rigter
Message 5 of 5 , Apr 4, 2005
• 0 Attachment
This is interesting. I can see (I think) how this works with an AC245 as a motor driver, for instance, but can it be done using an H-bridge?

Wilf Rigter <wrigter@...> wrote:
This feedback effect can be obtained in circuits with motor drivers (which normally isolate the bicore from the motor) by connecting the bicore capacitors, not to the 240 outputs, but to the motor driver outputs.

wilf

Find local movie times and trailers on Yahoo! Movies.

Your message has been successfully submitted and would be delivered to recipients shortly.