Diodes and dropping resistors

Diodes and dropping resistors

Post by Bill » Tue, 12 Apr 2005 20:10:21



Quote from a previous unrelated thread...
"BTW an old timer on another forum suggests placing the dropping
resistor between the transformer and diode, NOT placing the dropping
resistor between the rectifier and smoothing capacitor.  I don't agree!"

. . . .

It depends.

Often what people are calling a dropping resistor is not there for the
purpose of dropping voltage.  The classic app is "I'm replacing a
selenium rectifier with a silicon diode - what value resistor do I need
to add?."  The typical 50s-60s ac/dc table radio with a selenium drops
about 4 or 5 volts DC as compared to 0.7 volts DC with the silicon
diode.  This really doesn't need any compensation but there DOES need to
be a surge limiting resistor to protect the rectifier.  Odds are that
there was one in place already and usually on the input side of the diode.

The same type of set utilizing a power xfmr really doesn't need much if
any limiting because the power xfmr does this inherently.  It doesn't
hurt to have one though.

The resistor commonly found in such an application is typically in the
22-47 ohm range, sometimes as much as 100.  Do the math.  For a 40ma
load, a 100 ohm resistor is only dropping 4 volts.  This is hardly of
consequence on the B+ line which may vary that much given incoming ac
line conditions.

If I want to do massive voltage dropping I'd personally tend do it on
the DC side of the rectifier.  The is the situation often found in TV
sets, suitcase radio filament strings like the TO, etc.  I suppose it
could be done equally as well at the rectifier input.

I'm not sure why this convention is in place.  Maybe its just an
engineering 'habit' to do the AC-related chore of protecting the diode
on the AC-side and to do the DC-related task of voltage dropping on the
"DC side".  Maybe there are issues with a resistor's ratings on the AC
side?  Maybe the resistor AFTER the diode is more effective at limiting
outgoing DC voltage during the first couple of cycles while the filter
cap charges?  Maybe it really doesn't matter at all?

Comments, discussion?

-Bill

 
 
 

Diodes and dropping resistors

Post by Paul Sherw » Tue, 12 Apr 2005 21:30:19


There are 2 separate issues here.

Firstly, some people want to use a diode to replace a selenium
rectifier or (less often) a tube rectifier. Some series resistance
will be kinder to the smoothing electrolytics, and will also limit the
switchon current through the diode. The B+ voltage is less of an issue
though some metal rectifiers have quite high internal resistance even
when new. I've found that 100-220 ohms gives about the right B+. It
doesn't really matter whether the resistor is before or after the
diode. If half wave rectifying, two diodes in series reduce the
voltage stress on the diodes and provide some protection if one fails
short circuit.

Secondly, people use diodes to reduce heat dissipation by a series
string dropper resistor or to replace a resistive line cord.
Personally I use the following arrangement:  diode, fuse or panel
bulb, 'backwards' diode between fuse and chassis, dropper resistor.
The backwards diode will immediately blow the fuse if the main diode
fails short circuit.

There's lots of stuff on diode and capacitive droppers here:
http://www.vintage-radio.com/repair-restore-information/valve_dropper...
This is a British site so most of the examples are based on 100mA
heater strings and 240V power.

Best regards, Paul
--
Paul Sherwin Consulting     http://paulsherwin.co.uk

 
 
 

Diodes and dropping resistors

Post by Bill Sheppa » Tue, 12 Apr 2005 21:34:57


Quote:
>"...an old timer on another forum
>suggests placing the dropping resistor
>between the transformer and diode, NOT >placing the dropping resistor
between
>the rectifier and smoothing capacitor."

Seems like pure convention more than anything else, even where a
filament dropper comes off the top of the first filter cap (as in
AC/battery sets). Does it make any difference electrically which side of
the diode the "surge limiter" is on?
                Bill(oc)
 
 
 

Diodes and dropping resistors

Post by John Byr » Wed, 13 Apr 2005 02:22:02




Quote:
> The resistor commonly found in such an application is typically in the
> 22-47 ohm range, sometimes as much as 100.  Do the math.  For a 40ma
> load, a 100 ohm resistor is only dropping 4 volts.  This is hardly of
> consequence on the B+ line which may vary that much given incoming ac
> line conditions.

Actually the drop for a 100 Ohm resistor directly in series with the diode
would be considerably more than 4 volts because we are dealing with a peak
current in the rectifier considerably greater than the 40ma DC load
current.

Quote:
> If I want to do massive voltage dropping I'd personally tend do it on
> the DC side of the rectifier.  The is the situation often found in TV
> sets, suitcase radio filament strings like the TO, etc.  I suppose it
> could be done equally as well at the rectifier input.

I think voltage dropping is better done after the filter capacitor because
less heat is produced when the dropping resistor is after the filter
capacitor because the RMS current is lower at that point than in the
rectifier circuit.

Regards,

John byrns

Surf my web pages at,  http://users.rcn.com/jbyrns/

 
 
 

Diodes and dropping resistors

Post by robert case » Wed, 13 Apr 2005 01:59:31


Quote:
> Seems like pure convention more than anything else, even where a
> filament dropper comes off the top of the first filter cap (as in
> AC/battery sets). Does it make any difference electrically which side of
> the diode the "surge limiter" is on?
>                 Bill(oc)

Electrically it makes no difference.  Whichever is more
convenient to do in a set will be fine.
 
 
 

Diodes and dropping resistors

Post by Jeffrey D Angu » Wed, 13 Apr 2005 02:16:03


Quote:

>> Seems like pure convention more than anything else, even where a
>> filament dropper comes off the top of the first filter cap (as in
>> AC/battery sets). Does it make any difference electrically which side of
>> the diode the "surge limiter" is on?                 Bill(oc)

> Electrically it makes no difference.  Whichever is more
> convenient to do in a set will be fine.

In the sense of a simple series circuit, as in ONLY the
diode and resistor in series, it makes NO difference.

o--Resistor----Diode---o is equal to o---Diode----Resistor---o

However, if we're talking about placement of the resistor with
respect to filter capacitors....

o---Resistor----Diode---+---o
                         |
                     Capacitor
                         |
                         o

Is decidedly different from

o---Diode---+---Resistor---o
             |
         Capacitor
             |
             o

The first example has the resistor protecting the diode from
excessive charging currents.

The second example has the resistor just dropping the available
B+ based on load current, and not protecting the diode from the
charging current of the filter capacitor.

Jeff

--
"They that can give up essential liberty to obtain a little temporary
safety deserve neither liberty nor safety." Benjamin Franklin
"A life lived in fear is a life half lived."
Tara Morice as Fran, from the movie "Strictly Ballroom"

 
 
 

Diodes and dropping resistors

Post by Randy or Sherry Gutter » Wed, 13 Apr 2005 04:22:08


Quote:

> Actually the drop for a 100 Ohm resistor directly in series with the diode
> would be considerably more than 4 volts because we are dealing with a peak
> current in the rectifier considerably greater than the 40ma DC load
> current.

Only on initial power up, then the peak current becomes inconsequential,
and the RMS value is all that matters (as long as the peak voltage
across the resistor doesn't exceed the voltage rating of the resistor).
You can argue peak ***all you want - but that peak is over a (shorter
and shorter) duty cycle. That's why RMS is all important.  Why? what IS
RMS? -- and I don't mean it's mathematical derivation - what is RMS in
the practical world?

The AC equivalent of DC in the amount of energy converted - which in the
case of a resistor - that product is heat. That's why resistors are
rated in Watts. Peak current doesn't matter, peak voltage (to the point
of flash-over, anyway) doesn't matter. Just heat. Watts RMS.

And then he went on to say:

Quote:
> I think voltage dropping is better done after the filter capacitor because
> less heat is produced when the dropping resistor is after the filter
> capacitor because the RMS current is lower at that point than in the
> rectifier circuit.

How so? If it is then you have leakage somewhere- either in reverse
current in the rectifiers, leaky filter caps, or something - otherwise
Kerchkoff has it wrong.  Power in has to equal power out - and unless
you're generating a lot of loss in the rectifiers, filters, or such -
the RMS power value at any given point in a power supply is going to be
the same.

If you were to restate your comment above as "because the AC current is
lower at that point than in the rectifier circuit", then you would be
correct.  But the RMS value IS THE SAME, (just most of the component is
DC rather than AC at that point - just as the major component at the
rectifiers is AC rather than DC). Remember RMS does NOT imply nor mean
AC, though it's ***y rare to hear anyone refer to DC as RMS - but in
truth 100W DC is EXACTLY the same as 100W RMS (or volts or current, for
that matter).

best regards...
--
randy guttery

A Tender Tale - a page dedicated to those Ships and Crews
so vital to the United States Silent Service:
http://www.FoundCollection.com/

 
 
 

Diodes and dropping resistors

Post by Jeffrey D Angu » Wed, 13 Apr 2005 04:57:00



Quote:
> If you were to restate your comment above as "because the AC current is
> lower at that point than in the rectifier circuit", then you would be
> correct.  But the RMS value IS THE SAME, (just most of the component is
> DC rather than AC at that point - just as the major component at the
> rectifiers is AC rather than DC). Remember RMS does NOT imply nor mean
> AC, though it's ***y rare to hear anyone refer to DC as RMS - but in
> truth 100W DC is EXACTLY the same as 100W RMS (or volts or current, for
> that matter).

Well, the solution is obvious Randy.

Split the resistor.

Put some of it ahead of the diode to deal with the AC current,
and the rest of it after the diode to deal with the DC current.

Remember to make sure your resistors are rated accordingly for
the "Type of service" and don't reverse them.

Jeff

--
"They that can give up essential liberty to obtain a little temporary
safety deserve neither liberty nor safety." Benjamin Franklin
"A life lived in fear is a life half lived."
Tara Morice as Fran, from the movie "Strictly Ballroom"

 
 
 

Diodes and dropping resistors

Post by Michael Bla » Wed, 13 Apr 2005 05:25:13



Quote:

>> If you were to restate your comment above as "because the AC current is
>> lower at that point than in the rectifier circuit", then you would be
>> correct.  But the RMS value IS THE SAME, (just most of the component is
>> DC rather than AC at that point - just as the major component at the
>> rectifiers is AC rather than DC). Remember RMS does NOT imply nor mean
>> AC, though it's ***y rare to hear anyone refer to DC as RMS - but in
>> truth 100W DC is EXACTLY the same as 100W RMS (or volts or current, for
>> that matter).

> Well, the solution is obvious Randy.

> Split the resistor.

> Put some of it ahead of the diode to deal with the AC current,
> and the rest of it after the diode to deal with the DC current.

> Remember to make sure your resistors are rated accordingly for
> the "Type of service" and don't reverse them.

Doesn't the first take care of the second?  Once you are using
two resistors, you are likely to use the same wattage resistors
as before, and now they will be seeing half the current than before.
And since you have two resistors instead of one, there will be less
concern about the voltage arcing around the resistor.

  Micahel

 
 
 

Diodes and dropping resistors

Post by Mark Robinso » Wed, 13 Apr 2005 05:24:14


Hi Randy,

Sorry, but your wrong here.  The RMS current on the DC load side of the cap
is most certainly not equal to the RMS current at the rectifier cathode.
The average current through the rectifier is equal to the average load
current, however.  Kirchhoff  applies here, but you are not considering the
substantial current flowing through the cap to ground due to the AC
component of the current waveform.  If you take all 3 currents into account
then: Input_current(t) = Cap_current(t) + Load_Current(t).  This satisfies
Kirchhoff.  Note it is not valid to simply add the RMS values of load and
cap current to obtain the input current.  Also, note that there is no such
thing as RMS watts.  That term was created by audio amp manufacturers and
has no technical meaning.  Its a shorthand way referring to a power
generated by an RMS current or voltage into a pure resistive load (the
waveform may be non-sinusoidal). Only current or voltage can be expressed in
RMS units.  Well, mathematically you can compute an RMS value for Power(t),
but the result has no meaning in the real world.

Mark



Quote:

> > Actually the drop for a 100 Ohm resistor directly in series with the
diode
> > would be considerably more than 4 volts because we are dealing with a
peak
> > current in the rectifier considerably greater than the 40ma DC load
> > current.

> Only on initial power up, then the peak current becomes inconsequential,
> and the RMS value is all that matters (as long as the peak voltage
> across the resistor doesn't exceed the voltage rating of the resistor).
> You can argue peak ***all you want - but that peak is over a (shorter
> and shorter) duty cycle. That's why RMS is all important.  Why? what IS
> RMS? -- and I don't mean it's mathematical derivation - what is RMS in
> the practical world?

> The AC equivalent of DC in the amount of energy converted - which in the
> case of a resistor - that product is heat. That's why resistors are
> rated in Watts. Peak current doesn't matter, peak voltage (to the point
> of flash-over, anyway) doesn't matter. Just heat. Watts RMS.

> And then he went on to say:

> > I think voltage dropping is better done after the filter capacitor
because
> > less heat is produced when the dropping resistor is after the filter
> > capacitor because the RMS current is lower at that point than in the
> > rectifier circuit.

> How so? If it is then you have leakage somewhere- either in reverse
> current in the rectifiers, leaky filter caps, or something - otherwise
> Kerchkoff has it wrong.  Power in has to equal power out - and unless
> you're generating a lot of loss in the rectifiers, filters, or such -
> the RMS power value at any given point in a power supply is going to be
> the same.

> If you were to restate your comment above as "because the AC current is
> lower at that point than in the rectifier circuit", then you would be
> correct.  But the RMS value IS THE SAME, (just most of the component is
> DC rather than AC at that point - just as the major component at the
> rectifiers is AC rather than DC). Remember RMS does NOT imply nor mean
> AC, though it's ***y rare to hear anyone refer to DC as RMS - but in
> truth 100W DC is EXACTLY the same as 100W RMS (or volts or current, for
> that matter).

> best regards...
> --
> randy guttery

> A Tender Tale - a page dedicated to those Ships and Crews
> so vital to the United States Silent Service:
> http://www.FoundCollection.com/

 
 
 

Diodes and dropping resistors

Post by Bill Sheppa » Wed, 13 Apr 2005 06:00:57


Perhaps I'm missing sumpin', but I can't see why it should make any
difference whatsoever whether the resistor or diode comes first, _since
there is no capacitor to ground from the resistor/diode junction_.

I would certainly put the resistor on the input side of the diode out of
pure convention, but still can't see why it should make any difference
at all. But then I ain't the brightest bulb on the tree.
                Bill(oc)

 
 
 

Diodes and dropping resistors

Post by Randy or Sherry Gutter » Wed, 13 Apr 2005 07:57:42


Quote:

> Hi Randy,

> Sorry, but your wrong here.  The RMS current on the DC load side of the cap
> is most certainly not equal to the RMS current at the rectifier cathode.

Where does it go, then - unless your cap is defective?

Quote:
> The average current through the rectifier is equal to the average load
> current, however.  

Yes, now you're back on the right track - though your distinction of
"average" is no more accurate than RMS - just a different reference
point of measurement.

Quote:
> Kirchhoff  applies here,

Glad you agree...

  but you are not considering the

Quote:
> substantial current flowing through the cap to ground due to the AC
> component of the current waveform.  

You're right -  because it is an artifact that has nothing to do with
TOTAL current through the supply (i.e. from source through rectifiers to
the load, but note also the disclaimer about less than wonderful caps).

Explanation:
During rectifier conduction - the current flows through the rectifier to
both the load - and to (re)charge the cap.  This appears to "add" to the
current through the rectifier (and indeed this "peak" current can be
quite high).

BUT - while the rectifier is cutoff - the capacitor discharges (part) of
it's  charge into the load - to "hold up it's end" of the supply - and
it just so happens that this "discharge" current is just what's needed
to supply the load - and winds up being equal to (and opposite in
direction) the charge current.

The net result is 0 current flowing - or "lost" -- through the cap.

This "artifact" or apparent AC current through the cap is called ripple
current - and does not (in a reasonably good cap) cause any additional
loss (or additional current).

(here's the flaky cap disclaimer):
Again - a leaky or high ESR cap is another matter - and the ONLY time
ripple current adds to the overall current of the rectifiers over and
above the load.

Quote:
> If you take all 3 currents into account
> then: Input_current(t) = Cap_current(t) + Load_Current(t).  This satisfies
> Kirchhoff.

Nope - not unless your cap is lossy - either by leaking or high ESR (in
which case it's loosing power as heat). See above.

Quote:
> Note it is not valid to simply add the RMS values of load and
> cap current to obtain the input current.  

What Cap current? Again - if your cap is leaking (or has high ESR)
replace the thing - ain't healthy. See above!

Quote:
> Also, note that there is no such
> thing as RMS watts.  

yeah - I'll grant you this one - oh - technically there is such a thing
but it's a mathematical curiosity (and a monster of calculus to compute).

Quote:
>  Only current or voltage can be expressed in
> RMS units.  Well, mathematically you can compute an RMS value for Power(t),
> but the result has no meaning in the real world.

Actually - as a mathematical curiosity - CurrentRMS X VoltageRMS =
Average power -- and NOT RMSPower.    And yes - you are correct that in
the real world - it has no practical meaning - other than (as you
pointed out as done by the audio amp PR people) to convey a sense of AC
equivalent power with a non sinusoidal waveform - at least in audio amps.

In RF amps (transmitters) - you can calculate it out to PEV - but that's
WAY beyond the scope of this discussion).

Best regards...
--
randy guttery

A Tender Tale - a page dedicated to those Ships and Crews
so vital to the United States Silent Service:
http://tendertale.com

 
 
 

Diodes and dropping resistors

Post by Randy or Sherry Gutter » Wed, 13 Apr 2005 07:59:16


Quote:

> Doesn't the first take care of the second?  Once you are using
> two resistors, you are likely to use the same wattage resistors
> as before, and now they will be seeing half the current than before.
> And since you have two resistors instead of one, there will be less
> concern about the voltage arcing around the resistor.

Jeff's tongue was "firmly in cheek" on that post...

best regards...
--
randy guttery

A Tender Tale - a page dedicated to those Ships and Crews
so vital to the United States Silent Service:
http://tendertale.com

 
 
 

Diodes and dropping resistors

Post by Randy or Sherry Gutter » Wed, 13 Apr 2005 08:00:21


Quote:

> Perhaps I'm missing sumpin', but I can't see why it should make any
> difference whatsoever whether the resistor or diode comes first, _since
> there is no capacitor to ground from the resistor/diode junction_.

> I would certainly put the resistor on the input side of the diode out of
> pure convention, but still can't see why it should make any difference
> at all. But then I ain't the brightest bulb on the tree.

You're not missing a thing - just some trying to "over think" a circuit...

best regards...
--
randy guttery

A Tender Tale - a page dedicated to those Ships and Crews
so vital to the United States Silent Service:
http://tendertale.com

 
 
 

Diodes and dropping resistors

Post by Mark Robinso » Wed, 13 Apr 2005 10:45:08


Hi Randy,

I went back and re-read the old posts to make sure I was not misreading
anything.  To be sure we are on the same page, let me re-state things.

1. A dropping resistor located ahead of the filter cap will see a larger RMS
current then a dropping resistor located after the filter cap.

2.  There is no difference in the effect of the dropping resistor if located
on the anode or cathode side of the diode (still in front of the filter cap
in this case).

As to where the current goes, you provided the answer.  The current flows
through the cap as ripple current.  As you point out, the average current
through the cap is zero, but the RMS current is significant.

I said:

Quote:
> > If you take all 3 currents into account
> > then: Input_current(t) = Cap_current(t) + Load_Current(t).  This
satisfies
> > Kirchhoff.

You replied:

Quote:
> Nope - not unless your cap is lossy - either by leaking or high ESR (in
> which case it's loosing power as heat). See above.

This is wrong.  Note the time variable in the above equation.  Kirchhoff
applies at any instant in time.  You need to look at the waveforms to see
law hold.  Note that RMS is a scalar value, so there is no time factor
involved.  If you are not very careful, you can get hosed using RMS or
average values.  For example,  Average Power is not equal to V_RMS * I_RMS
except in very special cases.  To get the correct average power, you need to
average the result of V(t) * I(t) = P(t) over a suitable interval.

I said:

Quote:
> > The average current through the rectifier is equal to the average load
> > current, however.

You replied:

Quote:
> Yes, now you're back on the right track - though your distinction of
> "average" is no more accurate than RMS - just a different reference
> point of measurement.

This is also incorrect.  There is no direct relationship between the average
value and RMS value of arbitrary waveforms.  Different waveforms have
different ratios between the average and RMS values.  This is referred to in
some circles as crest factor.  For narrow pulses, the crest factor, or ratio
between the RMS and average value is high.  For DC, the ratio is one. For
half wave rectified sinusoids, the ratio is 1.57.  For a 10% duty cycle
pulse, the ratio is 3.16.

Just to be sure I was not crazy, I ran a simple quick Spice simulation and
found the rectifier RMS current was almost 3X's the DC load current.  The
average rectifier current measured the same as the DC current at the load.

Mark