[Boatanchors] Amp Supply LK500-ZC Amplifier Question

[Jim Wiley]

Actually, there is another way, and in my opinion, better:

No filter capacitor at all, but instead 2 diodes, one connected as a 
"half-wave" rectifier, in the normal manner, with  the other across the 
relay coil, connected so as to absorb the "reverse EMF" generated by the 
collapsing magnetic field.

This configuration works because the current created by the collapsing 
field, shunted through the diode across the field, maintains the field 
until the next half cycle of the driving voltage arrives.

Because the half-wave rectifier generates almost 18 volts of peak DC 
from a 12.6 VAC source, or 28.3 volts DC peak from a 20 VAC source, 
either should be enough to activate a 24-volt (nominal) DC relay.

In a practical sense, it is helpful to remember that it takes much less 
magnetic field strength to hold a relay closed than is required to 
initiate closure.

Also, the 2-diode configuration is much less expensive (parts wise) than 
the diode plus electrolytic capacitor, and it  has the advantage of not 
having to worry about the capacitor drying out at some future date.

Parts budget:  2 diodes (1N4007) between 1.1 and 20 cents each, 
depending on initial quantity purchased, vs, 1 diode plus one capacitor: 
60 cents to perhaps a dollar or more.   Since many hams find purchases 
of  1N4007 diodes in 1000 quantities (about $11 worth) makes sense, the 
2-diode approach becomes a more practical alternative.

-  Jim, KL7CC



On 8/2/2015 7:26 PM, WA5CAB--- via Boatanchors wrote:
> I think both of you are forgetting something else.  20 V RMS can only yield
> 28 VDC from either a half wave or a full wave bridge rectifier (full wave
> will produce zero) under two circumstances.  Either (a) there is a filter
> capacitor of infinite capacitance that has been charging for an infinite time
> (relatively unlikely) or (b) there is no load on the circuit.  Once you put a
> reasonable load on it, the DC will droop, worse for the half wave than for
> the bridge.
>
> In a message dated 08/02/2015 19:56:31 PM Central Daylight Time,
> mark.k3msb at gmail.com writes:
>> I was wondering about that.   20 VAC is an RMS value,  so  20 * 1.414 =
>> 28
>> peak.
>>
>> If the filter cap is good, it may (should be?) be charging to 28V (or
>> close
>> to it) giving around 26 Volts for the relay.
>>
>> My error in my original analysis was forgetting about the filter
>> capacitor.
>>
>> Mark K3MSB
>>
>>
>>
>>
>> On Sun, Aug 2, 2015 at 8:48 PM, Rob Atkinson <ranchorobbo at gmail.com>
>> wrote:
>>
>>> On Sun, Aug 2, 2015 at 6:13 PM, Mark K3MSB <mark.k3msb at gmail.com> wrote:
>>>
>>>> The coil is fed from 20 VAC through a half-wave rectifier, so the
>>> measured
>>>> reading of 12 VDC at the input end  of the coil is in the ball park.
>>>>
>>> That doesn't seem to be in the ballpark to me at all.  Maybe if it were
>>> 20VCT.
>>>
>>> I'd expect 20 V. AC through a diode to give me pulses of DC peaking at
>>> around 28 v.
>>>
>>> I must be missing something.
>>>
>>> 73
>>>
>>> Rob
>>> K5UJ
>>>
> Robert & Susan Downs - Houston
> wa5cab dot com (Web Store)
> MVPA 9480
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