Re: gEDA-user: OT: help needed; asymmetric load after rectifier seems to disrupt its working.

[Andy Fierman <andyfierman@xxxxxxxxxxxxxxx>]

Ooops,

Just missed the Undo Send window ...

Typo in (i):

"if the source has a peak to peak swing of x volts but a dc offset of y
then (neglecting the diode drops) vcc = x/2+y and vss = x/2-y."

:)

         Andy



On 19 June 2011 11:01, Andy Fierman <andyfierman@xxxxxxxxxxxxxxx> wrote:
> Rick is spot on.
>
> However, there are more things you need to consider:
>
> i) Does your signal source have a mean DC level of zero? Without C1,
> if the source has a peak to peak swing of x volts but a dc offset of y
> then (neglecting the diode drops) vcc = x+y and vss = x-y.
>
> If you have to remove a DC offset then you'll have to put a
> transformer between C1 and the recirfiers. Then C1 keeps DC out of the
> transformer primary and the transformer secondary provides the
> necessary dc path for the rectifiers to work as required.
>
> ii) I note that C1 and L1 form a series resonant circuit with a centre
> frequency at about 22.9kHz. So is your source really bandlimited 15kHz
> - 28kHz by the time it gets to your circuit or are you trying to do
> the bandpass filtering as part of your circuit? If the latter then you
> will have to keep C1, L1 but add the transformer as described in (i)
> above.
>
> The you will have to model that transformer to include at least the
> leakage inductance to get the bandpass response right.
>
> Such transformers are not difficult to design and source.
>
> iii) What is the source impedance? Does the 8 ohms represent all of
> your source impedance or is there more hidden in the source itself?
> You will need to allow for all of it to see how the rectified outputs
> drop and ripple increases with load current.
>
> iv) Don't forget that a SMPS represents a constant power or negative
> resistance load. As the input voltage drops the current it draws from
> the source increases. The actual behaviour of a real SMPS is
> complicated by any input undervoltage lockout and soft start features.
> This may or may not play well with your source.
>
> I'd like to make a general point here.
>
> This isn't a criticism but an important observation: when asking a
> question about how to do something, it saves everyone a lot of time,
> guesswork and blind alleys if the problem that is to be solved is
> clearly stated alongside whatever attempt at a solution that the
> specific question may be about.
>
> Essentially, include the design specification in the original question
> otherwise no-one knows the whole story so the question doesn't get a
> proper answer in a timely manner.
>
> Clearly in some instances the design spec may not be something that
> can be given openly but usually the part relevant to a question can be
> reframed so as to not give away too much. However, there has to be
> enough information so that the boundaries of the problem in question
> can be understood.
>
> This question is a classic example. Several people have discussed
> removing a part of the circuit that I now strongly suspect (C1 and L1)
> is an essential (if inappropriately implemented) part of the circuit
> because it wasn't clear what the overall function or scope of the
> circuit was.
>
> Cheers,
>
>          Andy.
>
> signality.co.uk
>
>
>
> On 18 June 2011 21:19, rickman <gnuarm.geda@xxxxxxxxx> wrote:
>> What is the purpose of C1 and L1?  If you want to filter anything, it should
>> be AFTER you rectify the signal to DC.  A series cap is going to remove low
>> frequencies... like DC which is attenuated very highly.  So much in fact
>> that you can't draw a DC signal through a capacitor.  That is why your
>> circuit is not working.
>>
>> If you remove C1 and L1 the circuit will work the way you want it to I
>> believe.   Also, with an input frequency of 15 kHz or higher, you won't be
>> needing 100 uF output filter capacitors for a light load.  How many mA  is
>> your load?  How much ripple can you allow?  Use those two values to
>> calculate the value of output filter capacitor you need.  Once I fix your
>> circuit by removing the input "filter" I measure 19.14 volts out and 38.2 mA
>> of current into a 500 ohm load.  Is that what you are shooting for?  The 100
>> uF cap gives around 10 mV of ripple.  With lighter loads or more ripple the
>> cap can be smaller.
>>
>> Rick
>>
>>
>> On 6/17/2011 4:44 AM, myken wrote:
>>>
>>> Yeap, it should be a very low power power supply. Vx is not important Vcc
>>> and Vss are.
>>> Vin can be anything from 15Khz to 28Khz so a transformer is not the most
>>> desired option.
>>> I have designed two SMPS for Vcc and Vss but there load to the rectifier
>>> are not the same, with the described result.
>>> I will try the options suggested in this list today.
>>> Robert.
>>>
>>> On 17/06/11 04:13, gene glick wrote:
>>>>
>>>> On 06/16/2011 02:30 PM, myken wrote:
>>>>>
>>>>> Hello all,
>>>>>
>>>>> I would appreciate some expert advice.
>>>>
>>>> Are you trying to make a low current power supply?
>>>>
>>>> I agree with DJ - the unequal loading on + and - cycle will average to
>>>> something other than zero (unequal capacitors, unequal diodes, etc) If Vx
>>>> must always be average zero - you'll need to do something else.
>>>>
>>>> If you can handle a little voltage drop, don't care what happens to Vx,
>>>> and don't mind adding a few parts, make a cheapo regulator with a zener and
>>>> BJT? (Or maybe use TL31 instead of zener)
>>>>
>>>> What about a small transformer, one winding on primary, center tapped on
>>>> secondary.  Add a diode and a cap for each leg - and there you go!
>>>>
>>>> Anyway, there's lots of ways to do this.  If regulated output is what you
>>>> want, a little more work is required.
>>>>
>>>>
>>>> gene
>>>>
>>>
>>>
>>>
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>>
>>
>>
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>


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