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

[Andy Fierman <andyfierman@xxxxxxxxxxxxxxx>]

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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