Electronics > Projects, Designs, and Technical Stuff
Linear power supply advice
<< < (3/4) > >>
not1xor1:

--- Quote from: tunk on January 23, 2020, 05:38:44 pm ---This setup should give you around 11-12Vdc after rectification and smoothing.
It may be better with a 7.5Vac or maybe 6Vac transformer.

--- End quote ---

???
it should be:
9V * sqrt(2) - (2 diodes drop at around 20A) - (max ripple)

if we are lucky and the AC line is not much below the nominal voltage, since the transformer is 50VA and should provide a bit more than 9V we get:

13 - 1.8 - 4 = 7.2V
RogerThat:
not1xor1, thanks for your input.

Trying to learn here, how do you arrive at 4v? 'With 8800µF caps the peak ripple voltage should be around 4V.'

At the moment I'm looking at MC7805 for regulation and it states 2v drop...very little head room if the unregulated is around 7-7.5.
tunk:
I stand corrected - what not1xor1 is saying is right.
You could do a web search for "rectifier ripple calculator", e.g.:
https://www.changpuak.ch/electronics/power_supply_design.php
https://www.fxsolver.com/browse/formulas/Ripple+voltage+%28full-wave+rectifier%29%29
mariush:
When using a bridge rectifier  :

V dc peak = sqrt(2) x Vac - 2 x Vdiode  (Vdiode = voltage drop in a diode inside the bridge rectifier at the current desired)

Idc = ~ 0.62 x Iac  (0.62 is an approximation that usually works well for 50-200VA range transformers)

ex a 9v AC 50VA transformer will have :
Iac = 50 VA / 9 = 5.5 A 
Idc  = ~ 0.62 x 5.5A = 3.41A 

Vdc peak = 1.414 x  9 - 2 x 0.8v =  ~12.75v - 1.6v = ~ 11.15v

(so 11.15v x 3.41A = 38w .. maybe a bit more ... 0.8v x 2 x 4A = 6.5w lost in the bridge rectifier)

Keep in mind that most transformers (but esp. the low VA ones) will have a higher output voltage at low load... for example at  50-100mA the 50VA transformer may output 10..11v AC instead of 9v AC
Also keep in mind that your AC voltage may fluctuate depending on network load, time of day etc ... in a 230v country, at 2-3 AM I often see 245v AC at my outlet. So, account for these fluctuations on the primary side, which will affect the output on the secondary side.
For example, in the example above the formula says 10.75v, but in all further calculations it would be smart to consider 10v or 10.25v as minimum voltage, and around 12-14v as peak DC voltage (+1-2v at low load, ~+1v if your AC input is high)

Not applicable to you, as it would be too expensive (this would be useful for 20A+ rectification and higher voltages) but it's worth mentioning:
 
Instead of a bridge rectifier, you can use ideal diode controller and four mosfets : https://www.analog.com/media/en/technical-documentation/data-sheets/4320fb.pdf
This way you have less heat produced.. in your example, at 4A of current, the bridge rectifier would produce approx. 2 x 1v x 4A = 8w of heat.
The ideal diode controller would cause a voltage drop of a few tenths of a volt and maybe just 1w of heat (it really depends on the mosfets chosen)

Anyway... the size of capacitor can be estimated with formula :  Capacitance (in Farads) =  Current / [ 2 x AC frequency x (Vdc peak - Vdc min)]

So for example, let's say your "safe" Vdc peak is 10v and Vdc min is 7v (to have 2v above the linear regulator's output voltage) and you're in a country with 60 Hz mains frequency

C = 4A / [2 x 60 x (10-7)] = 4 / 360 = 0.011111 Farads or 11000 uF ... that's the minimum estimated.  So, I'd say 2 x 6800uF 25v caps in parallel would work, or 3 4700uF capacitors.

You can use more capacitance, but too much will cause a very high current when you plug the power supply (capacitors act as black holes sucking loads of energy when they're empty) and that can cause the fuse in front of your transformer to blow up, if you don't calculate the right value for it.

There are better linear regulators than the classic 7805, for example LM1084 is a good example: https://www.digikey.com/products/en?keywords=lm1084


It's also available in packages easy to heatsink, and also adjustable (use 2 resistors - or resistor + potentiometer - to configure output voltage): https://www.digikey.com/product-detail/en/texas-instruments/LM1084IT-ADJ-NOPB/LM1084IT-ADJ-NOPB-ND/363557
See the datasheet on the page for example circuits and detailed description of what components to use.
 
 
not1xor1:

--- Quote from: mariush on January 23, 2020, 11:45:04 pm ---[...]
Anyway... the size of capacitor can be estimated with formula :  Capacitance (in Farads) =  Current / [ 2 x AC frequency x (Vdc peak - Vdc min)]

--- End quote ---

2 x AC frequency is for full wave rectification otherwise (half wave) it is just x AC frequency
that comes from Vpeak=f(C) => Vripple = I / (f * C) where f is the frequency of the rectified wave (double for full wave rectification)
for various reasons the real ripple is a bit less than the theoretical one so even with 8800µF you would get less than 4.5V...


--- Quote ---You can use more capacitance, but too much will cause a very high current when you plug the power supply (capacitors act as black holes sucking loads of energy when they're empty) and that can cause the fuse in front of your transformer to blow up, if you don't calculate the right value for it.

--- End quote ---

AFAIK the problem with larger capacitance is a decrease in the power factor, with larger and shorter current peaks

BTW Rod Elliot wrote lots of useful information on his site... (about this and other subjects). I think that RogerThat may find those quite useful.
Navigation
Message Index
Next page
Previous page
There was an error while thanking
Thanking...

Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod