transformless power supply or explosives? :D

[grim]

i remember a good few years back that maplin had a kit for a sound to light disco machine, that was mains powered using a dropper resisitor. always seemed very iffy to me, especially as it was a kit, and kits have a habit of being assembled wrongly.

they brought out a mkII after a few years, with a transformer........


i know you don't want to use the dropper resistor either.

i know how i would do it, but that's just me. tx all the way.

 

[Nigel Goodwin]

i know how i would do it, but that's just me. tx all the way.

I fully agree! - if someone has to ask how to do it, they shouldn't even be attempting it!.

 

[mojat]

Hai there, I recently used4.5v from a normal power adaptor(which shows around 7v volts) when u measure and gave to 7805 and used it for ttl circuit using counters and it worked properly and the output from the 7805 was approx.4.85v. I even connected the adaptor directly to the circuit by changing the voltage to 6v and it worked quite well, but there was a slight raise in the current. I think u can try.

 

[ikalogic]

I fully agree! - if someone has to ask how to do it, they shouldn't even be attempting it!.

I don't fully agree - if i would have never attempted new/risky projects, i would be still playing with leds and resistors.

 

[ikalogic]

great idea.. i found in my junk box an old PALM charger.. it outputs 2A 6Vdc..

Can i feed the 6V into a 7805 to get clean 5Vdc, i mean, doesn't the 7805 require at least 7.5 Vdc input?

Can i attache 6V directly to a TTL uController? (it says 5.5 Vdc max in the AT89S52 datasheet... )

any simple solution to get rid of this extra 0.5 volt..?

A 1N4004 Diode in series will do the tric?

And i still don't know, will a diode drop the extra 0.5 volt from a 6V powersupply? is it a good solution?

 

[eng1]

You may consider a 5 V low drop-out regulator, instead.

 

[ikalogic]

You may consider a 5 V low drop-out regulator, instead.

examples? (7805?)

and why not use a diode to drop this extra 0.5V?? .. even if the Vd is more than 0.5.. its ok as long as the voltage drop is between 0.5 and 1.5 ...

 

[eng1]

LDO regulators have a drop-out voltage around 0.7 V. The 7805 requires around 3 V. If you browse the web, you'll find datasheets and applications of LDO regulators.

 

[eng1]

and why not use a diode to drop this extra 0.5V?? .. even if the Vd is more than 0.5.. its ok as long as the voltage drop is between 0.5 and 1.5 ...

If you planned to use a 7805, this means that you want a regulated output?
If you don't, use the diode, that will drop around 0.7 V, but this value depends on the current. What about the input voltage? is it noisy or clean?

 

[ericgibbs]

Hi,
The expensive, very large capacitors are in fact less than $3
for a 440vac 4mF metallised polyprop. Dimensions are:-
30mm Dia by 75mm long.

Good quality, high power resistors are more expensive.

I am puzzled that the previous claims of 'explosive' or
'blowing up' have suddenly disappeared ??. The colour of the cap I have
suggested is 'blue', perhaps that might be cause for concern for someone.??

I believe that its vital that some of the more 'experienced' members
read carefully what has been posted and do their homework before replying.
After all the more junior members are guided by your answers and they
deserve your informed response.

During the lifetime of an active posting, its understandable that
postings may drift away from the original topic. Thats not a bad thing, as
it opens up alternatives that we may not have considered.

Regards
EricG

 

[ikalogic]

If you planned to use a 7805, this means that you want a regulated output?
If you don't, use the diode, that will drop around 0.7 V, but this value depends on the current. What about the input voltage? is it noisy or clean?

no, the output of my supply is clean, i just need to get rid of some extra voltage.

 

[ikalogic]

If you don't, use the diode, that will drop around 0.7 V, but this value depends on the current.

I know, and in the datasheet it will vary from 0.6 to 1.3 depending on the current. and this is ok for me.

are there any other problems in using a diode to drop the voltage?

 

[nura100]

Hello IKA

YES, you can drop that extra 0.6 V using a 1N4004 series diode,

But first you need to Check the output Voltage of your Charger, without considerable Amount of loading , the O/P volatge may be higher than specified as its rated to supply 2Amps. you can connect an LED with Series Resistor to the O/P and check the Voltage.

Why don't you use a Cap in place of high Watt Resistor which wastes a lot of power.

i have seen one of the members giving you the Formula for deriving the Cap's Value, forget all those Lenghty Maths steps,

here i'll give you a simple, easy to remember formula for calculating the Cap's Value.

uF(Cap Value in Micro Farads) = mA( Required Current in Milliamps) divide by 75

For Example : if you need 50 mA ,
then Cap value will be = 50/75 i,e =0.66uF, 650 volts Rating to to be on safe side.

arun

 

[Sceadwian]

Using a diode to drop the voltage would probably work, but it's not going to filter it for you. The noise on a switcher output can be pretty bad, so the LDO is probably your best bet.

 

[on1aag]

5V / 100mA power supply.

Hi Ikalogic,

This is only an example how things could be done.
When no load is present, the 100mA will have to be dissipated by the
12V/5W zener diode, at full load the regulator will have to dissipate
about 7V multiplied by the maximum load current. So both require a
suitable heatsink. It would have been better to use a 10V zener diode
and a low drop regulator, but I had to use what was on hand.
The multimeter measures the voltage accross the zener diode and you
can see that is has dropped below it's zener voltage indicating that the
maximum load current has been reached.
I used six blue led's each with a 100 hm: resistor in series to load
the regulator. Total supply current is 6 times 19mA.
If Nigel sees this circuit I think he's gonna blow a gasket.

on1aag.