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[Reply #15 on:]

It does cause more ripple that's why I pointed out that the input filter had to be bigger and in this case since there is a linear regulator the added ripple makes no difference . Even with a 470uF instead if the 2200 uF min I had mentioned the ripple could Never be big enough to cause the regulator to even be close to it's dropout voltage. You will remember that the original issue was the reg shutting down because of overvoltage... So I listed several different workable strategies to fix that problem.

Doc

[Reply #16 on:]

As far as I understand a common 7805 has no 'automatic' shutdown due to input voltage being too high. There is automatic shutdown for output over current draw and for maximum heat dissipation, but over voltage of the input would just subject the chip to damage, not any kind of automatic response.

Lefty

[Reply #17 on:]

thanks  all for your help think i will pick it off the centre tap
just one more question as im here anybody know a good website to explain the correct way of working out the correct va transformer for your application?

[Reply #18 on:]

Calculate or Measure the current drawn and double it... Just for luck and multiply that by the voltage realizing that control/signal transformers come in standard values. A 24 V transformer, a 10 ohm resistor and a single diode with a 2200 uF - 4700 uF input filter is the simplest method to deal with 24V transformers... Because of the required high value of the input filter for a single diode (half wave) it is recommended to use a 5 to 10 ohm 'Surge' protection resistor in series with and before the diode to reduce the inrush current to the diode.
It is also recommended to use a diode of 3A forward current, typically a 1N5407 and the input filter after the diode.
This is for a supply for a 5V regulator and a low current load. This is from memory and I am certain of it's validity, It's just that right now I am too busy to go find the text and quote it here and for that I apologize...
However the loaded voltage should be about 11-13V which, while being a little high for heat considerations in the regulator should be OK.
The Best method was already mentioned and that was to use a switcher to begin with.
The methods I outlined earlier were only "Fixes" for the immediate issue which was the regulator shutting down for an over-voltage condition.
There was a comment about "Not Believing???" that a regulator would shut down... YES VERY MUCH SO...
Read the Data Sheet before you consider yourself enough an expert to quote opinions here.
The Shut down is done for several reasons primarily to control the device Power dissipation.
Ex. 1. 50 V in 5 V out @ 1 A = 5W load to the regulator and 45 Watts wasted as heat... 50W total device dissipation
Ex. 2. 35 V in 5 V out @ 1 A = 5W load to the regulator and 35 Watts wasted as heat... 40W total device dissipation
50W/40W = 1.25~ X increase in power dissipation for a 15V increase in supply voltage or simply by making the Max voltage 35 V instead of 50 V makes a 25% decrease in required power dissipation.
Under any circumstances the final shutdown is due to heat and limiting the input voltage to a 'reasonable level' increases the available current at high voltage inputs because the final control in the regulator is it's internal temperature... Not current as much as heat.
Internal to the regulator is a comparator that measures the supply voltage against a reference voltage and if high merely shuts down the drive to the pass transistor to prevent exceeding the total device dissipation.

Doc

[Reply #19 on:]

thankyou for that information
so if i have my boubled load at 1A at 24V i should have a 24VA transformer? sounds a bit high
could you do a quick drawing on what you describe to give me and others a better understanding
thanks

[Reply #20 on:]

All of what I wrote in my last post is I believe correct... for a device that needs 200 - 300 mA Max current from a 24VA transformer and yes you do need a 24VA transformer to supply 24 VAC @ 1A load current. A "Half Wave Rectifier will give you 1/2 (approx) of the 24V ac or about 12 Vdc. I suggested a large enough diode that the surge resistor might not be needed and that there would be 9 to 10 Vdc available for the regulator.
The best method (IMO) is a switcher and I had very good results with the part i mention next... I made about 100 of them (Not me I just did the design and PCB layout) but they were really quiet and trouble free.
 I built a lot of 24Vac based switchers in 2003 - 2006 I used an LT1170 because I needed about 3A for 10 - 15 seconds to power a 10W transmitter. It was an easy device to use... all the information you will need is in the data sheet. There are three varieties of this device, LT1170 - 1 - 2 for 5A @60 V max input voltage (LT1170 2.5 A (LT1171 and 1.25 A for the LT1172 I used the 5 lead TO-220 and the 5 Lead SMT package and all worked very well. This is the link to the data sheet.
http://www.stanford.edu/class/ee122/Parts_Info/datasheets/LT1170.pdf
an ac transformer and a bridge rectifier work very well with this device for a stable and adjustable buck mode power supply. If you need a lot of current. If you are just doing a small control device then my last post should have more than enough information. I would add that a Tranzorb should be included for transient protection on the secondary of the transformer to protect the diodes. I also included poly fuses to prevent fires in case of component malfunction. I also used tranzorbs on the rectifier output as these were used in Kansas on a Rural Power Grid where KV level spikes were common and some of long duration... 1/2 second spikes were measured... Usually though when the Power company did a "Load Drop", A frequent nightly occurrence as most of the big Agricultural companies used power at night when rates per KW were cheaper.

Doc 

[Reply #21 on:]

Whats a load drop?

[Reply #22 on:]

ok and if the current in secondary windings is 1A for example how can you calculate the current in the primary winding at 230V?

[Reply #23 on:]

Power is always conserved so power out equals power in, minus any losses
For simplicitys sake say 23v secondary drawing 1 amp, so 23w power used
230v/23w=.1 amp
In reality it will be a little bit more do to various losses