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7805 high input voltage problem.

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hi;

i am using 7805 and 7812 in my circuit. the DC input voltage out of Bridge and Capacitor is 40V DC but the maximum input voltage range is 35V for 78xx series.

when i connect voltage divider circuit to apply 9V at the input of 7805; The voltage divider circuit normally shown 9V but when i connect that node of voltage divider network to the input of 7805, the voltage dropped down to 3.1V :-( ... i repeatedly perform this experiment with many voltage divider networks consisting of different value combinations like 10k and 30k ... 1.8k and 5.6 k and 10M and 30M ohms but all the time it gave same problem.

after that i connect the 10V zener diode in place of voltage divider network.. but it also gave the same effect..... :-(

please help me i want to provide 8 to 10V at the input of 7805 and 15 to 20v at the input of 7812 ICs.. But i have
the out from bridge and capacitor is 40V????

please tell me the solution.

Thank u.
 
What's the current sourcing ability of the DC power supply you're using? You can't feed power to a regulator from the middle of a divider network like that you're putting a series resistance to the regulator which explains why the voltage drops, how are you connecting the zener diode?
 
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What's the current sourcing ability of the DC power supply you're using? You can't feed power to a regulator from the middle of a divider network like that you're putting a series resistance to the regulator which explains why the voltage drops, how are you connecting the zener diode?

current soursing ability is 5A, but my circuit may require not more than 1A.

i am connecting the components as follows,,,,

View attachment 61855


View attachment 61856
 
As I stated, you can't feed the input of a voltage regulator from a resistor divider like that, it's resistance will be in series with the regulator's input which limits current destroying the ability of he regulator to regulate. You could try tacking on a zener regulator beforehand to help get the max voltage in line but dropping from 40 to 5 volts is an incredible waste of power, you're going to wasting almost 90% of the power being used as pure heat.

If your 5V circuit draws 1 amp that will be 35watts of power lost as heat in the regulator.
 
As I stated, you can't feed the input of a voltage regulator from a resistor divider like that, it's resistance will be in series with the regulator's input which limits current destroying the ability of he regulator to regulate. You could try tacking on a zener regulator beforehand to help get the max voltage in line but dropping from 40 to 5 volts is an incredible waste of power, you're going to wasting almost 90% of the power being used as pure heat.

If your 5V circuit draws 1 amp that will be 35watts of power lost as heat in the regulator.
i have tried 10v zener with 10k resistance in series with it as i have post the immage in last post....
but the problem remains the same..... i think i should reduce the resistance.... plz guide me about that...
Thank you.
 
You have some very serious misunderstandings of how basic circuit components work.

Take your 40V source and figure out what the current will be through a 10k resistor directly connected to ground.

You're placing this in series with the regulator, no more current can ever go through that regulator than can go through the 10k resistor at the supply voltage. I hope that helps your understanding a little bit.

Just a suggestion, why not simply use a half wave bridge rectifier instead of a full wave? You'll increase the ripple but the voltage will be inline with what the 7805 needs and won't waste nearly as much power as heat.
 
Try adding a NPN transistor (D400 etc...) to the output of the Zener junction.

Base to Zener/Resister junction
Collector to 40V
Emitter driving the 7805.

If the D400 cannot withstand with 40V to its CE voltage use a higher rating NPN transistor.
 
Replace the transformer with a lower output voltage (say 15V RMS) one. Any other solution will need big heatsinks (which may cost nearly as much as a tranny :))
 
.....................

Just a suggestion, why not simply use a half wave bridge rectifier instead of a full wave? You'll increase the ripple but the voltage will be inline with what the 7805 needs and won't waste nearly as much power as heat.
Unless you use a low filter capacitance to give a large amount of ripple, the average DC voltage will be similar to that from a full-wave rectifier. It still charges to near the peak AC voltage at the peak of the sine wave input. It just does it at 1/2 the frequency.
 
Replace the transformer with a lower output voltage (say 15V RMS) one. Any other solution will need big heatsinks (which may cost nearly as much as a tranny :))

actually such a large transformer is the requirement of my project related to power electronics i am acctually using 30V, 20A transformer and these volts boosted up to nearly 40V after passing through 3300uF 50v capacitor.. all i want to do is to use the that output for small and low power part of that high power project. i don't want to use separate transformers and bridge and capacitor for that purpose as it will increase the cost of the circuit that will be very redundant.... i want low cost and effective solution ...

Can i use this circuit to reduce voltage i.e; to divide input voltage by 3 ???

View attachment 61871
 
It will divide 15V by 3, but that's not what you want. You said you needed up to 1A output current. Check the datasheet for the 741. It can provide only a few mA !
The lowest cost solution is probably to use a SMPS buck converter run from the 40V to get a 12V supply and use the buck converter to give you a 5V supply too (or else use the 12V to feed a 7805).
 
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Go to the store and buy one of these. Using that as an example of a DC / DC converter. The input being 36 to 76 VDC with 12 VDC out (regulated) at 2.5 Amps for the MGS304812 version. It also affords In / Out isolation. Then use your LM7805 or whatever to get 5 VDC. If you only want or need 5 VDC then get a 5 Volt output version. Units like this are quite common and versions from China carry lower prices. The unit I mention cost about $50 (USD) and I doubt materials, making the board and etc I could build one of discreet components for that price. Less current has a lower price also. So my suggestion would be to just buy a DC to DC converter ready to use.

Ron
 
i don't want to use separate transformers and bridge and capacitor for that purpose as it will increase the cost of the circuit that will be very redundant.... i want low cost and effective solution ...
The cost of the wasted power using a linear regulator to drop that much power will over the life of the project cost far more than the components to do it right.
 
...but the OP wants 12V and 5V. The 10V zener would need to be replaced with a 15V (at least) one.
The 2N3055 will have to dissipate up to 26W, so will require a massive heatsink.
 
Something like this should work.

i have tested your given circuit using BD135 npn transistor. i replaced 2k resistor with 2.2k and 12v zener diode and 1k in place of 100R and i get 11.8V at the output but when i connect a small load (PC power supply fan) the voltages dropped down to 11.2V which lead to an unregulated solution...... that is high loads can not be driven using this circuit.
but when i drived my circuit consist of ann adc controller and 16X4 lcd .............. BD135 blown out......... now i am thinking that LM 2576 can be a good solution...
 
You need to use a transistor that can supply 1 amp while dropping about 35 volts. That is about 35 watts. So it is a big one on a heatsink. Think how hot a light bulb gets.
 
yes i am thinking to use LM2576 switching regulator for that purpose. i will get 12V and also 5v from 7805 then safely. i will perform the experiment tomorrow and post the result here..
 
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