Qaisar Azeemi
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You need to show your circuit to fix the problem. If it is simply a resistive voltage divider then that is a poor way of doing it. It will be very inefficient and waste a lot of power, depending on how much current you need. Mind you, a linear regulator from 40V down to 5V is never going to be very efficient!
I guess you have forgotten the current required by the regulator even when it is doing nothing - around 5mA or more. How much current do you need from the 5V supply?
A simple circuit like this would do. It would also help you move some of the power dissipation away from the 7805.
Keith.
Dear Qaisar,
First of all try to get some basics and read ohms law... please don't take it otherwise as i have no intention to offend you but to bring to your notice what basics you are missing .. actually voltage dividers are mainly used to scale down a high level voltage to some small voltage so that it can be sampled .. usually in voltage dividers the resistors used are several hundreds or kilo ohms in value... so the current the divider can supply is very low..so if you are going to load this output from divider then definitely voltages will drop to a voltage which is not at per the calculated voltage divider output voltages ( as per ohms law you can see that voltage drop across a resistor in series is directly proportional to current drawn through the resistor ) ...
SOLUTION :- here you can use a raw pre regulator to step down the input voltages to a level under 7805 permissible input voltage specification.. try using a 180- 330 ohm 2 watt resistor and a 12 volts 1 watt zener at input of 7805.... proposed solution is what we uses in many of our designs.. hope that helps.
Best Regards,
Saifullah Khalid
You can use also LM2575-5 or LM2576-5 Switching Regulators :
View attachment 70349
View attachment 70350
can it provide two outputs at a time.. i need 5V and 12V... or i have to use separate regulators for that???
and what is the formula to select program resistor??
Looking at your circuits you need to go back to basics and check your calculations.
Your circuit with a zener has 5.6 ohms to drop 40 down to 10V - that will take 5.3A and something will go up in smoke.
Your circuit using dropper resistors has 5.6 ohms & 1.8k and says 9V output but in reality the 5.6 ohms will drop almost nothing and you will still have nearly 40V on the 7805 input.
I am not 100% certain what you were trying to do with the transistor circuit but if you were driving the base with 5V then you will never get more than 5V out of the emitter no matter what the collector voltage is - that is the point of that sort of circuit.
Keith.
Qaisar
Just use separate regulator like LM2575-5 and LM2575-12 or LM2576-5-12,....
With that your zener or resistor you will make just good source of heating element for colder days with smoke, nothing more.
above zener + resistor combination figure should work ok.. try 100/220/330 ohm 2 watt resistors (for R1) and zener (D1) can be any between 12-14 volts zener with at least one watt rating...
My advice use some glasses to protect eyes when experimenting with this circuit.
As I pointed out earlier, it would take 5.3A, assuming it worked long enough for you to see it. That is 159W into the resistor and 53W into the 1W zener diode.
With 5.6k instead of 5.6 ohms ... again, see my earlier posts.
Keith.
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