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schematic

simulate this circuit – Schematic created using CircuitLab

Please help.... I have an input of 7.5V but when it goes out of the circuit the voltage drops in 2.5V but i want 5V of output voltage thats why i borrow 7805 transistor, what is the problem in my circuit. It will be a big help if you help me so please.. thank you..

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closed as unclear what you're asking by Leon Heller, Daniel Grillo, Voltage Spike, Armandas, pipe Sep 24 '16 at 10:44

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    \$\begingroup\$ If your input comes from a battery like in the schematic drop the bridge rectifier (you might want to use the bridge rectifier symbol for it). 7805 is not a transistor but a linear voltage regulator. \$\endgroup\$ – Arsenal Sep 23 '16 at 13:45
  • \$\begingroup\$ JohnGaleViray - There is not enough explanation in your question (and may be an X-Y problem) e.g. (a) What type of battery is your 7.5V battery? (b) Why do you then say (in a later comment) that it might be only 5V? You need to explain much more about your power source and what its maximum & minimum voltages are. (c) What is attached to the 5V output and what current does it require? (d) Why are you using a bridge rectifier circuit, after the battery? (e) Why are you using 2 capacitors (C1 & C2) in series? (f) What is the purpose of Q2? (g) Overall, what exactly are you trying to do? \$\endgroup\$ – SamGibson Sep 23 '16 at 14:22
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A 7805 voltage regulator needs at least 7V into the input in order to regulate correctly. If you are feeding the bridge rectifier with 7.5 volts then, due to diode volt drops (about 1 volt on light load and more on heavier loads) the input voltage to the 7805 will be below 7 volts. That's your main problem as I see it.

Here's what the diode volt drop looks like for a 1N4148 and please note that this will be similar on any silicon diode: -

enter image description here

A 7805 takes about 4 mA unloaded so, as you can see, the diode forward volt drop will be about 0.7 volts. Because there are two diodes conducting in series in a bridge, the total volt drop will be at least 1.4 volts and suddenly your 7.5 volt battery is 6.1 volts when connected to the 7805. If the loading on the 7805 causes a current of 100 mA to be taken, the diode volt drop will be 2x ~1V = 2V and your supply voltage to the 7805 is only 5.5 volts.

Maybe try searching for a low drop-out voltage regulator - you should be able to find one that works down to an input voltage of about 5.5 volts or maybe a shade lower. If your load current is in the order of 1A then you will likely have to use schottky diodes and a low drop-out regulator.

You should also use decoupling capacitors on both the input and output lines of the regulator - read the data sheet.

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  • \$\begingroup\$ what if my input voltage is only 5V and it need to pass through a circuit without dropping any volts what must be done? \$\endgroup\$ – John Gale Viray Sep 23 '16 at 13:38
  • \$\begingroup\$ That is too general/loose to make a sensible answer. Dropping zero volts generally means taking zero current. \$\endgroup\$ – Andy aka Sep 23 '16 at 13:42
  • \$\begingroup\$ But is there a way to make a circuit that will not drop a voltage if there is show me please.. \$\endgroup\$ – John Gale Viray Sep 23 '16 at 13:49
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    \$\begingroup\$ Listen - this question is just repeated - I don't understand what you mean and I can't comment on something that I clearly don't comprehend. \$\endgroup\$ – Andy aka Sep 23 '16 at 13:51
  • \$\begingroup\$ ANY linear voltage regulator WILL drop some voltage. The minimum voltage drop may be called "headroom" or "drop-out voltage" in the datasheet. The headroom for th 78xx regulators is about 2 volts, but some Low Drop Out regulators have a dropout voltage of less than 0.5 volt. \$\endgroup\$ – Peter Bennett Sep 23 '16 at 17:10

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