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I am trying to make a power supply of 5 V; as we all know it uses a bridge circuit, then capacitors and LM7805.

But I want to know how the value of capacitance is calculated. What is the formula for it?

If the formula is \$C = I_t / V\$

Then what will be the value of \$I\$ and \$V\$? What is \$t\$?

I have 220V, 60Hz main supply. What are the calculations?

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    \$\begingroup\$ Huh? How are we all supposed to know it's a "bridge circuit"? \$\endgroup\$ – Olin Lathrop Jul 23 '14 at 11:18
  • \$\begingroup\$ I think the OP means that it uses a bridge rectifier, but that still leaves the question of whether it is full or half wave. \$\endgroup\$ – JYelton Jul 23 '14 at 17:08
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Answer to this question depends on a few things:

First of all in addition to these components you would need a step down transformer in order to make the full-wave rectified output voltage of your bridge rectifier small enough. You can't feed the rectified 220 V, directly to LM7805, because LM7805 operates in the input ranges of 7 V to 20 V (and has a maximum input rating of 35 V).

If we assume that your step down transformer reduces the amplitude of 60 Hz sine wave from 220 V to 15 V, and if we assume that your 5 V power supply will need to output at most I_max = 1 A current, then we can start making some calculations. Now, as in this wikipedia article , your reservoir capacitor, which you will place after the bridge rectifier, will have V_max = 15 V on it, which is the amplitude of your sine wave. In the image:

figure1

you see that capacitor discharges during almost the whole period of half-wave rectified wave (in our case this discharge is caused by the I_max = 1 A load current going into LM7805). The discharge time of reservoir capacitor in the case of half-wave rectifier is T_discharge = T = (1/f) = (1/60 Hz) = 16.6 ms, however, notice that, in our case we have a more sophisticated rectifier (Diode bridge) which gives a full-wave rectified output. So, the discharge time will be T_discharge = T/2 = (1/2*f) = 8.3 ms.

Now, at the beginning of each discharge period our capacitor is charged up to V_max = 15 V. In order to prevent our capacitor voltage going below V_min = 7 V (which is the lowest input operating point for LM7805 voltage regulator) in the end of the discharge period, our capacitor value should be chosen with the equation:

C >= (I_max*T_discharge)/(V_beforedischarge-V_afterdischarge)

Using the values; V_beforedischarge = V_max = 15 V and V_afterdischarge = V_min = 7 V and I_max = 1 A and T_discharge = 8.3 ms, we can calculate that:

C_min = (1 A)*(8.3 ms)/(15 V - 7 V) = 1 mF. You can see that if you use a step down transformer which reduces the 220 V input into 20 V instead of 15 V and if your power supply will require at most I_max = 0.5 A current, you can use an even smaller capacitance with the value:

C_min = (0.5 A)*(8.3 ms)/(20 V - 7 V) = 0.32 mF.

Here

figure2

you can see an example design which uses LM7805 just like you are, and they picked a capacitor value of 0.47 mF, which is close to the values we calculated above.

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Capacitor value should be large enough that it can provide enough voltage(+2 volts means 7v for 7805) to the regulator IC, means voltage across capacitor should not go below 7v. I have found a article where Capacitance calculation has been explained well,it may be useful for others,

Capacitance calculation for 5v DC

Thanks

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