# Voltage Drop Across the circuit

I am providing a 5 V power supply to play with 3 Relays, they are 2 channel 5 V relays, when I turn on one bulb other relays do not get the sufficient voltage to operate. How would I prevent or overcome on such voltage drop?

If I do not get the sufficient voltage then other relays will not work, so I just want to keep the voltage above 4.5 V at all ends.

Here is the circuit,

Note :

1. I am taking readings from multi-meter at the ends shown in the image (bottom left corner).

2. To turn on any bulb we need to turn on the switch , in other words the IN pin of the relay needs LOW signal to make the bulb ON.

• Enlighten means: give (someone) greater knowledge and understanding about a subject or situation this is not so applicable to light bulbs. Better use: on / off – Bimpelrekkie Dec 22 '16 at 9:32
• Your input source seems a symmetric -6 - 0 - +6 V supply. Are you using that as 6 V or 12 V ? 6V will not be enough for that 7805, it needs 7 V or more to make a proper 5 V above a certain load. – Bimpelrekkie Dec 22 '16 at 9:39
• If you have a specification sheet for the relays check to see if you can drive them direct from 6V. – RoyC Dec 22 '16 at 10:42
• @FakeMoustache : thanks for the suggestion. i will edit it. : 6-0-6V double winding transformer i am using. its a 6Volt. – KarmaCoding Dec 22 '16 at 10:50
• 6-0-6V double winding transformer Hmm, transformers generally only work on AC but the 7805 and the relays need DC, so you also forgot to include to describe how you convert that AC to DC. You must describe your setup properly as not to leave us guessing. Also you appear to use a relay module instead of separate relays. That is also relevant, link to a datasheet or description of that relay board (no only a photo is not enough). – Bimpelrekkie Dec 22 '16 at 10:57

either you're putting 6v into the 7805 which is not enough input voltage, or it's overheating in which case you'll need a (bigger) heatsink.

see if you can get a cheap phone charger from somwhere and cut up a USB cable to get 5V from it.

• tried it ,, the same voltage drops i am facing. – KarmaCoding Dec 23 '16 at 9:31

You may have a superposition of two problems:

1. Vcc and GND wires too thin, too long or both. If you're using breadboard jumper wires or the like then you may get a significative voltage drop at high currents due to their resistance. This problem may persist even after you bypass the 7805 regulator and connect a 5v source directly to Vcc. Try using thicker, shorter wires.

2. Insufficient filtering after rectifier. From your schematic it seems that you're using the input capacitor of the 7805 regulator to filter the output of the rectifier. Check if the capacitor is big enough for that purpose.

3. Excess dropout in 7805. You seem to have ruled it out as the problem didn't go away when you bypassed the regulator connected a 5V source directly to Vcc. However, this potential problem may appear when you solve the one that's apparent now (inadequate wires). Keep an eye on it because, as you can see in the curve below, you should expect a 1.5-2.0 volts dropout. Try using a low dropout (LDO) regulator like MIC39101-5.0 instead.

• nice idea. will surely try this one. – KarmaCoding Feb 24 '17 at 10:37
• Do you have a pic of your assembly? – Enric Blanco Feb 24 '17 at 10:39
• nope. i am not from electronics background. whatever i could create i have uploaded in the post. – KarmaCoding Feb 24 '17 at 12:58
• I mean a photo. Your link tiikoni.com/tis/view/?id=aefb20f doesn't work anymore. – Enric Blanco Feb 24 '17 at 13:04
• here is the image of the power supply.. tiikoni.com/tis/view/?id=b715ae4 – KarmaCoding Mar 8 '17 at 7:26

You are attempting to draw more current from your supply than it can provide and still remain in regulation.

Fixing your supply might fix the problem. The circuit you show is a mess:

1. The two bottom diodes in each of the (intended, apparently) full wave bridges do nothing. Stop and actually think about it. They are two back to back diodes in series. One of them will always be reverse biased, and no current will flow thru them. You would get the same effect by simply removing them.
2. The third diode in each bridge that is connected between the DC output and ground also does nothing. This is the top right one in both bridges. Again, think about it. It is directly connected between the DC output and ground in reverse. It will always be reverse biased, so it won't conduct. You would get the same effect by simply removing it.
3. The remaining diode, the top left in both cases, makes a half-wave rectifier.
4. Two separate unregulated DC outputs from two separate rectifiers makes no sense in this case. You could use the two diodes that are actually doing something, together with the center tapped secondary to make a single full-wave rectified unregulated supply. That can feed as many 7805 regulators as its current can support.

You don't get more current by using multiple rectifiers. In this case, you get less because each of your unregulated supplies is only half-wave, and therefore has lower current capability.

5. You don't show the values of the caps around the 7805s, but a single cap on the input can't be right. You need a decent amount to store significant energy to ride out the half power line cycle between peaks (in the fixed design). That would take a electrolytic with current technology.

However, the regulators need their inputs to be low impedance at high frequencies for stability. That would take a not-electrolytic with current technology. Something like a 10 µF 20 V ceramic would do.

In other words, what you show obviously isn't right because there would need to be two caps on the input of the 7805.

6. If you want to make 3.3 V, it's simpler to do it directly from the unregulated supply. Using a 7805 intermediary does spread out the power dissipation, but there are other simpler ways to do that.

Even better would be to replace the bottom 5 V and 3.3 V regulators with a single buck switcher. That gets around the problem of getting rid of the heat.

Do some math. You say each end of the secondary puts out 6 V. That is presumably RMS, so 8.5 V peaks. Using just the two diodes to make a full wave bridge looses one diode drop in the rectifier. Figure 700 mV for that, so now you're left with 7.8 V peak out of the diodes.

Read the 7805 datasheet. See the minimum dropout figure, or maybe its shows as the minimum required input voltage. Some variants need up to 2.5 V headroom, so require a minimum of 7.5 V in to be able to keep the output regulated at 5 V.

There is very little room between the 7.8 V coming out of the rectifiers and the 7.5 V the 7805 might require. Just for laughs, let's see how large a capacitor it takes to provide 1 A while not dipping below 7.5 V. (1 A)(8.3 ms)/(300 mV) = 28 mF. That's a lot, but it could be done.

However, that is assuming the transformer still puts out 6 V under 1 A load, and that the diodes still only drop 700 mV.

A much better answer is to replace the 7805 regulator with a buck switcher. Those can work with less headroom, and can also work with higher input voltage without extra power dissipation. Now you could use the 12 V out of the ends of the secondary with a real 4-diode full wave bridge. That would cause too high of a voltage drop in the 7805s for them to dissipate as heat. The buck switcher, however, will be much more efficient, so it can use the extra voltage to give you a lot of headroom.