Regulator Circuit

Understanding How Voltage Regulators Work | Analog Devices

It's impossible to help without a circuit diagram.

I have tried with this circuit..

That is a wrong diagram (its too tiny to make real sense of) if that is a regulator chip between the Caps.

All reg chips that I have seen need a connection to ground.....

Even when correct, its only a single phase connected power supply!! How could it possibly sense 3 phases??????

In a previous post I told you how it could be simply (and correctly) done without using a reg chip. Did you understand my post?

I'm guessing that you have 3 of these circuits, one for each phase and what you are trying to do is to generate a TTL logic 1 (+5 v) to feed to some sort of a digital circuit.

The 7805 is a 3 terminal device. You show 2 connections. It should have an input, output, and ground (negative side). It won't work if not hooked up right. Look at the data sheet.

I think what you are seeing is residual voltage on the capacitors after removing the voltage to test the circuit. Get rid of the capacitors on the output side of the regulator and put in a bleeder resistor across the output.

Bharati S. Aravatagi:

Ground reference for the regulator is missing but that looks like a drawing error.

You have no discharge path for the output capacitors so if they are large enough then they can hold charge for a very long period and that is why to are getting voltage when there is no power at input.

Take the following steps:

1) Place a diode from output to input of the regulator to discharge the voltage to the input on power off. This is also required for the safety of the regulator.

2) Add some load resistor 1K ohms 1W in parallel to the capacitors. This will have RC time constant for discharging the output capacitor.

if you are deriving 9V from one phase of a 3 phase transformer then that is what you get. You will have to provide separate input phase sensing to control the output if you want it phase sensitive. Otherwise I do not understand the reference to PHASE.

Design a circuit that does what you want it to.

I have used bridge rectifier.

By "one phase" do you mean one side of a center-tapped secondary transformer? By "all phases" do you mean both ends of a center tapped secondary transformer? Are you attempting to feed the 7805 directly from the transformer or do you have a rectifier, such as a full wave or full wave bridge rectifier, converting the secondary from AC to DC? What is the schematic and components of the circuit?

Having never done it I don't know what would happen if you were driving a 7805 with AC. Perhaps smoke signals saying it can't.

Driving the 7805 with one side of the secondary of a center tapped transformer and rectifier will give you one half of the voltage of from both ends of the secondary up to the output voltage of the voltage regulator. for example 4v ct-end will give 3.2v out of the regulator (losses considered). Use end to end of the secondary winding and you are pushing 8v into the regulator which limits the output voltage to 5 volts.

Also, incomplete questions get incomplete answers. This answer is based on what I thought you were asking. Please resubmit your question with more details, components and schematic. We can all then give you better answers. Please don't tell my friends that I was assuming (ass / u / me) this. It could ruin my reputation.

Good Luck, Old Salt

I have used bridge before providing input to regulator. and three phase means 420v, 3 phase power supply. 2 inputs of transformer are connected to 2 phases ,lets say R phase and B phase . output of this regulator is given to phase reversal circuit. so when i remove either R or B phase, output of regulator should give me 0volts but instead its giving around 3.2v which is effecting my phase reversal circuit.

Bharati S. Aravatagi-

What you have illustrated is a single phase system with a full wave bridge rectifier supplying voltage to a 7805 voltage rectifier. The supply to the transformer(s) and the secondary voltage(s) are what determine the voltage supply to the voltage regulator.

If by stating that you remove one phase of the transformer you really mean that you remove one wire from the transformer all that does is shut down the supply of voltage to your rectifier. Removing one wire from the secondary will also have the same effect.

I don't know why you state 420v in post #17 but 9 volts supply in your original post. With a supply of 9volts from the transformer, if one terminal of the bridge rectifier is disconnected you now have a half-wave rectifier supplying voltage during only half of the voltage during the cycle. This reduces you effective source of dc supply to 9volts*50% = 4.5 volts average. The 7805 has an inherent loss of voltage across it due to the need for the input voltage to be greater than the output voltage. This is then represented by ( 9 volts input to the rectifier) * (0.5 due to only using half of the full wave rectifier as a half-wave rectifier) - ( 1.3 volts internal loss within 7805 ) = output voltage of 3.2 volts.

You would have been able to get this answer if the question was presented with full and complete information with a schematic of what you are attempting to do.

Good Question ---> Good Answer

Poor Question ---> Poor Answer

Remember, we can only work with what you give us.

Good Luck, Old Salt

Post #20 should read "What you have illustrated is a single phase system with a full wave bridge rectifier supplying voltage to a 7805 voltage regulator."

Sorry about the keyboarding error. Must have been brain flatulence!

Good Luck, Old Salt

Me, too! Took me forever to figure out what error you were talking about. Thanks for clarifying it.

You cannot do this in this manner.

7805 is a already a designed regulator IC so what is that you have designed? Tell to everyone that part here to let other understand of what is the wrong part in the design.

9V from a "transformer"?

Which one of these are you talking about?....

OR THIS......

Second row, first item.

LOL!!

top right corner

The truth of the matter is that you simply do not have a strong enough grasp on electronics, if, as in this example, of not understanding what is needed to be produced to allow the people fron CR4 to give you proper help!!

You copied a circuit, you did not design it. It was also apparently (lack of proper infos) the wrong circuit!!!

If, as I believe, you have a 3 phase output and you need to identify when a phase is missing, there is a very simple way to do with without using a regulator or even any "proper" electronics. (for proper read chips/active components for example.)

Assuming (because you are simply not clever enough with electrics/electronics to provide us with proper information) that you have 3 x 9 volt transformers output, if any single phase drops out, that 9 volt output will be missing. If you only have one transformer, buy and install 2 more, so that each phase has a 9 volt output.

DO NOT CONNECT THE OUTPUTS TOGETHER, EVER!!

Then all you need is 3 relays, each one attached to a single (one of the three!) 9 volt output, via a full wave rectifier. Note which output is + or -.

When all three phases are available, all three relays are activated, closing at least one contact (depending upon relay design of course!) on each phase.

Then you need a circuit that is switched on via all three contacts (serially), that causes say a LED to glow (which could be supplied by any of the three outputs) to show everything is OK. Feed the LED through a resistor to only allow correctly biased DC though the LED.

If any contact is open (phase missing) this breaks the serial circuit and the single LED does not glow.

3 more LEDs, one on each phase after the rectifier (do not forget the resistor), will also show which phase is actually missing.

It would even be possible to use the three relay contacts to switch say a contactor off if needed (or better said, to not allow it to go on!), if a phase is missing.

I have only covered this in a very simple fashion, just to give you the idea. You will notice that no regulator is needed......

Details of rectifier type, relay coil values, resistor values etc., you should work out for yourself!!!! If possible!!

DO REMEMBER THAT ANY WORK WITH MAINS VOLTAGE IS DANGEROUS, IF YOU ARE NOT PROPERLY TRAINED, GET A TRAINED/KNOWLEDGEABLE MENTOR TO GUIDE YOU FURTHER.

I dont need to do this much of circuitary to know about which phase is missing because that is not what I requir. I intentionally removed one phase (I know which i am removing so phase sensing circuit is not required) to varify regulator circuit. please suggest me if you have any idea about regulator circuit(not phase sensing circuit)

Tell us again what you are trying to do, your previous attempt was completely misleading.

Use short sentences only.

it's unclear if you're designing a commercial product or if it's a school task to introduce sharing loads on 3-phase systems

the component names and voltages are arbitary

1) what you should gain/design?

2)what of it generates a headace?

From your cryptic question I conclude you complain that the regulated output does not go immediately to zero when you stop applying input voltage. That is possibly happening because somewhere in your circuit a capacitor stores some charge. If you want immediate voltage cut-off, put a 9v coil relay fed from ac input to disconnect output when AC is off. If you don't need that fast disconnection, just use a bleeder resistor at regulated output. (This last one should have been there already to provide minimum regulator load) S.M.

Bharati S. Aravatagi:

If you are getting 3.2V out of nothing connected as power source to your circuit then there is some serious problem somewhere.
Can you draw your circuit in paint brush or something that I can see and tell you something about this strange thing?
Are you using three phase transformer or three transformers?