Voltage Regulator Drops Voltage too Much

[JohnKElectroTech]

Hi. I have 4 new AA batteries in series that result in about 6V. It passes through a 5V voltage regulator (IC3) and a few capacitors to remove high frequencies as shown in the attached image. When I attach the third electrode of the voltage electrode to ground, I smell something burning. Why is this? I remove the ground connection and when I remove the four resistors, the voltage at V1 comes close to 5V as I would expect. When I include the resistors, the voltage at V1 drops to 3V. Would you explain to me why and how I can avoid this? I thought that the resistors were very large compared to the battery's internal resistance, so V1 would not drop.



Thanks in advance for your help and patience with these beginner questions.

 

[MikeMl]

Get rid of the input diode. What are you doing with the "TURN-ON PIN"?

 

[Dx3]

All voltage regulators use up some of the voltage to do their regulating. There are "low dropout" regulators that only use half a volt for themselves. That might work for you.
I'm not sure what the third electrode of the voltage electrode is, but only one part in this drawing that has 3 electrodes is the voltage regulator. When you connect the third electrode of the voltage regulator to ground, the regulator delivers all the current it can deliver, to ground. That makes smoke because you have shorted the power to ground.

 

[Thunderchild]

at 6 V in you won't get proper regulation, chances are that you need at least 6.8/7 volts, what exactly is the circuit for ?, it does not really make any sense

 

[JohnKElectroTech]

Thanks for your quick replies.

Mike - what is a "turn-on pin"? The voltage regulator has three electrodes - input, output and the third that is shown attached to ground. If you're referring to the third, I've left it unattached - it produces a burning smell otherwise. I also tried removing the diode. No effect.

DX3 - I should have been more specific - the voltage device is a Seiko Low Dropout Regulator, S-812C50AY-B. Your explanation about the burning makes sense.

Thunderchild - I'll try a 9V battery. The circuit is part of a motion detector. I've only built part of it because I'm testing it piece by piece.

 

[Thunderchild]

that circuit is fine, you can remove the diode and make sure you have min 7 V, what is the other pin your referring to ? are there 3 or 4 pins ? if your smelling burning with the attached diagram then something is going very wrong

 

[JohnKElectroTech]

Thunderchild - I've attached a circuit image of the voltage regulator and there are three pins. I applied 5V to V_in, I measure voltage at V_out and V_ss is the third pin.

 

[Thunderchild]

Vss is earth and should be permanently connected to earth (negative) if connecting it causes burning you have a serious problem, check there are not accidental shorts

 

[Dx3]

I'm beginning to suspect that you have looked at a drawing for a different regulator and gotten the pins mixed up. The manufacturers have not been so kind as to make the pinouts the same from model to model. If the pin you think is Vss is really the output pin, that would explain everything.

 

[MikeMl]

Are you using this pin out?

 

[JohnKElectroTech]

Thanks, Mike, Dx3 and Thunderchild. I reordered what I connected to the pins and it worked. I'm at a loss as to what diagram I originally looked at.

 

[Thunderchild]

Yea I was thinking, only time I ever saw this was when i got a reg back to front and thats the only thing they are not protected against