Voltage Regulator

[2camjohn]

I have a L7806CV voltage regulator linked to a battery which outputs 6.7V

The L7806CV is only giving me 5.3V


Is it because my input voltage is too low?

Is there any way I can fix this?

 

[Exo]

Yes, a 78XX regulator always needs a couple of volts extra at the input...
What are you trying to power with it? perhaps it can be powered from the battery directly.

 

[2camjohn]

I am powering a 6V solenoid from a 6V battery.

But at full charge the battery gives out 6.8V, im worried that might be too much. 5.3v is definately too low.

Are there any voltage regulators which will provide 6V from a 6.8V supply?

Thanks

 

[Exo]

You could put a diode in series with your solenoid...
There will be a 0.6V voltage drop across the diode, leaving about 6V

 

[2camjohn]

That would work when the battery has just been charged,

but the battery goes down as low as 6V when empty. This would put me back on 5.3V

 

[2camjohn]

Would replacing the voltage regulator with a 6V zener diode be any better?

 

[Styx]

a stupid question but why cant you run the solenoid directly from the 6.7V. It must have some tolerance on it

 

[2camjohn]

I have been doing that.

But 6.8V instead of 6 is an increase of over 10%. I dont want to break my solenoind as its quite expensive to replace.

 

[Styx]

The solenoid is not going to have a problem with 10% extra voltage. If anything it might have a problem with the 10% extra current.

However, you are running it from a battery so the current level isnt going to be that high. anyway 10% tolerance isnt that much, resistor have upto 5% tolerance on their value thus 5% tolerance on the current through them.

If you are really that concerned you have 2 options

1) Use zener diodes. There is avaible 3V zeners so 2x in series will give you 6V, but you will want to put a burn resistor in series to limit the current so you will have to use a slightly lower value zener. next is a 2.7 not good enough.
However as the voltage of the battery drops the current through the burn resistor will also drop and there will be a voltage where the zeners will no longer hold the 6V since they need a minimum reverse-current to establish a zener voltage.

2) forget about the zeners, measure the imedance of your solenoid and add in series with it a resistor with a value 10% that of the solenoid.
However as the battery voltage drops...

 

[andrew2022]

connect to batterys together to give a higher voltage and use a regulator

 

[laroche73]

Solenoid

I doubt running the solenoid at 6.8V will do any harm. It's a coil of wire, and will heat up a little more, but most likely doesn't warrant the concern you're having. A semiconductor device could be a different matter, though many of those are designed with a +/- 10% supply tolerence. If you're still concerned, you could try using a low-dropout (LDO) regulator like this one

**broken link removed**

It has a typical loss of 280mV @1A. Many LDO regulators are designed to maintain the input voltage (with a small loss & no regulation) at their output when it falls below the minimum input LDO voltage.

There was a design in EDN (design notes) some years ago that used an external PNP pass transistor along with a low-current LDO regulator to achieve a 50mV dropout. Again, it's probably unnecessary to use a regulator at all.

 

[2camjohn]

It does seem simple just to forget the regulator, and as you say it wont break my solenoid, which is good.


But my circuit is designed to be left connected to the battery as it is being charged, so I need to protect my circuit (a pic and the 6v noid) from damage as the battery is given up to 12V during fast charge.


If i run the pic off the regulator, but the solenoid off the battery it is allright to put 12v across the battery as long as i dont turn my solenoid on..right?

 

[Nigel Goodwin]

It does seem simple just to forget the regulator, and as you say it wont break my solenoid, which is good.


But my circuit is designed to be left connected to the battery as it is being charged, so I need to protect my circuit (a pic and the 6v noid) from damage as the battery is given up to 12V during fast charge.


If i run the pic off the regulator, but the solenoid off the battery it is allright to put 12v across the battery as long as i dont turn my solenoid on..right?

Yes that's fine, but the voltage across the 6V battery won't reach anywhere near 12V, you charge a battery via some kind of current limiting device - this will effectively have 12V going in, and 6V coming out (actually a bit more than 6V as you charge the battery, it might reach 7.5V perhaps?).

 

[2camjohn]

It does seem simple just to forget the regulator, and as you say it wont break my solenoid, which is good.


But my circuit is designed to be left connected to the battery as it is being charged, so I need to protect my circuit (a pic and the 6v noid) from damage as the battery is given up to 12V during fast charge.


If i run the pic off the regulator, but the solenoid off the battery it is allright to put 12v across the battery as long as i dont turn my solenoid on..right?

Yes that's fine, but the voltage across the 6V battery won't reach anywhere near 12V, you charge a battery via some kind of current limiting device - this will effectively have 12V going in, and 6V coming out (actually a bit more than 6V as you charge the battery, it might reach 7.5V perhaps?).

It does get quite high as it has some kind of fast charge initially. This is from the existing charger circuit which I havent measured.

When I charged it without a regulator I had to remove the pic from its socket as it was getting quite hot.



I was just woried that if the voltage regulator sank all the current which is heading for the battery then it wouldnt charge properly.

 

[Nigel Goodwin]

It does get quite high as it has some kind of fast charge initially. This is from the existing charger circuit which I havent measured.

If the voltage across a 6V battery is reaching 12V on charge, then it looks like the battery is probably knackered :lol:

What sort of battery is it?.

 

[2camjohn]

NiMH.

If it is broken then its not showing, it lasts just as long as it did when brand new.

12V is probably wrong, it was enough to get my pic very hot! I will measure it next time I need to charge the battery.

Can you tell me if putting a battery in series with a voltage regulator or zener diode will impede the ability to charge the circuit?
In my understanding a zener works by breaking down at the zener voltage, allowing current to flow thus stabilising the voltage, so wont all the excess current go though the zener instead of into the battery?


Thanks for your time
John

 

[Nigel Goodwin]

Can you tell me if putting a battery in series with a voltage regulator or zener diode will impede the ability to charge the circuit?
In my understanding a zener works by breaking down at the zener voltage, allowing current to flow thus stabilising the voltage, so wont all the excess current go though the zener instead of into the battery?

Try posting a drawing of the circuit you mean, I'm not very clear on what you are suggesting.

 

[2camjohn]

Sorry for not being clear. I am trying :?

**broken link removed**

Sorry for my artwork too!


As the voltage from the charger increases wont the excess current go through the zener diode instead of into the battery?

 

[Nigel Goodwin]

Sorry for not being clear. I am trying :?

**broken link removed**

Sorry for my artwork too!


As the voltage from the charger increases wont the excess current go through the zener diode instead of into the battery?

That's a simple zener voltage regulator, it will regulate the voltage to the electronics, but won't affect the battery charging at all.

 

[2camjohn]

Thanks Nigel I have one more question:

I have moved the + wire from the solenoid to the +v of the battery instead of the output of the voltage regulator, I have not moved the protection diode.

I went from this

**broken link removed**

to this

**broken link removed**

Am i risking blowing the transistor? Cos it seems so far so good.