Simple circuitry question
 

Ok so I have a circuit consisting of three components, their +'s and -'s connected to jumper strips so that I can connect the terminals of the battery to each strip. The 12V battery puts out ~12.6 volts when hooked up but as is drains, it drops slightly and the output of the components have to be recalibrated (output is dependent on input voltage). To remedy this I got a small 12v regulator chip that when hooked up to the battery alone puts out 11.65V which is fine (my thinking is that the circuit will get a constant 11.65v until the battery drops to below 11.65v). My problem is when I connect the regulator between the battery and the jumper strips with the circuit ground connected to the regulator ground and that connected to the neg battery terminal, I no longer get 11.65v but instead 10.7V at the jumper strips.

Why would I get the same voltage at the terminals and entrance to the circuit with the battery alone, but a drop in voltage (11.65v to 10.7v) at the output of the regulator as soon as the circuit is connected?

Did that make sense? I'll snap a pic if I need to. Thanks

There are a few things that come to mind:
• What type of a voltmeter is being used? If this is a low amperage circuit and the voltmeter requires a healthy amount of wattage to work, then the overall voltage will go down because the voltmeter is sucking out the power.
• Considered a larger battery? If space is not a problem, then a larger battery may be better able to maintain the desired voltage over time.
• What type of battery is being used? A deep cycle battery will maintain maximum voltage over a longer period of time than a starter battery.
• What is the application?
• Attaching another battery in parallel may be a thought. This is, again, based on the idea that space is not an issue.

Pic.
What are you trying to do?
Does the 12v reg require any support circuitry?

Voltemeter is a craftsman and althought the amps are low (each component drawing a few hundred mA I believe) I dont think thats the problem bc it read correctly when i did not have the regulator in place... 12.65 at the terminals and 12.65 in the circuit.

Battery is 12v 7Ah "alarm type" (?) battery which should be fine considering Im only drawing less than an amp total and there will also be a solar charger connected at all times. The battery is brand new and is fully charged and drops is voltage VERY slowly over a few hrs.

Il snap a pic, give me a sec.

In this picture, 12.6v going into the circuit with no voltage drop from the battery.

http://forums.pelicanparts.com/uploa...1227136682.jpg


Here the regulator is connected. When the short wires connecting it to the strips are not hooked up, the output is 11.6V. As soon as those wires are connected, output from regulator drops to 10.7v.


http://forums.pelicanparts.com/uploa...1227136777.jpg

I would believe that the circuit load could drop the voltage but how come that doesnt happen with the battery alone. Im not very good with circuitry so theres something Im not understanding correctly Im sure.

The voltage difference is the ohmic drop caused by the load of all the components connected to the battery. That is normal. The battery draws more current when connected to something than to nothing. The problem is that it draws the voltage lower than what your regulator can regulate, so I would use a regulator with a lower limit, like 10V, if that is possible. You can also double the battery (same model, in parallel), it will decrease your ohmic drop by half.

Aurel

I think I understand but why is there is no ohmic drop WITHOUT the regulator? Thats whats confusing me. The circuit components can have variable voltages (one is 7-16V, one is 9-24v, and I forget the other but everything seems to work on the 10.7v I am getting, I just have to calibrate outputs accordingly). I might just use it as is with the regulator with 10.7v. That being said, am I correct in assuming that I will have a constant 10.7v as long as the battery can output at least 10.7v?

The voltage drop is higher with the regulator because its own impedance is high.
It may be designed to operate in lower power circuits. You are correct in your assumption though.

Aurel

You need to input more than 12.6 volts to a regulator to get 12 volts out.

It appears that you have 2 wireless voltage transmitters and 1 wireless current transmitter. If this is correct, each of these units has an internal battery to power it. If the lower unit is a current transmitter, it is not hooked up correctly. The input to the unit should be tied in series with the positive(+12vdc) lead to detect the current flow. Also, the voltage regulator is likely capable of 1amp max. You could be drawing more than that if the current transmitter is hooked up incorrectly. What is the spec for current draw on the oxygen sensor?

Im unclear about one thing from the 2 above answers. If the regulator's impedance or the lack of more inital voltage is accounting for the voltage drop from the regulator to the circuit, how come if I just hook the regulator up to the 12.6v battery (no circuit) I DO get a regulated 12V (11.6) from its output? ONLY when the circuit is connected do I see the drop from 11.6 to 10.7.

Sorry if this has already been answered and I just dont understand. I feel like I am being really thick about this, but its not my expertise.

You are exactly right about the transmitters but I believe its hooked up correctly. Whats not clear in the picture is that there are 3 connectors there not two. It is hooked up as the O2 sensor directions state: one lead to 9-24v DC power, one 4-20mA output lead to the transmitter (+), and one lead that has a common ground for the transmitter and the power source. Am I right?

Sound right. What is the current draw of the oxygen sensor at 12vdc?
Also, what is the part number for the thermocouple signal conditioner?

When you measure the voltage to the regulator not hooked to the circuit, there is very little current flowing, so no ohmic drop. When you connect the circuit, you cause more current to flow, so bigger ohmic drop.

Aurel

Is there a spec for headroom on the regulator like this

VI Input voltage μA7812C MIN 14.5 MAX 30

edit: Even a Low Dropout Regulator will require about 1.5 volts of headroom.

Aaahhhhh.....There we go. Thank you. If this chip were in a very low power circuit (for which it is designed I think) there would be much less ohmic drop and it would really give 12v to the system, even more accurately if the battery were, say, 24v.




As to the part number: Omega DRG-SC-TC It has a couple dip switches for configuration. This one I set up for a "T" thermocouple, 0-257 degree F range, and 0-5v output.

I did not see anything like this on the package. I do realize that a 12v regulator is meant for a power source of more than 12V but my main goal was to create a constant voltage that is adequate for the components. 10.7v is not what I expected but I think it will work and at least now it is constant. In fact it gives me almost 2 volts of battery droppage before my compenent outputs gets out of wack.

Thanks everyone!

What is the part # of the regulator you are using?

If the other circuits will work with a lower voltage, like 9 volts, use a 7809 regulator.

Need to know the current & voltage requirements of the other components that you want to supply regulated power to.

Pretty sure that 10.7 volts will vary directly with the battery voltage. Usually those 3 terminal regulators have a darlington output and what you are seeing is the drop across that.

whatcha bil'den?