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I am currently working on a project that uses a 9.8V 1000mAh battery to power a solenoid, and as I understand it (please correct me if I got this wrong.) a battery capable of delivering 1000mAh is also capable of delivering 1Ah, or 2A for 1/2 hour, 4A for 1/4 hour, etc.

So, assuming I got the above correct, I need to know how to limit the battery output current to 1.0A (My circuit would get really hot otherwise.)

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    \$\begingroup\$ The current demand is a function of your circuit layout. That is, just to say you are powering your circuit with 9.8V does not mean that is will draw 1A or even 2A. In other words, the current sourced from the battery is not an intrinsic characteristic of the battery you choose. If I am completely missing your intent then please forgive me, but considering the words you chose it came across that you are a bit confused with this concept. \$\endgroup\$ – sherrellbc Aug 15 '14 at 18:40
  • \$\begingroup\$ I probably am. Like I said, I am still very new to this. :) \$\endgroup\$ – CoilKid Aug 15 '14 at 18:48
  • \$\begingroup\$ Honestly, as I am using a solenoid as my entire circuit, I have no idea what it will draw. I assume/d the coil would basically draw whatever I could throw at it until it melts. As my force calculations rely on the current being between 0.7 and 1.0A, I do not want it to under/over power the coils. \$\endgroup\$ – CoilKid Aug 15 '14 at 18:58
  • \$\begingroup\$ The solenoid will only 'use' finite current. It would melt itself if its resistance is low enough. So it should be easy to fix. Have you any test equipment, like a multi-meter? \$\endgroup\$ – gbulmer Aug 15 '14 at 19:16
  • \$\begingroup\$ Yes, but I have not built the coils yet. I am trying to design them ATM. I have a target current and target number of turns in my design, so I'd like to keep the number of turns and the gauge set. \$\endgroup\$ – CoilKid Aug 15 '14 at 19:25
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Your first assumption "battery capable of delivering 1000mAh is also capable of delivering 1Ah, or 2A for 1/2 hour, 4A for 1/4 hour" is incorrect. (That has been answered here already, I will have a look for it.)

The battery capacity vs discharge is far from linear, and the mAh rating is quoted against a low discharge rate (~0.1*capacity).

Secondly your circuit will use as much current as it needs. Trying to limit the current is likely to stop it working. To use less current, redesign the circuit. (You might want to write a new question about that, if you need help.)

Answer: So, if the current must be 1A, and the voltage is 9.8V, wind a coil with a resistance of 9.8ohms.

Edit: Clarified that the circuit is only the solenoid, and it is made by OP.

To use less current, either reduce the voltage, or increase the resistance of the coil.

Increase the resistance of the coil by using more turns, or thinner wire. Thinner wire will have a higher resistance, but you might not have any to hand. So more turns is easier to do.

Edit:
The relationship between resistance (R), current (I) and voltage (V) is Ohm's Law, and is: V = I x R

The battery provides a fixed voltage, in this case 9.8V

So I x R is fixed, it equals 9.8

Let's assume R is 2ohms
So I x 2ohm = 9.8V
I = 9.8V / 2ohm
I = 4.9A

If the resistance increases, the only possible way to ensure I x R stays the same, is to reduce I. Double the resistance, then I must half. increase R by 10, and that changes I by 1/10, or multiples I by 0.1.

Let's do it, R is 10 x bigger, so R become 2ohm x 10 = 20ohm
I = 9.8V / 20ohm
I = 0.49A

Further statements:
"If I run the same current through a wire with higher resistance ..."
Well how could you cause that to happen?

We know that the voltage (the force, if you like), the battery can exert is fixed at 9.8V. So if the resistance is increased, then the current must be decreased. Why? That is how the universe works.

So if we take the initial calcuation, 2ohm resistor allows 4.8A to flow
Power = V x I
Power = 9.8V x 4.9A
Power = 48.02W

Let's double the resitance to 4ohm
9.8V = 4ohm * new-I
new-I = 9.8 / 4
new-I = 2.45A

new-Power = 9.8V * 2.45A
new-Power = 24.01W - so a wire with twice the resistance will heat up more slowly

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  • \$\begingroup\$ Well, the circuit is a loop of wire, so... If I need to redesign the coils I can. I guess I need to research battery discharge rates better. :) \$\endgroup\$ – CoilKid Aug 15 '14 at 19:01
  • \$\begingroup\$ Okay. I did't interpret your question that the circuit is only a solenoid. I'd assumed it would be doing other stuff too. If it is a home-made coil, can you use lots more turns of thinner wire? Double the turns, or half the cross-sectional area. \$\endgroup\$ – gbulmer Aug 15 '14 at 19:07
  • \$\begingroup\$ Yup, its home-made. The only other thing on the battery would be a switch. I can use more turns if I need to. As of now, I was going to use AWG #26 magnet wire. \$\endgroup\$ – CoilKid Aug 15 '14 at 19:12
  • \$\begingroup\$ Well, if my coil has a set number of coils @ set current, I could reduce voltage using a voltage regulator? (Or am I still not understanding?) \$\endgroup\$ – CoilKid Aug 15 '14 at 19:16
  • \$\begingroup\$ A coil of wire will (after the turn-on transient settles to steady state) appear to the voltage source as a resistor, with resistance of the wire used in the coil. A short piece of thick wire will look like a short circuit (a few milliohms). A very long, thin wire will look like a resistance which you can calculate by multiplying (ohms per meter of that gauge wire) x (meters of wire). For your 9.6V battery you get current less than 1A (1C rate) if the resistance is more than 9.6 ohms. If resistance is less than 3 ohms you are probably discharging your battery at too high a rate. \$\endgroup\$ – Matt B. Aug 16 '14 at 3:40
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The circuit below allows you to limit the coil current to the maximum value of your choice. It was designed to allow relays to be operated from voltage above their rated value.

This circuit was provided by Richard Prosser for driving a relay or solenoid at constant current from a supply voltage higher than it is rated for.

Circuit from here - PICList, 2001, Richard Prosser.

Q1 & Q2 form a self oscillating buck regulator where the relay or solenoid is the target load and the inductor simultaneously. Q1 base is DC biased by D2 and R10, R2. When V on R7 due to Irelay causes Vbe Q1 to fall below about 0.6V, Q1 turns off and turns off Q2 and coil current circulates in L1, R7, D3.
R8, R9 provide hysteresis by changing Q1 base voltage so Irelay must fall slightly before Q1 turns on again and restarts cycle.
C3 provides positive feedback during switching to get squarer waveform.
The 'World famous on the internet' "Black Regulator" was derived from this circuit by Roman Black in response to a design challenge that I initiated.

enter image description here

At 1st impression it may seem "a little complex" but it's not too bad in practice.
Components
2 jellybean transistors,
8 resistors
2 Capacitors
2 diodes,
1 zener.

R10, R2, D1 simply provide a roughly regulated voltage at Q1 base.

When I out gets too high Q1 and Q2 turn off,
when the current drops a little Q1 & Q2 turn on again.

The difference between turn off and on is affected by feedback via R8 R9 D2 - this injects some low going voltage into (or "sucks out current" from) Q1 base which lowers the reference voltage from R10 etc so the current has to fall somewhat to restart.

If Vin was reasonably constant or if you don't mind a degree of variability in Iout as Vin varies you could simplify by :

Remove D1 and increase R2. Needs playing. D1 is meant to offset the Vbe drop of Q1 so the R10, R2 divider voltage is about the same as the R7 voltage caused by Inductor current.

Remove R8 and D2. Presently the zener stabilises the high swing at this point when Q2 is on regardless of Vin. If Vin is fixed you may be able to remove the zener and one resistor. The ratio of R9 to R2 affects how much hysteresis is caused.


Related

Black switching regulator voltage regulator based on the above circuit.

  • Ground the output with a current sense resistor and use a solenoid or relay as the inductor and it is similar to the original circuit above.

  • Ground the output with a current sense resistor and place an LED in series with the inductor and you have a constant current LED driver.

See above link for discussion and a number of predesigned variants for various applications. enter image description here

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  • \$\begingroup\$ Whoa, that's a complex. I think I'm gonna study that drawing for a while before I understand it... ;) \$\endgroup\$ – CoilKid Aug 17 '14 at 4:49
  • \$\begingroup\$ @CoilKid 1st impression is complexity - but it's not too bad in practice. 2 jellybean transistors, 8 resistors as shown, 2 Cs, 2 diodes, 1 zener. | R10, R2, D1 simply provide a roughly regulated voltage at Q1 base. | When I out gets too high Q1 and Q2 turn off, when the current drops a little Q1 & Q2 turn on again. | The difference between turn off and on is affected by feedback via R8 R9 D2 - this injects some low going voltage into (or "sucks out current" from) Q1 base which lowers the reference voltage from R10 etc so the current has to fall somewhat to restart. You could simplify by .... \$\endgroup\$ – Russell McMahon Aug 17 '14 at 8:22
  • \$\begingroup\$ .... doing the following if desired: Remove D1 and increase R2. Needs playing. | Remove R9 and D2. Needs playing. \$\endgroup\$ – Russell McMahon Aug 17 '14 at 8:23

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