I’m designing my first circuit that uses a battery as the supply voltage. The required voltage is 3.6V and the battery I’ll be using supplies 3.6V, so no regulator is necessary. Is it still a good idea to put smoothing capacitors near the battery leads on the PCB? Or will the decoupling capacitors at the power pins of each of the ICs be enough?
-Dan
What would these be "smoothing"? A battery output does not have any 'ripple' that you would have from rectifying AC. So no these caps would not be needed.put smoothing capacitors near the battery leads
Decoupling caps are needed near each IC but some additional bulk capacitance on the PCB would not hurt.
3.6V is the NOMINAL voltage of a single cell Lithium battery. Beware that the Peak voltage at full charge is closer to 4.2 volts and needs to run down to about 3.0 volts to use the full capacity of the battery. Will your circuit work properly with this voltage range?
Hm, no, I think not. 3.6V is the upper end of the allowable voltage range (this is for a pic18f67j60). Looks like I’ll have to use two coin cells and put a regulator in.
Thanks for the advice, waltr.
Using a regular ol' linear regulator with a couple of coin cells and you'll likely lose more power in regulator waste than the PIC would use. Easier to use just a single coin cell with an inline diode (1N5820 for example) to drop the voltage by about .25volt. The '67J60 can handle that slight overvoltage...and with the PIC operating, the coin cell voltage will likely be a bit lower anyways.rusttree:
Hm, no, I think not. 3.6V is the upper end of the allowable voltage range (this is for a pic18f67j60). Looks like I’ll have to use two coin cells and put a regulator in.Thanks for the advice, waltr.
Exactly what kind of batteries are you wanting to use?
And what components are in your circuit? A schematic would be really handy here ya know!
Unless your circuit only uses an average of (say) a milliamp a "micropower" low-dropout (LDO) regulator is probably the best option. A diode drops a few hundred millivolts, as opposed to an LDO which only drops a few millivolts when lightly loaded.skimask:
rusttree:
Using a regular ol’ linear regulator with a couple of coin cells and you’ll likely lose more power in regulator waste than the PIC would use. Easier to use just a single coin cell with an inline diode (1N5820 for example) to drop the voltage by about .25volt.
I like to use 3V regulators instead of 3.3V when running on a lithium battery, as it generally gives a bit more life out of the battery.
True...absolutely. But I was going the 'dirt simple' route.MichaelN:
Unless your circuit only uses an average of (say) a milliamp a “micropower” low-dropout (LDO) regulator is probably the best option. A diode drops a few hundred millivolts, as opposed to an LDO which only drops a few millivolts when lightly loaded.I like to use 3V regulators instead of 3.3V when running on a lithium battery, as it generally gives a bit more life out of the battery.
Thanks for the suggestions, skimask and MichaelN.
I’m using the ethernet functionality in the 67j60. The circuit is the recommended one provided by Microchip in the datasheet:
http://ww1.microchip.com/downloads/en/D … 39762e.pdf page 211. There are no other additions to the circuit that are worth noting.
My setup is:
25MHz crystal
HSPLL mode
41.667MHz internal frequency
ENVREG tied to VDD (internal regulator on)
According to the electrical specifications, this setup will consume a maximum of 21.6mA. The allowable VDD supply range is 3.1 - 3.6 when using the ethernet module. Now, my final product will have an extremely limited on-time. It may be used for only a few minutes one time and then never activated again (it’s a demonstration piece with no real practical purpose). So I’m OK with using just a coin cell or two.
I haven’t picked out a coin cell yet, but it’s looking like my best option is two 3V’ers in series with a 3.3V LDO regulator. Does that seem reasonable?
Looks like this 3.3V LDO regulator will do the trick:
http://www.mouser.com/Search/ProductDet … 551SN33T1G
The only thing I don’t get is the reference to a minimum output current (in this case, 150mA). I can’t imagine this regulator requires the circuit to pull a minimum of 150mA at all times. What does that actually mean?
Yea, that spec is a little confusing. What it means is:
150mA is the Min Iout(nom) with the listed conditions. 3.1 V−4.0 V (Vin = 6.0 V) is for a 3.3V output.
Remember that there is a thermal shut-down circuit so if the dissipation is higher then the regulator will Current Limit. For that spec the dissipation is (6 - 3.3) * 0.15 = 0.405Watt. If you raised the input voltage then the Iout(nom) could be lower.
This is affected by the temperature of thermal shut-down and also by the drop-out voltage. Read the Thermal section in APPLICATIONS INFORMATION for more hints.
The graph in Figure 3 clearly shows Iout down to 0mA and regulated at 3.3V so you do not need to draw 150mA.
That's not the manufacturer saying "minimum", but rather whoever entered the data at Mouser. It's really a *maximum* current, but in reality the maximum current is generally limited by the temperature rise. You don't need a minimum load for the regulator to work properly.rusttree:
Looks like this 3.3V LDO regulator will do the trick:http://www.mouser.com/Search/ProductDet … 551SN33T1G
The only thing I don’t get is the reference to a minimum output current (in this case, 150mA). I can’t imagine this regulator requires the circuit to pull a minimum of 150mA at all times. What does that actually mean?
BTW, looks like a nice regulator.
Huh, that is confusing. But I think I understand now. Thank you both (again).