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lmz10501 and supercaps
Hi to all,
regarding the lmz10501, is possible to use it to charge 2 supercaps of 3V with esr of 100mohms?
With large output capacitance, the LMZ10501 will likely enter hiccup mode (see page 13 and fig 3 of the datasheet) as it is starting up.
However, even if it is in hiccup mode, the regulator will still pump current into the output cap (with hiccup pauses). The pauses will disappear once the output voltage exceeds 0.375V. If there is no load attached during the cap charging, the part will eventually get out of hiccup mode and start up. Here is a plot to illustrate this behavior with excessive output capacitance:
You see that initially the part is in hiccup more (see the VCON voltage charging up and discharging). Current is still supplied to the output in small portions during that time. Since there is no load attached, the output cap charges up and reaches above 0.375V. Then normal operation resumes. The 0.375V is the internal threshold on FB which tells the regulator that there is a short circuit event on the output. If the output voltage is below that threshold the regulator works in hiccup mode.
I happen to have some supercaps and LMZ10501 board. If you want to share your conditions, I will be happy to try this out in the lab and let you know.
What is the supercap value?
You mentioned 2 capacitors. Are they in parallel?
What input voltage are you using?
Do you have a time requirement on how fast the super capacitor must be charged?
Thanks Denislav for the quick reply,
Frankly, i used to charge the spercaps. But as the datasheet says about the hiccup, I was wondering if this could damage the lmz10501. I notice that when the supercaps is fully loaded, they begin to discharge while they still are connected to the lmz10501(no load connected). I was thinking that maybe is necessary to disconnect the lmz10501 from the supercaps to avoid the issue (EN pin low).
DenislavWhat is the supercap value?
I have two supercaps of 50F (taiyo supercaps)
DenislavAre they in parallel?
Yes, they are in parallel to rich 100F.
DenislavWhat input voltage are you using?
Alkaline batteries of 4.2V.
DenislavDo you have a time requirement on how fast the super capacitor must be charged?
I used the equation from page 15 of datasheet to get the values of Rb and Rt for a fixed Vout. But when i used to recalculate the values to the fixed version of 2.5V or 3.3V, the values that I get to Rb are not the same by choosing the value of Rt as it says in the schematics.
RB = ( VOUT / (5.875V – VOUT) ) x RT
Choose RT to be between 80kΩ and 300kΩ.
2) Shutting down the lmz10501 will prevent the discharge of the supercaps?
Here are my comments.
1. The resistor values you see are close standard values. It is possible to find a slightly better combination depending on which resistor you pick for the top and what resistor tolerance you choose to have.
2. On the capacitor discharge question, is the source (battery) still connected to the input of the LMZ when you see the discharging? I am thinking that it is better to have the LMZ always connected so that the voltage on the supercap stays up and regulated.
Here is a plot of charging 150F capacitor I had available. You can see the hiccup operation and then normal operation after VOUT is above the 0.375V threshold. At the end, the supercapacitor is charged to the regulation voltage (1.8V in this case) and the LMZ10501 maintains that voltage level.
In terms of the LMZ10501 hiccup mode operation, there is no problem as long as your application is OK with the charge-up time. One suggestion would be to place a schottky diode from VOUT to VIN. This would help in case of short circuit events on the input of the LMZ10501. If the input gets shorted, the supercap can discharge through the high side body diode and cause damage to the IC if the current is large. A diode would steer that potentially large current away from the IC.
The purpose of the supercaps is to extend the battery life. In this case i was thinking of disconnect the battery from the circuit and use the supercaps till they reach a low voltage, in wich case I will restart the lmz10501. Is this correct or is the wrong approach?
Thanks for the support.
This approach should work.
Another thing I wanted to mention, if the LMZ10501 is disabled but still connected to the supercap there will be a low current discharge path to GND through the internal feedback resistors. The sum of the internal resistors from FB to GND is 198kohm. This would result in several micro-amps of discharge current when the LMZ10501 is disabled. I don't know how critical this discharge current is for your application. If it is critical, there may be ways to reduce it.
How can I reduce this discharge current? I have an msp430 attached and consumes 0.7uA in stand by. Any help will be appreciated.
One way to cut the feedback divider current is to insert a resistor in series with the FB pin and adjust the VCON voltage down so that VOUT is set for the same target. This approach works, but has some drawbacks. The part will stay in hiccup mode longer so it will take longer to charge the supercap. A small capacitor will be necessary in parallel with the resistor for stability. Also, the current will not go down to zero. The best you can do is around 3uA.
A better approach in this case would be to use a small logic level NMOS switch and disconnect the FB connection when the part is disabled. You mentioned you are driving the EN pin already. See the partial schematic below:
In normal circumstances you would want to have controlled timing between the time the LMZ gets enabled/disabled and the time when the FB disconnect switch turns on/off, so that there is no open loop operation. For example, during power-up the NMOS gate of the FB switch should be high before the LMZ gets enables so that feedback is available when the part starts up. On the power down sequence, the EN signal would have to become low first before the switch is open.
In this case however this critical timing is not necessary because of the supercapacitor. Even if the part operates in open loop for some short duration, the output voltage is not going to overshoot because the supercap will not allow the voltage change in short period of time.
I tried this approach and the current consumption out of the supercap when EN is pulled low is 0.3uA (vs. 13uA when the switch is not in place). I would suggest using this method if you want to decrease the current consumption.
Let me know if you have any questions.
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