If the battery is positive voltage and negative referred to ground, I feel that the way the P-MOSFET is applied is wrong and for a better understanding, please study a primer as below.Hi
Please see the attached file for my circuit.
Circuit description:
I am trying to charge a battery through a PFET switch, Q2. When charger is connected, it pulls Vcharger high, which causes the inverter output to be low and turns Q2 ON. The problem is that when the charger is removed, Q2 switch will feed back VBAT to Vcharger, therefore, reinforcing the inverter output to be low and keeping Q2 ON forever.
Any ideas on how to solve this issue?
Thanks
Jeet
Post it as a propriate picture format, Then I might help.
Try to add pull low current at VCHARGER and use resistor divider for INVERTER gate.
please see Fig3 on page2 of attached pdf file . this is a way how to use p-mosfet for power switching.
the battery charging current has some how to be limited by you.
Hello Erikl
I am currently in the process of the prototyping the following circuit:
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However, my hunch is that, even with nFET (used as open collector type), PFET will feed back and keep itself on when the charger is removed.
Thanks
Jeet
You didn't insert the resistor between the Drain and GND which I suggested. Perhaps it might help?So I have tested the following circuit:
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It didn't work. The PFET would keep itself ON even when the VCharger was removed, as I had expected.
Any more ideas?
You didn't insert the resistor between the Drain and GND which I suggested. Perhaps it might help?
it becomes a load and not pull down for gate. i suggested you to follow the fig3 of the pdf, I had uploaded. your sch is still wrong i fear. a gate can be terminated by a smaller resistor like 2.K, 3.3K etc between Gate and source. later the switch could be turned ON by grounding the gate by say 220ohms., this could be done using a N mosfet and triggering with positive voltage like 12V, with a resistor in series, and a high value resistor to ground for that N mosfet gate.
then, when the N fet is ON, the ground goes to pFET input.and switches it ON.
VGSS will never exceed 4.2V in my application [Jeet]No, from the Drain of Q3. But 1MΩ might be too large for pull-down; try 1..10kΩ.
I have already tried 10k pull down from Drain of Q3 to Gnd, although the sch image shows 1M connected [Jeet]
On the other hand, you cannot expect an effective charge control from this circuit, because - during charging - Q3 operates in inverse configuration, and the charge current flows via the parasitic drain-source-diode. The only purpose of this circuit is to avoid possible discharge of the battery via the charger, if it is connected but not powered. However, this could be achieved more easily by a simple diode.
The problem with using a simple diode is that the voltage drop across diode will be around 0.7V or something. I can not afford that kind of voltage drop across diode. I am trying to charge a 4.2V/25mAh Li/ion battery. Its a small battery with 4.2V maximum voltage. The voltage drop across the diode will be high. [Jeet]
And more: I checked the FDN306P dataSheet: VDSS(max)=-12V is right on -- resp. over -- the limit if you want to charge a 12V accumulator (during the required shut-down phase), because such one will easily reach 14V or more. May be this is the reason that you can't pull down the input node, even with a low-ohmic resistor.
VDSS will never exceed 5V [Jeet]
Moreover, VGSS is specified as ±8V, so your FDN306P (or your NMOS) could already be damaged (if you charge an accu ≥8V).
In your configuration, during charging, Q3 is operated in inverse mode (drain more positive than source), as I already told you above, i.e. the charge current flows via the parasitic diode, which is a Si diode, hence has a voltage drop of ≈0.7V :!:Hi Erikl, Please see my response in red below [Jeet]
The problem with using a simple diode is that the voltage drop across diode will be around 0.7V or something. I can not afford that kind of voltage drop across diode. I am trying to charge a 4.2V/25mAh Li/ion battery. Its a small battery with 4.2V maximum voltage. The voltage drop across the diode will be high. [Jeet]
VDSS will never exceed 5V. [Jeet]
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