mosfet of constant current battery charger of cheap UPS-design keeps frying - Page 1
Forum section: electronics / projects

author: brammekeuh987
author_rank: Newbie
timestamp: June 09, 2021, 04:34:36 pm
content: DearI based my circuit on this proposal of MK14:https://www.eevblog.com/forum/projects/minimalist-design-challenge-200ma-constant-current/msg3425690/#msg3425690I adapted this circuit so that the battery that is charged can also deliver power to the circuit. So this would be a cheap-charly UPS with the disadvantage that you can overload the battery when you're loading too much. The components as in the schematic are the actual components used.I simulated this circuit with spice, and tried to approximate the spice models for the components. Since KiCad gives some syntax errors with different spice models I couldn't use the 'real' models.Explanation of circuit:INPUT_voltage is a climbing voltage from 10 to 18V with some added noise at 10kHz, there is a diode behind it (D4), which prevents that the battery would supply power to the source (since this won't happen)R_circuit is an equivalent resistance for a circuitR1, Vbat1, Q1, Q2, R4 are as per design of MK14, but Q1 is not a Germanium transistor. R4 is determined by trial and error with the simulation in order to get +- 200mAC1 is added to dampen peaksQ3 is added in order to control when the battery will be charged (When Q3 is conducting the battery will not be charging)D3 is added so that the current, when supplying to the system, doesn't need to come through Q2 and R4R_test is added to know the current through Q1 when preforming the simulationR_dumpLoad = 10.7 Ohm is there to simulate the test bench when I tested the circuitIn the test bench I replaced the Vbat1 by the R_dumpload. So when the R_dumpload is getting "charged" I measure the voltage over the R_dumpload and calculate the current passing through, in order to check if it complies with design.During the test I saw that the Ups mode works: so when INPUT_voltage is zero, the battery is delivering the current. But when I try to charge the R_dumpload, the Q2 burns through. I checked following things:in the simulation is already some noise added, and the overshoot isn't that big at all --> OKin the simulation the current from Q2 to Q1 is way smaller than the current through R4 --> OKQ2 has a maximum current of 350mA, the design is for 200mA --> OKreplaced transistor Q1 (if this would be faulty it could be the cause of burning --> OKremoved C1, since this could be adding too much delay in the feedback loop and cause an overshoot. But it still burnsdoubled R4 to 6.6 ohms, than the mosfet would only smoke a little (maybe from the leftover flux?). The voltage over the R_dumpload is 0.75V, so the current is 70mA, which doesn't really explain why this little current already produces some smoke. In simulation the current should be around 100mA. So I guess the simulation is a good approximationCan anyone explain why Q2 keeps lighting up like a brittle star at the summer evening sky?

author: TheMG
author_rank: Frequent Contributor
timestamp: June 09, 2021, 07:25:23 pm
content: You need to consider the power dissipation in the transistor!(Supply_Voltage - Battery_Voltage) * Current = Power dissipated in Q2In your example, base on 18V supply, 0.75V across the test resistor, current of 70mA, that's about 1.2W dissipated in that poor little 2N7002! Too much!Also, what type of battery are you charging?

author: brammekeuh987
author_rank: Newbie
timestamp: June 10, 2021, 07:21:15 am
content: Oooh, right, makes sense! The battery is a 12V lead-acid battery.So actually I have two problems: the limiting factor ain't the maximum Amps, but the power rating; the test bench with only the 10 ohm resistor makes a larger voltage drop across Q2, which causes more power dissipation.The max power dissipation of a 2N7002 is only 0.37W.So in normal operation (as per design) we would have 15V supply, so with the 12V battery this leaves 3 Volts. Voltage drop over 3.3 resistor @ 0.2A = 0.66V, so a 2.33V*0.2A = 0.446W, which is still too much.So I could use a larger mosfet or reduce the current by a larger R4. So when I put 6.6 resistor in I get about a 0.1A. In practice this would still mean a 0.66V voltage drop across R4, but only half (0.223W) of the original power dissipation in the 2N7002. So no smoke!In order to check the current on the testbench, it's probably better to just have the normal setup with the battery instead of the R_dumpload. I can get the current when measuring the voltage across R4.In order of efficiency: I guess since the 3V gap between the 15V supply and 12V battery always needs to get dissipated it, the above changes don't do anything about the overall efficiency of this charger. The 3V needs to be 'bridged' eitherway. The voltage drop across the R4 remains the same, so the relative contributions to power dissipation of Q2 and R4 remain the same in this circuit. So the measure of increasing R4 only lowers the delivered power (amps) to the battery and doesn't change anything else. For my application this has no disadvantages. The charger only needs to charge the self-dissipation of the battery and recharge the battery after a power outage. The outage time is measured and circuit dissipation is conservatively guessed. Power outage is highly unlikely since the application is connected to the grid. So 80% efficiency is very acceptable.Thanks TheMG!

author: TheMG
author_rank: Frequent Contributor
timestamp: June 10, 2021, 01:39:47 pm
content: Lead-acid batteries DO NOT like constant-current charging! The battery will fail in a short amount of time.You can use a current limiting circuit to limit the maximum charge current when the battery is significantly discharged, but it is very important that the final float charge voltage not exceed about 13.5 to 13.7V.I would suggest reading up on lead acid battery chemistry and charging methodology.