I am using TPS61026, BQ24070 and BQ27501 to
build my Battery management circuit. My AC-DC
is 5V 2A, the battery is 3.7V Lithium Ion Battery 2250mAH. The input of
TPS61026 is 4.4V and output is 5V. My load is an ARM CPU board (around 240mA)
and a 3.5 inch LCD touch screen (around 350-400mA). My problem is I can boot up
CPU board but can’t boot up CPU board and LCD at same time. The input of
TPS61026 is 4.4V but VOUT is only 1.8V when boot up with LCD. If I power up PM
circuit first and put resister as a load, I can get 800mA to 1000mA constantly
from TPS61026 VOUT with AC-DC or battery. In my circuit FB pin is NC.
I suspect your problem is a known limitation of the TPS6102x family during startup. The short circuit protection and soft-start circuitry are essentially combined in this part. Therefore, during startup, if the load that the IC sees is excessive, the TPS6102x will stop in one of its 4 startup phases. Figure 21 and the paragraph before it gives you some insite into why this happens. Two possible solutions are either to stagger the statup of the load circuitry or place a FET in series with the TPS61026 output voltage to isolate it from the load until the IC has fully started.
Thanks Jeff. I will try with a FET and let you know.
Can you confirm that the problem is startup related. Measure the inductor current on an oscilloscope. If the peak current is 40% of the nromal peak current (Isw Switch Current Limit), then it is a simple startup problem.
I tried power-on PM circuit and get 5V from VOUT of TPS61026 first and then plug in my load(CPU board end LCD). The VOUT becomes 1.8V too. On the LCD drive board,
I found three inductors for boost up the voltage for the LCD back light. Maybe LCD drive board has big peak current? If so what I can do?
How can I catch the peak current by an oscilloscope. I got the voltage only. By put a 1 ohm load in series?
A clip on current probe is the best way to measure inductor current. If you don't have a current probe, you can connect a resistor between the inductor and the output capacitor. This resistor should be in the range of 100mohms. Use the smallest resistor you can use and still see the proper waveform on the oscilloscope. 1 ohm is too large and affect performance too much.
Here are three things to consider:
1. Most LCD back lights that I have seen need a much higher voltage than 5V to power them. Does the backlight have its own boost converter to boost up the 5V to a higher voltage? Stand alone inductors that don't have a switching converter to drive them will not boost up the voltage.
2. Does the LCD back light have a large capacitor to which you are tying the 5V output of the TPS61026? This type of hot plugging could be loading the TPS61026, causing it to go into over current protection and then causing it to get trapped in soft start.
3. Check your LCD back light datasheet for the maximum voltage and current that it requires. Then choose a boost conveter which can provide that voltage and current. The maximum output current for a boost converter is much less than its switch current limit specification.You will have to use the datasheet equations to determine the required input current, and therefore minimum switch current limit, for the boost converter. For example, if the TPS61026 is the correct boost converter then use the equations in the datasheet (equations 4 and 5) to calculate the peak inductor current. The peak inductor current is going to be the average inductor current (equation 4) plus 1/2 the ripple current. Rearrange equation 5 to calculate the ripple current, delta I. This peak current must always below the TPS61026 minimum peak switch current (1500mA). You must design against the specified minimum of 1500mA, not the typical of 1800mA. If the required current reaches the peak switch current limit, the IC terminates the switching pulse and the output voltage drops. You must either accept a droop in output voltage or choose a boost converter with a higher peak switch current.
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