Application of MOS tube in low power consumption system

qewnja

Application of MOS tube in low power consumption system [Copy link]

 

Here is another case about low-power applications. This time it is a small problem caused by the characteristics of MOS tubes. Haha, I prefer to post the problem analysis process for your reference. It is better to teach a man to fish than to teach him how to fish. I share the whole process with you in the hope that you can find the knowledge points that suit you. At the same time, I am not telling you directly that you can’t use it this way or that way. What I wrote is similar to a short story. After reading it patiently, it’s as if you have done such an experiment yourself. This can deepen your impression and avoid detours again. At the same time, shouldn’t a good engineer first have enough patience? This case is the same as the one I posted last time (a case about low-power design). It is also a problem that occurs in low-power products. However, everyone should pay attention to this phenomenon (let's call it that because it is not a fault). This is a characteristic of the MOS tube, which will cause some special faults under certain circumstances. The description is as follows: According to the customer's requirements, we designed a terminal product with a standby time of 90 days. This is really a bit too much to ask. Also, due to the size limitation of the terminal shell, a large-capacity battery cannot be used. Finally, a 3.7V battery was selected. A 2400mah lithium battery. Since the customer requires this terminal to be activated once every 24 hours and work normally for 10 minutes, it is calculated that the current of this terminal must be below 100ua in sleep mode to meet the requirements. After a long period of research and demonstration, a hardware solution was obtained. In theory, the terminal current can drop to about 50ua in deep sleep mode to meet customer requirements. So the normal process was followed: schematic diagram-PCB-prototype. After the prototype was produced, it was tested and some small design bugs were eliminated (n jumpers flew, haha). It basically worked normally, with a sleep current of 43ua, which was successful at one time. However,"The first prototype cannot be 100% successful at one time" played its power. The prototype encountered an unexpected situation when it was heated to 75 degrees. The tester responded that when the high temperature was heated to 75 degrees, the terminal could not enter deep sleep. Specifically, it restarted once every 10 minutes and was cooled down at room temperature for a while. The first reaction to hearing this news was that there must be a component with poor soldering. The solder joints cracked at high temperatures and closed at room temperatures. Because it was a hand-welded sample, this phenomenon was not surprising at all, so the relevant deep sleep circuit part was re-soldered with a soldering iron and put into the high temperature box for testing again. The phenomenon remained the same. It seems that it is not a problem of cold soldering. Another possibility for this failure phenomenon is that the thermal stability of a component is not good and it fails at high temperature. So the hot air q1an9 was blown to the relevant circuits one by one, and the current and voltage were monitored at the same time. Sure enough, when the machine entered sleep mode, as the temperature of the hot air q1an9 continued to rise, the leakage current of the deep sleep switch Q10 gradually increased. When it reached about 70 degrees, the switch failed. As shown in the figure above, the hardware engineer used more field effect tubes when designing the circuit because of low power consumption. Due to the characteristics of MOS, its leakage current will increase with the increase of temperature. As the leakage current of Q10 increases, the voltage drop on R22 becomes larger (because the MOS tube is a voltage control device and its internal resistance is very large, it can be ignored). Therefore, Q11, the main low-power control device, slowly entered the saturated conduction state from the cut-off state, so the original Q11 was replaced with an ordinary NPN transistor to eliminate the fault (Q10 leakage current is insufficient to turn on the transistor).

topwon

Is it OK to reduce the values of R20 and R22 (for example, 1/10)?

qewnja

topwon posted on 2018-6-27 18:09 Is it OK to reduce the value of R20 and R22 (for example, 1/10)?

Yes, but the resistance of R20 and R22 should be reduced to a very small value, because after all, MOS tubes are voltage-driven, and I think this design is unreasonable. It is more reasonable and reliable to use triodes to drive MOS.

yyhhgg

I would like to ask, does a low-power MOS tube have low internal resistance? I used a low-resistance MOS tube and a high-resistance MOS tube for comparison testing. On the contrary, the high-resistance MOS tube has low power consumption, while the low-resistance MOS tube has high power consumption. We tested them in the same circuit, but I don't understand why.

maychang

yyhhgg posted on 2019-7-24 14:20 May I ask if low-power MOS tubes have low internal resistance? I used low-resistance MOS tubes and high-resistance MOS tubes for comparison testing, but the high-resistance MOS tubes have lower power consumption...

When you say "MOS tube with low power consumption", which indicator is particularly outstanding? Does "MOS tube with low internal resistance" mean that Ron is particularly small?

伟林电源

Are you talking about Rds(ON)? This parameter is the positive temperature coefficient, and silicon carbide will be better.

PowerAnts

The main concern for low power consumption in standby mode is leakage current. The problem encountered by the OP is that the negative temperature effect of Vth causes Q10 to turn on slightly. R22 is too large, and a few microamperes will generate enough voltage for Q11 to turn on to a large extent.

PowerAnts

At least "low internal resistance MOS has large loss", that's because the Vth of a small rdson is relatively low, which is more prone to "leakage".

伟林电源

You are right. Properly reducing R22 (or adding a series resistor at the D pole of Q10) can reduce the voltage value established by the leakage current at the G pole, thereby avoiding the misleading conduction of Q11.

PowerAnts

Wei Lin Power published on 2019-7-27 11:23 You are right. Properly reducing R22 (or adding a series resistor to the D pole of Q10) can reduce the voltage value established by the leakage current at the G pole, thereby avoiding the misleading conduction of Q11...

Can the series resistor of the drain of Q10 reduce the leakage current? Only reducing R22 is effective. I have never seen such a large resistance value added.

伟林电源

Adding a series resistor at the Q10D pole actually increases the input resistance of Q11, and the Q11G pole requires a larger current to establish the conduction voltage.

PowerAnts

Wei Lin Power published on 2019-7-27 16:56 Adding a series resistor to the Q10D pole actually increases the input resistance of Q11. The Q11G pole requires a larger current to establish a conduction voltage.

Think about it again

qewnja

Please give me a thumbs up, and invite all the big guys to discuss

maychang

Wei Lin Power published on 2019-7-27 16:56 Adding a series resistor to the Q10D pole actually increases the input resistance of Q11. The Q11G pole requires a larger current to establish a conduction voltage.

R22 should be reduced instead of the series resistance at the drain of Q10.

The purpose of reducing R22 is to obtain a smaller voltage on the gate of Q11 under the same Q10 leakage current.

伟林电源

First, I am not against the method of reducing R22;

Secondly, the series resistor actually adds a current limiting resistor, which also reduces the voltage value generated by the gate of Q11 under the same leakage current condition.

伟林电源

Although the resistance of R22 is indeed a bit large, can this equivalent circuit be different to reduce the impact of Q10 leakage current on Q11?

maychang

Weilin Power Supply published on 2019-7-27 21:23 Although the resistance of R22 is indeed a bit large, can this different equivalent circuit reduce the impact of Q10 leakage current on Q11? & ...

This connection method will reduce the voltage received by the gate of Q11 when Q10 is turned on.

In fact, when reducing R22, R20 must also be reduced, otherwise the voltage divider ratio when Q10 is turned on will change.

PowerAnts

Wei Lin Power published on 2019-7-27 21:23 Although the resistance of R22 is indeed a bit large, can this equivalent circuit be different to reduce the impact of Q10 leakage current on Q11? & ...

The leakage current of mosfet belongs to the reverse saturation current of the pn junction. In the application, it is only affected by temperature but not by ds voltage. It is equivalent to a UA-level temperature-controlled current source. The additional series resistance does not affect the voltage drop on R22

伟林电源

Let's put it this way, the leakage current of Q10 needs to charge the Cgs of Q11. Does its charging current need to pass through RX? If so, then VRX will be established. Is the generated IRX part of the leakage current?

PowerAnts

Wei Lin Power published on 2019-7-27 21:55 Let's put it this way, the leakage current of Q10 needs to charge the Cgs of Q11. Does its charging current need to pass through RX? If so, then VRX will be established, and the generated IRX ...

Oh my, we are not on the same channel