用于电源的I/O引脚上的电压下降

不应该发生，是的：我做了很多次…我只是添加了一些去耦，比如一个10欧姆串联电阻和470N/1UF盖并联：可能有帮助。



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Shouldn't happen, yeah: I did that some many times...
I just added some decoupling, say a 10ohm resistor in series and a 470n / 1uF cap in parallel: might help.

数据表中的规范D370表示，在输出电流为6Ma的情况下，Voh将处于VDD - 0.7之内。在3Ma时，您看到的电压下降为0.5V，所以它没有超出标准。如果需要较低的电压降，您确实应该使用外部FET。



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Specification D370 in the datasheet says the Voh will be within Vdd - 0.7 at an output current of 6mA.
You are seeing a voltage drop of 0.5V at 3mA, so it's not out of spec.
You really should use an external FET if you need a lower voltage drop.

对于I2C设备，时钟和数据线上的上拉电阻使用什么？



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What are you using for the pullup resistors on the clock and data lines for the I2C device?

4.7K拉上电源。当直接从3.3伏电源供电时，在拉柱上有一个0.5伏的下降。当从IO引脚供电时，在拉流器上有1伏的下降。



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4.7K pullups.
When powered directly from a 3.3 volt supply there is a 0.5 volt drop across the pullups.
When powered from the IO pin, there is a 1.0 volt drop across the pullups.

我希望得到相反的结果，但是在第36章中，如果没有特征图，似乎没有足够的信息。这些图表在16F18323的数据表中应该是完整的。图35-35：VoH/IOH为3.0V。



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I kinda expected the opposite result but any case it looks like there is not enough info without the characteristic graphs in chapter 36. The graphs are complete in the data sheet for 16F18323 which should be close. Figure 36-35: has VOH/IOH at 3.0V.

您的意思是与提供I2C设备的IO引脚内联吗？然后一个帽子从那里到地面？我不知道盖子能做什么，但是电阻器会产生电压降，不是吗？



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Do you mean inline with the IO pin supplying the I2C device? and then a cap from there to ground? I'm not sure what the cap would do, but the resistor would create a voltage drop, wouldn't it?

这就是诀窍。我假设你指的是一个当前缓冲区，就像附图中的“FET电流缓冲器”。现在几乎没有任何电压下降。



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That's done the trick. I assume you meant a current buffer, like in the attached picture "FET current buffer". Now there is hardly any voltage drop.



Attached Image(s)

当然，缓冲器是更好的，但从我的经验（与一些PIC18和24，或多或少相同的电流消耗，我没有看到这个问题）。至于过滤，它的目的是防止振荡和类似。小电阻不会降低任何电流率。



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Of course a buffer is better But from my experience (with some PIC18 and 24, at more or less the same current drain, I did not see that issue).

As for the filtering, it was meant for preventing oscillations and alike. The small resistor would not drop anything significant at this current rate.

第一个图是不正确的，第二个是切换低侧。如果低侧开关是可接受的，那么你可以用PIC GPIO Pin来做同样的事情。GPIOS通常可以比他们能驱动的更低地驱动低电平。或者，使用P沟道FET来切换高侧，并且反转S。TPIE的GPIO引脚。



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The first diagram is incorrect, and the second one is switching the low side.
If low side switching is acceptable, then you could do the same with the PIC GPIO pin.
GPIOs can usually drive low harder than they can drive high.
alternatively, use a p-channel FET to switch the high side, and invert the state of the GPIO pin.

有趣的是，现在我看它，图1是怎么做的？”高频放大器通常使用公共基座的结构。低输入阻抗，低输出阻抗…但这是一个MOSFET，所以…



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Funny, now that I look at it, what is diagram #1 supposed to do?
"common base" configurations were used usually for high-frequency amplifiers... low input impedance, low output impedance... but this is a mosfet so...

嗯，我真的不想要缓冲区，因为我买不起这个空间，我会进一步研究它。谢谢



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hmmm well I don't really want the buffer coz I can't afford the space, I'll look into it further. thanks

图1，我不知道。我错误地上传了它，无法解决如何删除它。



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Diagram #1, I don't know. I uploaded it by mistake, and can't work out how to remove it.

我已经尝试了我能想到的一切。不幸的是，串联电阻/盖没有显示任何东西。我尝试了10K I2C拉普拉斯，而不是4.7K。检查端口的无意弱拉，开漏控制和转换率。双检查外围引脚选择映射，因为我不使用默认引脚。没有差别。SPI设备本身会吸引10mA，不会造成电压下降。I2C器件只吸引2mA，但在IO电源引脚上造成了500毫伏的下降。我使用的是16LF，而不是16F，尽管我已经尝试过这两种方法。我能节省几百毫伏的唯一方法是使事情慢下来。把I2C从400千赫降到100kHz给了我一个额外的150毫伏减慢内部微控制器振荡器——主程序中的130MVA 5ms延迟。慢I2C例程被称为-250MVE他们不合并，但它不像我节省500毫伏通过做这三件事。这是正常的吗？我可以看到当I2C设备被提供电压时，IO引脚电源下降500毫伏。然后当它断开时，IO引脚电源返回满。但是当它重新连接时，它只下降了300毫伏。我猜这意味着它是Micro内部的I2C功能，导致了一些下降，并且I2C设备也造成了一些下降。或者有些地方漏水。我不知道。我想我得带着缓冲器去，虽然我觉得有些东西不对。



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Well I have tried everything I can think of. Unfortunately the series resistor/cap did not show up anything. I tried 10K I2C pullups rather than 4.7K. Checked the ports for unintentional weak pull ups, open-drain control and slew rate. Double checked the peripheral pin select mappings as I am not using the default pins. No difference.

The SPI device on its own which actually draws 10mA, does not cause a voltage drop. The I2C device draws only 2mA but causes a 500mV drop on the IO supply pin. I am using a 16LF rather than a 16F, though I have tried both anyway.

The only way I could save a few hundred millivolts was to slow things down.
Dropping I2C from 400KHz to 100KHz gave me an extra 150mVolts
Slowing the internal microcontroller oscillator - 130mV
A 5mS delay in the main routine, which slows how often the I2C routines are called - 250mV
They don't combine though, so its not like I am saving 500mV by doing all three things.

Is that normal?

I can see when the I2C device is supplied with voltage the IO pin supply drops by 500mV. Then when it is disconnected, the IO pin supply is back to full. But when it is reconnected it only drops by 300mV. I guess this means it is the micro's internal I2C functions that are causing some of the drop and the I2C device is also causing some drop. Or some leakage somewhere. I don't know.

I guess I'll have to go with the buffer, though I feel there is something not right.

你如何测量电压？你是用一个高速的范围观察它，并且确保地电流不影响测量吗？



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How are you measuring the voltage?
Are you observing it with a hi speed scope, and making sure ground currents are not affecting the measurement?

请你解释一下好吗？



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Can you explain that please?

你先来。你如何衡量？也就是说，它和什么联系在一起？



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You first. HOW are you doing the measurment?
i.e. with what, and where is it connected?

请参阅附件。Micro的+VE和-VE连接到A3伏板凳电源。I2C器件的VE连接连接到微器件的-VE，并且I2C器件的+VE连接到微器件上的IO引脚。我用万用表测量IO引脚和-VE之间，它测量2.73伏特。当没有连接到任何东西时，IO引脚测量3伏特。我将SPI器件的+VE和-VE与I2C器件连接起来，然后在-VE和IO引脚之间有2.43伏。断开I2C器件，只留下SPI器件，在-VE和IO引脚之间有2.97伏。I2C上拉是介于IO引脚和-VE.谢谢



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Please see attachment.

The micro's +ve and -ve are connected to a 3 volt bench power supply. The -ve connection of I2C device is connected to the -ve of the micro, and the +ve of the I2C device is connected to an IO pin on the micro. I am using a multimeter to measure between the IO pin and -ve, and it measures 2.73 volts. The IO pin when not connected to anything measures 3.0 volts.

I connect the +ve and -ve of the SPI device along with the I2C device and then there is 2.43 volts between -ve and the IO pin.
I disconnect the I2C device, leaving only the SPI device, and there is 2.97 volts between -ve and the IO pin.

The I2C pull ups are between the IO pin and -ve.

thanks
Attached Image(s)

一些新信息。首先感谢你的帮助，我缺乏知识，所以我从这个论坛得到的信息是很棒的。用示波器测量在系统中显示了一个巨大的缺陷，但我不知道为什么。每次I2C和SPI例程运行时，我都能看到电压的下降。即使在使用工作台电源电压而不是IO电源电压时。当使用IO电源电压时，结果明显放大。我想万用表只是读数平均值。所以，这里是我的目标。我不再使用IO电源，而是从板凳电源供电。3伏。当I2C例程运行图2MA时，电源电压下降5MV。当SPI例程运行时，绘制10mA的电源倾角40mV。如果我使用IO电源，放大大约十倍。SPI设备是一个射频发射器，我想知道瞬时的曲线是什么？在每个器件上有10uF钽和100nF陶瓷。我尝试了100UF电解跨越供应，减少了一半的倾角和470UF大部分去除所有纹波。但是这么大的价值看起来很奇怪，不是吗？重要的是，虽然它大大减少了纹波，但对万用表在IO引脚上读取的平均值没有任何差别。



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some new information. thanks for your help first, I lack knowledge so the info I get from this forum is great.

Measuring with the oscilloscope showed a huge flaw in the system, but I don't know why. I can see a dip in voltage every time the I2C and SPI routines are ran. Even when using the bench supply voltage rather than the IO supply voltage. Though when using the IO supply voltage the results are magnified significantly.
The multimeter i guess was just reading the average.

So, here is where I am up to. I am no longer using the IO supply, but rather am powering everything from the bench supply. 3.0 volts.

When the I2C routine runs drawing 2mA, the supply dips 5mV. When the SPI routine runs, drawing 10mA the supply dips 40mV. If I do use the IO supply this is magnified about ten fold. The SPI device is a RF transmitter, I wonder what the instantaneous curent draw is?

There is a 10uF tantlum and a 100nf ceramic across each device. I tried a 100uF electrolytic across the supply which reduced the dip by half and 470uF mostly removes all the ripple. But is seems odd to have such a large value, doesn't it? And importantly, although it reduces the ripple significantly, it does not make any difference to the average value read by the multimeter on the IO pin.

让我们确定你想要使用的PIN实际上是在寻找你认为是当前的（并且规格表说它会）。把所有东西从那个引脚断开，把一个已知的固定电阻器“供电”到接地怎么样？电压是什么，欧姆定律告诉你电流是什么？检查针的行为范围，而不是一米。让你的代码切换PIN，看看你的固定电阻器的上升/下降时间是怎样的。第二步：加上你期望你的“负载”消耗的电流，并计算一个等效电阻值。把它固定在引脚上，接地，看看电压是什么。如果计算出的电流低于规范表所说的电流，并且你对你的电阻值有信心，那么你可以关注PIN配置。任何一个在这个设备系列中的PIN都有一个打开的漏选项，可能会被打开？您是否启用了某种弱的上拉选项而不是真正的主动高端驱动程序？（我不熟悉这个家庭，但这些是其他PIC设备的选择）。



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Let's make sure the pin you want to use is actually sourcing the current you think it is (and that the spec sheet says it will). How about disconnecting everything from that pin and "powering" a known fixed resistor to ground? What does the voltage do then, and what does Ohm's Law tell you the current is?

Examine the pin behavior with a scope, not a meter. Have your code toggle the pin and see what the rise/fall times look like with your fixed resistors.

Second step: Add up the current you expect your "load" to consume and calculate an equivalent resistance value. Stick that on the pin, to ground, and see what the voltage does. If the calculated current is below what the spec sheet says should be available, and you're confident of your resistor value, then you can focus on pin configuration.

For example... any chance the pin in question on this family of devices has an open drain option that might be turned on? Have you enabled some sort of weak pullup option instead of a true active high side driver? (I'm not familiar with this family, but these are options on other PIC devices.)

Report back!