• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Answers 1 answer
  • Subscribers 34 subscribers
  • Views 246 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

MAX3232E: Why Is a Smaller C1 Capacitance Recommended at Vcc = 5 V?

teritama
Expert 2430 points

Part Number: MAX3232E

Tool/software:

Hi team,

In the datasheet, Table 9-1 shows the recommended capacitor values versus Vcc. It seems to suggest using different values for Vcc = 5.0 V and Vcc = 3.3 V.

My understanding is that the internal charge pump (doubler) is regulated at ±5.5 V. Therefore, when Vcc = 5 V, the overcharge voltage is higher compared to Vcc = 3.3 V.

This could lead to greater losses than at Vcc = 3.3 V. Is this the reason why a smaller capacitance value is recommended for C1? Is my understanding correct?

I would appreciate your opinion.

Best regards,

teritama

  • 0
    TI__Guru* 79731 points

    Teritama,

    The way I think of C1 and C2 is they are used like buckets to fill a larger bucket (V+/V-). The smaller the caps are the smaller the bucket is. When a smaller cap is used, it can be filled and emptied faster but you end up having to fill it and empty it more often (it affects the switching). If you use a larger cap, the bucket is larger and you fill and empty it less often but the ideal point of the last buckets (V+/V-) might overfill or be under the ideal voltage so there is some trade off between too small and too large. 

    Below is an app note that walks through this.

    https://www.ti.com/lit/ta/sszt674/sszt674.pdf?ts=1759866170934&ref_url=https%253A%252F%252Fwww.ti.com%252Fdocument-viewer%252Flit%252Fhtml%252FSSZT674

    -Bobby

  • 0 in reply to BOBBY
    Expert 2430 points

    Hi Bobby-san,

    Thank you for your reply. I understand the trade-off between using a capacitor that is too small and one that is too large.

    As I understand it, the difference in capacitance between Vcc = 5V and Vcc = 3V arises because, when using the same capacitance for both, the load transient characteristics become more pronounced at Vcc = 5V. To avoid this, a smaller capacitance is chosen for C1 and C2. Is this correct?

    Also, C1 is selected to be smaller than C2. Is this because C1 is directly connected to Vcc and therefore more affected by Vcc than C2?

    Is my understanding correct?

    Best regards,

    teritama 

  • +1 in reply to teritama
    TI__Guru* 79731 points

    Teritama,

    I don't have much insight on the C2 circuit (designers treat this as a black box when I ask about and do not want to provide additional details). 

    I believe C1 looks similar to the one shared in the app note. C2 (my guess) will likely still see a connection to Vcc but it's job is to store more of the charge while C1's is to scoop up more water to keep C2 filled up. (C1 is going to switch faster than C2). C2 then uses it's water to fill up V+ and V- tanks. If C2 were smaller, I think it would affect the V+ and V- stability when the device is actually driving the bus/load. 

    -Bobby