• Locked Locked
  • Replies 4 replies
  • Subscribers 65 subscribers
  • Views 268 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

TLV61048: Low-noise bipolar boost power supply

stanley
Prodigy 60 points
Part Number: TLV61048
Other Parts Discussed in Thread: DAC80508, LM27762, , TPS65131

Tool/software:

Hi E2E,

I am working on a new design, and have a few clarifying questions.

The circuit is powered from a 5V USB-C connector, and contains a 16-bit DAC (DAC80508) outputting 0-5V, fed into an op-amp circuit to shift it into -5 to +5V.

Since the USB is not always guaranteed to have a stable 5V, and I would also like some head room to not operate the op-amp to rails, I would like a single-IC solution with as few components as possible, that converts 4V-5V into a low-noise, bipolar power supply with output >= ±7V, where +7V (>=100mA) and -7V (>=10mA). The positive output will also be further downregulated to 5V and 3.3V for the DAC and a microcontroller.

I have tried to look for solutions, and most of them have some limitations. For example, LM27762 is a nice, bipolar-output, inductor-less solution that can boost the voltage on positive output, but the negative output is limited to the input voltage magnitude. The only one I could find was TLV61048, so I decided to build a prototype using configuration in application note SLVAEJ3 (Figure 5).

Reference Qty MPN Value Manufacturer Package Datasheet
C1 1 CL05A105KA5NQNC 1uF Samsung Electro-Mechanics 0402 Link
C2,C4 2 GRM188R61H225KE11D 2.2uF Murata Electronics 0603 Link
C3 1 CC1206KKX5R8BB106 10uF YAGEO 1210 Link
D1,D2,D3 3 B120AF-13 B120AF-13 Diodes Incorporated DO-214AD Link
L1 1 XAL4030-472 4.7uH Coilcraft 4.0x4.0x3.1 Link
R1 1 HP03W5F100KT5E 1R UNI-ROYAL 0603 Link
R2 1 RK73G1ETTP1004F 1M KOA Speer Elec 0408 Link
R3 1 0402WGF1003TCE 100k UNI-ROYAL(Uniroyal Elec) 0402 Link
U1 1 TLV61048DBV TLV61048 Texas Instruments SOT-23-6 Link

It seems to work fine, but I was wondering:

Question 1:
With the FB pin connected to a 1M/100k voltage divider, I would expect the output voltage to be 8.8V. However, I measure the positive and negative outputs to be +8.9V and -8.8V respectively. Is the 100mV difference expected?

Question 2:
Can I substitute the recommended B120AF-13 Schottky diode with a different one, if they have similar Vf and Io? I would like to use the 1N5819W.

Question 3:
I would also like to have a connector through which I can connect an external bipolar power supply without causing backfeeding, even if connected at the same time as the USB. I made this circuit using 1N5819W diodes.

+V_EXT / -V_EXT External bipolar power supply
+8V / -8V On-board bipolar power supply generated from TLV61048
VSS / VDD Protected power domains - bipolar power is only taken from here

Initial tests indicate it works as it should, but I would like to hear if this is a good design choice. Can you spot any obvious mistakes or something that can go wrong?

Question 4:
Is there a better solution for my application, that would require less components (and preferably no inductor or diodes), and be better in terms of noise characteristics?
Or would you recommend a multi-ic solution, where the first circuit just boosts the voltage, and the second one inverts it?

I am looking forward to hearing your opinion.

Thank you for help,
Stanley

  • TI__Genius 14180 points

    Hi Stanley,

    stanley said:
    Question 1:

    The 100mV should be caused by the normal resistance tolerance of the feedback resistors and there is no issues.

    stanley said:
    Question 2:

    It's OK to use other diodes if they have similar Vf and Io with the original one.

    stanley said:
    Question 3:

    It's a good design. I don't see any concerns.

    stanley said:
    Question 4:

    I think the current solution is almost the best one. Other solutions, for example, TPS65131 needs two inductors. Or two chips(one boost, one inverting charge pump) are needed.

    Regards

    Lei

  • in reply to Lei Zhong
    Prodigy 60 points

    Hi Lei,

    Thank you for answering my questions. I appreciate it.

    Lei Zhong said:
    The 100mV should be caused by the normal resistance tolerance of the feedback resistors and there is no issues.

    I would understand if there is a 100mV deviation from the theoretical voltage output due to feedback resistors, but this is a difference between the positive and negative output, which are both controlled by the same feedback loop. Could you please clarify this?

    Other than that, I am happy to hear you don't see any other problems with my circuit.
    Since, the output voltages are used as references, would you maybe recommend filtering? If so, what kind of solution do you propose to make both outputs as clean as possible?

    Thank you.

    Best regards,
    Stanley

  • in reply to stanley
    TI__Genius 14180 points

    Hi Stanly,

    stanley said:
    but this is a difference between the positive and negative output, which are both controlled by the same feedback loop. Could you please clarify this?

    For the charge pump solution based on TLV61048, actually what the loop regulates is the positive voltage. The negative output is not directly regulated. And it relies on the positive voltage regulation. So it is expected that the negative Vout (its abs value) is lower than positive Vout.

    Well, a little surprised that the negative Vout is just 100mV lower than the positive Vout. For normally the gap should be a little higher than one feedforward voltage of a diode. 

    You may refer to the formula (7) in this app note (click here). Please note that there is a typo in this formula. It should be ICHG1, not VCHG1.

    stanley said:
    Since, the output voltages are used as references, would you maybe recommend filtering? If so, what kind of solution do you propose to make both outputs as clean as possible?

    It depends on how big the Vout AC ripples can be accepted in your application. If the acceptable Vout AC ripples are very small, the usual ways are adding bead, LC filter or LDO.

    Regards

    Lei

  • in reply to Lei Zhong
    Prodigy 60 points

    Hi Lei,

    Perfect. That makes sense.

    Thank you once again and have a nice day.