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TPS40210 Iq problem

Other Parts Discussed in Thread: TPS40210, SWITCHERPRO, LM3478, LM3488, TPS40210EVM

Hello,

I am designing a step-up DC/DC converter by using TPS40210 for input of 16V to 40V and 200V output. The output current can be 10mA. By using what WEBENCH and SwitcherPro recommended, I built the circuit and it works. My problem is: its Iq (idling current without load) is too high. It is about 70mA@16Vinpout and 30mA@40Vinput. Any suggestions?

Thanks you.

Ning

• Hi Ning,

Most of this extra current is likely due to what it takes to switch the MOSFET that is requried to support 200V output. You could estimate the switching current multiplying the total gate charge (Qg) by the switching frequency (fsw) or Qg * fsw. There will also be some conduction loss to account for. The Iq is measured into the VDD pin without the GDRV signal switching an external MOSFET and measures what it takes to keep the internal control circuits active only.

Best Regards,
Anthony

• Anthony,

So, you think it is normal for the current consumption like this. Then, do you have any alternative suggestions other than use this chip? One of other WEBENCH recommendations is to use LM3478/LM3488. Could they require much less idling current to sustain 200V? I just need 10mA output with load.

Thank you again for your help.

Ning

• Changing controllers won't reduce the current needed to switch the external MOSFET. What is the Qg * fsw for your design? As can be seen from this equation reducing the switching frequency will reduce the switching current.

• Anthony,

The FET is FDB28N30TM. Total gate charge is about 50nC. I am using 820K and 470pF for RC. So, Qg*Fsw is about 50nC*150K=7.5mW.

Thanks a lot.

Ning

• Anthony,

Sorry, the fsw is about 50KHz not 150KHz. So, Qg*fsw is about 2.5mW.

• Thanks Ning. I guess the switching current isn't causing all of the input current then. Changing the controller should not have much impact on the idling current since the TPS40210 only makes up about 1.5mA.

Some other sources of power loss we need to consider are:

1. RMS current in the DCR of the inductor
2. RMS current sense resistor
3. Leakage current of the schottky diode
4. Leakage current of the output capacitors
5. FB divider current

The leakage currents and current the FB divider will add some load to the output even if there is nothing added externally. These could very easily add several milliamps of input current.

• Anthony

As what you said TPS40210 only makes up about 1.5mA.

I checked the specs:

1. RMS current in the DCR of the inductor: Inductor Irms=1.7A with DCR 0.086ohm
2. RMS current sense resistor:  ?, 0.02ohm
3. Leakage current of the schottky diode: Ir=100uA
4. Leakage current of the output capacitors: 10uA
5. FB divider current: 200/51.2K = 2mA

I do not see that the idling current can go 40mA without load. Can it be related to layout? I modified the TPS40210EVM to have my circuit built.

Thanks,

Ning

• Can we see your schematic?  Maybe we are missing something.

• Hi Ning,

This leakage current and feedback divider current adds a load to the output causing a lot more input current. Through the conversion ratio of 16 V to 200 V, the input current would be 26.4 mA assuming 100% efficiency. This would then contribute to more conduction losses in the inductor, MOSFET and current sense resistor. If the efficiency of supplying this 2.11 mA parasitic load was 50% it would mean the input current is 52.75 mA. These numbers are definitely in the range of what you're measuring. Increasing the resistance of the feedback divider will help a lot to reduce the input current.

Best Regards,
Anthony

• Anthony,

Yes, very good point. Then, what is the largest value I may use in the feedback divider, 510K&1.78K  instead of 51.1K&178ohm?

Best Regards,

Ning

• I don't see any issue with this change. You may have to change the compensation after changing the FB resistors as this changes the gain of the error amplifier. You should be able to simulate this with SwitcherPro or the TINA average model.

A general rule to get a starting point would be to decrease R3 by the factor the feedback divider was changed by. So with 10x the value of resistances, R3 should be 10x less or 1.96 kOhm from your schematic.