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# TIDA-00349: Isolated DC TO DC converter

Part Number: TIDA-00349
Other Parts Discussed in Thread: TPS60402

I have built circuit as per TI reference design TIDA-00349.

According to reference design, isolated output voltage is 5 V , 10 mA

our circuit gives 5 V on no load condition.

But for load (require 5 mA current) it output voltage drops down to 0.5 V.

Pl. suggest the solution for this issue.

• Hello,

what is your Vin in your case and are you using a transformer with the specified parameter in the Design Guide?

Regards

Alex

•

Thanks Alex

In my case Vin is 2.5 v.

Transformer used in this reference TIDA-00349 is made by WURTH ELEKTRONIK  who has following details.

PART NUMBER : 750314839

NAME : MID-HB Half-Bridge Transformers

we have taken 5 samples from WE.

Pl. find attached Transformer details

regards,

shashikant

TXR_314839.pdf

• Hello Shashikant,

The design is specified for input voltages between 3V to 5.2V. The output voltage  is highly dependent on this input voltage, as well as the turning ratio of the transformer. In our Design Guide, Table 1 gives you very good overview what output voltage for a given input voltage and output current can be expected.

This means, to achieve an isolated output voltage of 5V at an output current of 10mA, you also would have to provide 5V at the input.

The reason for not achieving the 5V is basically due to the low input voltage.

Regards

Alex

• Thanks Alex,

I have given 5 V input , on no load, it gives 10 v output and on load ( 5 ma current), It gives 2.8 V.
One more thing, I want to tell you that Instead of diode RB520S30 , I used SS210, due to availability problem.
Whether it will affect output ?

regards,

shashikant
• Hello Shashikant,

is it possible for you to use the same BOM as in our TI design? The diodes on the isolated side have a big impact on the overall efficiency performance. This is due to the forward voltage (Vf) of the diode. The smaller Vf is, the less voltage drop you have, thus increased efficiency. Are all other component values in your design the same  as specified in our BOM? Otherwise it is complicated to compare against our results.

Basically, the output voltage is approximately Vout=Vin/n-2*Vf, where n is the xformer turn ratio (here: 0.8).

Please find below an extract of out test data. Here you see the input / output voltages and currents we have measured for the calculation of the efficiency.

 Vin [V] Iin [A] Vout [V] Iout [A] Eff [%] 5.2 0.00016 8.89399 8.00E-07 0.85306 5.2 0.00016 9.15826 9.00E-07 0.992538 5.2 0.00016 9.14399 1.00E-06 1.101102 5.2 0.000161 8.53529 2.00E-06 2.039009 5.2 0.000163 8.07451 3.00E-06 2.86669 5.2 0.000164 7.83973 4.00E-06 3.677172 5.2 0.000165 7.65269 5.00E-06 4.451516 5.2 0.000167 7.48816 6.00E-06 5.186185 5.2 0.000168 7.34713 7.00E-06 5.887123 5.2 0.000169 7.23095 8.00E-06 6.570903 5.2 0.000171 7.13005 9.00E-06 7.23357 5.2 0.000172 7.03892 1.00E-05 7.879153 5.2 0.000185 6.49347 2.00E-05 13.50724 5.2 0.000198 6.31684 3.00E-05 18.43364 5.2 0.00021 6.24488 4.00E-05 22.86413 5.2 0.000223 6.20485 5.00E-05 26.80234 5.2 0.000235 6.17866 6.00E-05 30.31132 5.2 0.000248 6.15943 7.00E-05 33.47412 5.2 0.00026 6.14459 8.00E-05 36.33057 5.2 0.000273 6.13229 9.00E-05 38.90608 5.2 0.000285 6.12207 0.0001 41.28055 5.2 0.00041 6.06417 0.0002 56.87328 5.2 0.000535 6.0345 0.0003 65.08585 5.2 0.00066 6.01419 0.0004 70.12733 5.2 0.000785 5.99852 0.0005 73.52209 5.2 0.000909 5.98556 0.0006 75.94475 5.2 0.001034 5.97455 0.0007 77.75948 5.2 0.001159 5.9649 0.0008 79.17834 5.2 0.001284 5.95613 0.0009 80.29197 5.2 0.001409 5.94818 0.001 81.1954 5.2 0.002658 5.88952 0.002 85.23159 5.2 0.003907 5.8474 0.003 86.35607 5.2 0.005155 5.81182 0.004 86.7208 5.2 0.006404 5.77976 0.005 86.77974 5.2 0.007653 5.74998 0.006 86.6961 5.2 0.008902 5.7216 0.007 86.52453 5.2 0.01015 5.6944 0.008 86.30808 5.2 0.011399 5.6681 0.009 86.05943 5.2 0.012648 5.64247 0.01 85.79078 5.2 0.02514 5.40484 0.02 82.68932
• Hello Shashikant,
Alex is right with his recommendation to use the components as given in the original BOM of the design. This is especially true for the components which have parasitic leakages and capacitances, as for example the diodes on the secondary side. The RB520S30,115 was especially selected for that reason. You might also test a PMEG4002ELD or a SD103AWS-7-F which perform similarly.
Nevertheless I would guess that there is another additional reason for getting only a VOUT of 2.8V when powering the design with 5V VIN. Can you please check whether the transformer is connected correctly? There is a dot on the transformer core as marking for terminal 1 which helps also to identify the primary side of the transformer. Please see also the transformers datasheet and the board image as given on the 1st page of the TIDA-00349 user guide.
For me it looks as the primary winding and secondary winding are interchanged in your design. This would change the primary-to-secondary turns ratio from 0.8 to a value of 1.25.

Using the equation
Vout=Vin/n-2*Vf
as given by Alex would result in

Vout = (5V/1.25) - (2*0.35V) = 3.3V when using a primary-to-secondary turns-ratio of 1.25.

The 0.35V is the assumed Vf of the SS210 diode used by you. The calculated 3.3 V differs still from the 2.8 V measured by you - but goes at least in the right direction of your test result. A small part of the difference is based on the resistive losses in the TPS60402 and transformer winding. A major part of the difference can be caused by the large junction capacitance of the used SS210 diodes, which is roughly 10 x larger than the junction capacitance of the RB520S30,115.

Regards
Bert