LM5025 and LM5026 application

Hi Sulyn,

I have asked one of our applications engineers to respond to your post. But I suspect that at light load the output is peak detecting the voltage from the secondary winding and charging to the higher output voltage level. One way to reduce the output variation is to add a pre-biasload to the output. Alternatively it may be possible to regulate from an auxiliary winding on the primary side.

Regards

Peter

Hi Sulyn,

1, How did you implement the open loop? Can you show the circuit connects to COMP pin?

2, Do you mean output voltage is 40V at 0A load, and it reduced to 24V at 0.1A?

Regards,

Bosheng

Hi Bosheng,

Thanks for your support! As for your questions:

1, How did you implement the open loop? Can you show the circuit connects to COMP pin?

[Sulyn]: Yes, customer use open loop as they can't use close loop with isolator here. Pls see below attached schematic now, the COMP pin is floated, if you have suggestion, pls let me know. Thanks.

2, Do you mean output voltage is 40V at 0A load, and it reduced to 24V at 0.1A?

[Sulyn]: It's about like this.

Best regards,

Sulyn Zhang

Hi Sulyn,

The customer wants to use LM5025 to generate a fixed duty cycle for the open loop, is this correct? The PWM duty is generated by comparing RAMP signal to COMP signal. Add a resistor from COMP to ground can program the duty cycle. With a fixed duty, the output voltage keep getting charged, if there is no load, the voltage will be charged up. To prevent this voltage from shooting up, you need either add some minimum load, or disable PWM during zero load condition.

Regards,

Bosheng

Hi Bosheng,

Many thanks for your support! Yes, customer want to use open loop. It seems we can realize this by using LM5025, so if customer want to realize fixed 60% duty-cycle(Hope the output voltage variation won't change too much, 28V~30V variation output are acceptable) with open loop configuration, So the questions are:

1. What the resistor value recommended from connecting COMP pin to GND? 

2. What the minimum load needed here(I can let customer add a dummy load) to make output voltage stable during operation?

Best regards,

Sulyn Zhang

Sulyn,

In normal operation, an external opto-coupler sinks current from COMP to control the duty cycle. Now you can use a resistor to do the similar function, you will need trial to find out the correct resistor value. Same as the minimum load, you need trial.

 

Regards,

Bosheng

   

Hi Bosheng,

We tried to add a pull down resistor at COMP pin, customer's target is 67% duty cycle. So we choose about 7.9kohm resistor here, calculation as below:

[R/(R+5k)*5V-1V]=2.5V(0.67/0.8), we get R=7.9kohm. Here the 0.67 means 67% duty cycle, 0.8 means 80% duty cycle. But with 7.9kohm pull down resistor at COMP pin and also adding dummy load resistor realize 5mA load current. But with this configuration, we got >50V output voltage!! These should be something wrong.

So Is above calculation be correct? If not, can you help let me know how to calculate the pull down resistor here and what's the recommended pull down resistor here?  Below I attached the schematic(the pull down resistor is COMP pin is not shown here):

As this is very important project at this big customer side and it's platform socket. So need  your great help on this asap, thanks a lot!

Best regards,

Sulyn Zhang

Hi Sulyn,

The PWM is generated by comparing COMP with RAMP. The slope of RAMP is proportional to Vin, plus the RAMP will also be reset if it exceeds the 2.5V volt x second clamp, therefore it is difficult to calculate the duty cycle. The customer is using the device in a very special way which it is not designed for, they need trial to find out the correct resistor value.

Since it is an open loop with a fixed duty cycle, the output capacitor is charged every time the switch turns on, you need to put enough load to consume this power, otherwise the output will be charged up.

Regards,

Bosheng

 

Hi Bosheng,

Thanks for the support. Knowing that customer is using it in a special way. But can you help me understanding how to calculate the R or give me a direction how to chose the R at COMP pin? As for the dummy load, how about load current is needed to make output stable? In this test, we are using 5mA, but what's about the dummy load current based on  your opinion?

Best regards,

Sulyn Zhang

Sulyn,

Please refer to the functional block diagram on page 11 of the datasheet, you need to measure the RAMP magnitude. Assuming it does not exceed 2.5V, the intersection of (COMP -1V) and RAMP determines the duty. Here -1V is because there is a 1V offset on COMP pin. Then you can figure out how much COMP signal needs to be. The COMP signal comes from the internal 5V through a voltage divider, the upper resistor of the divider is 5K, then you can calculate the value for bottom resistor.

The dummy load really needs to be determined by trial.

Regards,

Bosheng

Hi Bosheng,

Thanks a lot for your support, it seems we are close to the calculation result. I tested at customer side and we got the 67% duty cycle by using about 12kohm resistor. So based on your calculation method:

But strangely that we use 12kohm resistor, if we calculated it, it should be 12/(12+5)*5V=3.5294V, but we got 3.32~3.34V at the COMP pin(Is this caused by the internal 5kohm resistor? Which not accuaracy, the external 12kohm resistor is 0.1% accuracy). If so, then COMP-1V=2.32~2.34V,  and suppose that we know the voltage on RAMP pin, then we can get:

[(COMP-1V)-RAMP]*X=67%

So what is the "X" here? It's 80%? or 100%? Thank you.

Best regards,

Sulyn Zhang

Sulyn,

The difference between the measured COMP and your calculation might be caused by the tolerance of internal 5Kohm resistor.

The duty should be calculated as: duty = (COMP-1)/RAMP

Regards,

Bosheng

Hi Bosheng,

Thanks for your support. Strangely that we measured that the voltage of COMP pin is about 3.32~3.34V; The voltage of RAMP pin is about 2.6V, but the duty cycle is 67%. So

(Comp-1)/RAMP=(3.32-1)/2.6=89.23%, not 67%?!

Anything wrong here? Thanks a lot for your great support!

Best regards,

Sulyn Zhang

Sulyn,

The equation I gave you assuming the RAMP signal does not exceeds the volt*second limit, but yours already exceeds it. In this case you can get the correct duty by trying different resistors. You need also make sure CS1 is not triggered, it works as CBC and also affect the duty cycle.

Regards,

Bosheng

  

Bosheng,
Many thanks for clarify this, but with the RAMP voltage exceeds the Volt*second limit, can you pls help give some guidance and caculation method for calculating the duty cycle? I can't tell customer that you just try... and customer won't have the confidence to use if no caculation theory support, thanks a lot!

Best regards,
Sulyn Zhang

Bosheng or anyone,

Can you help give some guideline to calculate this duty cycle?  As Customer will use LM5025 as platform socket, so it's important to make clear why or why not, otherwise they won't take risk to use this solution. Thanks a lot for your support.

Best regards,

Sulyn Zhang

Sulyn,

You can try this: You have the RAMP waveform, extend it to the end of period, then use duty = (COMP-1)/RAMP to calculate the duty.

Regards,

Bosheng