Control design for PWM converter

[Hubert Tchio]

Hi Everyone,

here is Hubert again. Here attached is my design and the problems I am facing.

Can anyone be of help?

Thanks in advance.

Best regards
Urbain

 

[Svein]

The main difference is the removal of the reference voltage.

If V_LED<1.5V, no current flows through the LED, the output of the INA210 is 0V - and that is what you use as a reference voltage.

 

[Hubert Tchio]

Hi Svein,
thanks for your fast reply, so how can i add the INA210 in my circuit into for the circuit to be functional as required?

 

[Svein]

Start with your first circuit. This has V_LED as a feedback - when it increases, it is regulated down. If you look at the output of INA210, you have the same polarity - when the output increases, you want V_LED to go down. Thus - sum V_LED and the output of INA210 using a resistor in series with each. Experiment with the relation between the resistors. Or - insert a 10k potentiometer with V_LED and the output of INA210 at each end and the slider to the op-amp input.

 

[Hubert Tchio]

Hi Svein, thank but i don't get it.
please can you just give me a rough sketch of what you mean?

 

[Svein]

Like this. Use your first circuit, add the INA210 and connect the output as shown here.

 

[Hubert Tchio]

Hi Svein,
thank you for your response, but i believe adding a 10K in the circuit also increase power loss thereby reducing the overall efficiency of my design (correct me if I am wrong).
I wish to have a design with high efficiency.

 

[Svein]

thank you for your response, but i believe adding a 10K in the circuit also increase power loss thereby reducing the overall efficiency of my design (correct me if I am wrong).

1. The regulator system tends to minimize the voltage across the potentiometer.
2. Even with a voltage of 1V across the potentiometer, the current through it is negligible compared to the power draw of the rest of your circuit.

If you are really into high efficiency, read the application notes from Linear Technology - for example: http://cds.linear.com/docs/en/design-note/dn310f.pdf

 

[Hubert Tchio]

Hi Svein,
Thanks. I will look at the document and get back to you.

Best regards
Hubert

 

[Hubert Tchio]

@Svein,
thank you so much, your idea works perfectly and the problem is solved.

My second problem is still without a solution. Can anyone help?
how to use another current monitor to measure the LED current and use it as feedback to the temperature measurement? The Goal is to fix a maximum temperature for the LED.

Best regards
Hubert

 

[Baluncore]

Secondly, I indeed to add a temperature measurement as an additional input but I don’t know how to do this. What I mean is to put another current monitor to measure the LED current and use it as feedback to the temperature measurement. I don’t intern to use a temperature sensor.
Can anyone be of help?

As I understand it, you intend to use the LED current and LED forward voltage to identify the LED die temperature.
Will you then back-off the current to limit LED temperature rise?

What make and model LED are you using?
Can you please post your latest schematic model.

 

[Hubert Tchio]

Hi Baluncore,

Thank you once more for your time.
Yes you understood what my plans are. Indeed, i will then back off the current to limit LED temperature rise.
I am using an OSRAM LUW HWQP LED "3.25V and 1A "

Here attached is my lastest schematic. I can also feed VLED directly to the Error amp. Vref is my precison output voltage.

Thanks in Advance.

Best regards
Hubert

 

[Baluncore]

Re: OSRAM LUW HWQP data sheet;
http://www.osram-os.com/Graphics/XPic4/00151034_0.pdf/LUW HWQP - OSLON Black Flat.pdf

With I_fwd = 1A and V_fwd = 3.35V the LED will have 3.35 W power dissipation.
Knowing from the table on page 4 that the thermal resistance is 5.5°C / W.
We can expect a temperature rise of 5.5°C * 3.35W = 18.5°C
To limit junction temperature to 125°C requires the environment remain below about 105°C.

Relative Forward Voltage versus temperature is shown at the top left of page 10.
It shows that relative to 25°C, at the maximum operating temperature of 125°C, the voltage will have fallen by only 0.18V
This is less than the manufacturing tolerance grades, 8F, 8G & 8H listed at the bottom of page 5.
The Characteristics table on page 4, shows a 0.75 volt manufacturing spread.

It will therefore not be possible to determine temperature from V_fwd without a calibration process.
You will probably need to calibrate every LED with short "on" pulses to measure V_fwd at 25°C.Relative spectral emission, at the top of page 8, shows a peak at 440nm in the blue.
Knowing that " Energy = Plank's constant * frequency " we can compute V_fwd = 1240 / 440nm = 2.818V.
Ignoring the Gaussian thermal energy spreading expected, that should be the intercept of the I_fwd curve with V_fwd axis.

Looking at the graph of I_fwd versus V_fwd, (top left of page 9), we see it passes through the point 3.4V at 1.4A
The series resistance of the LED will be Rs = (3.4V - 2.818V) / 1.4A = 0.416 ohms
We can predict V_fwd at I_fwd = 1A as being 2.818V + ( 1.0A * 0.416R) = 3.234

The graph shows that it is a close enough model.
So the LED can be modeled as a 1.818V bandgap in series with a series resistor of about 0.416 ohms.

 

[Hubert Tchio]

Hi Baluncore,

Thank you for your time and effort put together to help me. But I must admit, i do not get it.
Are you saying my actual schematic is enough? should I just change the series Resitor to 0.416 ohms and the work is done?

Or is there anything to do next?

Best regards
Hubert

 

[Baluncore]

Are you saying my actual schematic is enough? should I just change the series Resitor to 0.416 ohms and the work is done?

No.A model of the LED you are using looks like a voltage step in series with a resistor having a value of about 0.416 ohms.
I am saying that it will be very difficult for you to implement LED die temperature estimation by monitoring LED V_fwd and I_fwd.
I would not take it on without a micro-controller to first calibrate the LED model and then monitor the model on the fly.

 

[Hubert Tchio]

Hi Baluncore,

thanks for your reply. Permit me to ask a foolish way. Do you think there is any other way to implement LED die temperature on the model of the LED i am using?
I am still trying to figure that out.
Thanks in advance

 

[Baluncore]

To limit junction temperature to 125°C requires the environment remain below about 105°C.

Do you think there is any other way to implement LED die temperature on the model of the LED i am using?

By independently monitoring LED heatsink or substrate temperature, you can reduce the duty cycle, the LED current and so LED power dissipation as the environmental temperature approaches the limit of 105°C. If the change begins only at 95°C and takes place so that current falls linearly to zero at 105°C, then the LED will be over-temperature protected. That might be achieved by reducing the current reference voltage at high temperatures.

 

[Baluncore]

@ Hubert Tchio. Looking at your circuit diagram, you have X2, a high-side P-channel MOSFET switch. N-channel MOSFETS have lower on-resistance, less drive requirements and switch faster for the same price. I believe you should invert your design to use the lower cost and more efficient switch.

You drive the gate of X2 with a voltage comparator, U39, hopefully it has some feed-forward to speed up the transition. But does it have the drive needed, (about 100mA), to switch X2 quickly through the inefficient transition?

 

[Hubert Tchio]

@Baluncore . Thanks for your feedback. I will try the circuit with a high-side N-channel MOSFET switch to see what i get.
So far the ciruit is ok. The voltage comparator compares my control voltage and my ramp input to provide a PWM signal at the output for driving the X2.

My main problem now is still on how to implement a temperature measurement. I have decided to go for an LM335 IC temperature sensor or themistor, I have built a thermal design, it seems ok but I still don't know how to integrate this thermal design to my current LED driver. It would have been easy if my thermal design switches off (input to ground) my LED driver when the temperature is too high but rather the thermal design should reduce the forward current instead.

Still looking for help.

Best regards
Hubert

 

[Baluncore]

To taper the LED current from a maximum at 95°C to zero at 105°C you can use a couple of LM334 Current Sources. They are available in 3 terminal TO-92 package. Get the data sheet from; http://www.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=lm334&fileType=pdf

Set up two in series, between the supply rails. (See the attached sketch). One configured as a temperature sensor, see page 6, fig 13; set it to 1uA per °K. The other as a zero tempco source, see page 8, fig 15; set it to (95°C+273°K) = 368uA.

Now place the two together in thermal contact with the LED mounting. A difference current is available at the junction node of the two. Take that current through a signal diode to a resistor connected to the current sensor output. Once the temperature reaches 95°C the diode will conduct and add voltage to the current sense voltage. The current will be reduced as a result. Select the resistor to give zero LED current when 10uA flows. For 1A LED current, (giving a 2 volt signal), use R = 2V / 10uA = 200k.

 

[Hubert Tchio]

Hi Baluncore,

thank you very much, i haven't had time to look at it. I will do that very soon and get back to you.
Once more thank you.

Best regards
Hubert

 

[Hubert Tchio]

@Baluncore,
here is Hubert again. I just want to make sure i undesrtood your last mail. Please when you have time can you go through the attached documents?
I will be waiting for your feedback.
Have a great day!

Best regards
Hubert

 

[Baluncore]

You seem to have understood the concept. Now just to fix the numbers.

The LM334 temp sensor setting resistor is usually specified as 226 ohms for 1uA / °K. I don't follow your 83.4K.
At 25°C = 298°K, the reference voltage of the LM334 is effectively 67.7mV.
For 1uA / °K, R = 67.7mV / 298uA = 227 ohms.

The two LM334s and the zero tempco diode should all be kept at the LED heatsink temperature, calibration then only requires adjustment of the temp sensor resistor to match the reference current at 95°C. The choice of 1uA / °K is arbitrary, but is probably sufficient.

A temperature of 10°C over the balance point at 95°C must fake the equivalent of 1A LED current. The INA210 has a gain of 200. If your LED current sense resistor is still 0R025 then 1A will give 0.025 * 200 = 5.0V which is too close to the rails. I would make it closer to 2.5V where the LM334s will be balanced when needed. You need to halve the sense resistor or change to the INA214 which has a gain of 100. Anyhow, back to the value of R. 10uA must drop 2.5V so R = 2.5 / 10uA = 250k.

 

[Hubert Tchio]

@Baluncore
Thank you so much, here is an update.
I hope i get it this time around. Have a great evening.

Best regards
Hubert

 

[Baluncore]

(A). You need the current summing node of the circuit to be away from the supply rails so as to give the LM334 some operating voltage. That is why I suggest 2.5V. The lower LM334 is in series with a diode so it requires about 1.5V to operate. The current summing diode gives a little more headroom by moving the voltage of the summing node up. Maybe use two or three signal diodes in series to raise the LM334 voltage? The current summing node is only important while operating in balance. Your R_sense 0R004 is a very low resistance to use as a shunt. It will need to be an expensive 4 terminal resistor so as to separate the effects of resistance in the main current terminals. You could make your own 0R004 by soldering two voltage taps onto a shunt wire. I would use R_sense = 0R0125.

(B). That is better, but you will adjust the value to trim the system to cut-off at 105°C once it is built.

( C). Close enough. Any variation in the individual circuit will be canceled by the adjustment as in (B). It does not really matter if I_LED cuts off over a range of 5°C or 15°C, so long as the final shut-down occurs at or below 105°C.

(D). Correct logic, questionable current summing reference voltage.

(E). Summing up the current difference.
You have drawn your “two terminal zero tempco circuit” upside down.
I would use 1N4148 signal diodes. 1N4001 power diodes are not a good choice for either diode in that circuit as it may have a few uA of temperature variable leakage and a lower forward voltage. Also ask yourself why the ti datasheet shows a 1N457 signal diode in the zero tempco circuit example. You need to use a diode with low reverse leakage for the 95°C current polarity switch, it will be in a regulated temperature environment.

 

[Hubert Tchio]

@Baluncore
I am completely confused. I don't get a bit of it anymore.
I started my design with 0R0250 but was not working until I had to regulate Rsense to 4m ohm. I have tried to change Rsense to 0R0125 or more but my signals are unstable.
I don't undestand what type of current summing node you mean.

here attached is my design, i don't know you can if you can do something on it.
The thermal design is not working.

Thanks in advance
Hubert

 

[Baluncore]

What are manufacturer and part numbers for U61, U62, X2, D1, D2 ?
What is frequency, amplitude and offset of the “Triangle” input to U62 ?

 

[Hubert Tchio]

U61 and U61 are LMH6622 from TI
D1 is LUW_HWQP LED from OSRAM
D2 is a normal doide from Pspice
Vramp = 1.8V (3.5V - 1.7V) with 455KHz frequency

 

[Hubert Tchio]

sorry U61 and U62 are the same. Vramp = Triangle

 

[Baluncore]

Attached is the switch, LED current sense and temperature protection sections only. The output will be 2.5V when operating correctly. The LM334 model used has an error, so ignore my resistor value detail in setting my over temperature protection currents.
I will try to get the rest of your circuit into the model later but I'm very busy today.

 

[Hubert Tchio]

@Baluncore,
Thank you so much for your time and the assistance. I am very grateful. I also understand you are very busy.
I will be waiting to hear from you when you are less busy. I will try to simulate this circuit and see the outcome.
Once again thank you.

Best regards
Hubert

 

[Hubert Tchio]

@Baluncore
Hope you had a nice weekend, just wish to know if you were able to complete the model as desirer.

 

[Baluncore]

@ Hubert.
I have got my spice model of the LM334 operating close enough to the data sheet. We need to have similar models.
What spice model are you using for the LM334 ? Can you please post the text in your LM334.sub file?

 

[Hubert Tchio]

@Baluncore
*Selfmade model of the LM334 from National Semiconductor or ST
* Helmut Sennewald
*
* Pin order: v+ v- R
.SUBCKT LM334 v+ v- R
Q4 N002 N001 v+ 0 pnp1
Q5 N003 N001 v+ 0 pnp1
Q6 R N001 v+ 0 pnp1 14
Q1 N003 N003 R 0 npn1 14
Q2 N002 N003 N004 0 npn1
Q3 N001 N002 R 0 npn1
C1 N002 N004 50p
R2 N004 v- 1
R8 v+ v- 1G
C2 v+ v- 10p
C3 N003 v- 1p
C4 R v- 1p
.model npn1 NPN(Is=1e-15 BF=200 TF=1e-9 Cjc=0.5e-12 Cje=1e-12 VAF=100 Rb=100 Re=5)
.model pnp1 PNP(Is=1e-15 BF=100 TF=1e-7 Cjc=0.5e-12 Cje=1e-12 VAF=100 Rb=100 Re=5)
.ENDS LM334
*

 

[Baluncore]

Q6 R N001 v+ 0 pnp1 14
Q1 N003 N003 R 0 npn1 14

The area parameter of "14" at the end of the Q6 and Q1 lines is wrong. That 14 is from a different ratio parameter specified in the data sheet.
You should change the Q6 line to end in an area of 16. You should change the Q1 line to end in an area of 12. The model will then be much more real.

What INA210.sub model do you use ?

Now you can design your temperature compensation circuit. Start here:
1. Make a simple circuit that uses an LM334 that runs on a 2.5V supply. It must source a current of about 1uA per °K. i.e. 273.15uA at 0°C.
You will need to use the .TEMP 0 spice directive while checking current at 0°C, .TEMP 25 at 25°C = 298uA and .TEMP 95 at 95°C = 368uA.
You will only need one resistor, what value does it need to have?
2. Make another LM334 circuit, with a 1N4148 diode, also on 2.5V supply. It should sink a temperature stable current of 368uA at 95°C, i.e. the same as the temp sensor, but it should be reasonably fixed at 368uA across the temperature range. Use the .STEP TEMP 0 100 10 to plot the current, trim the two resistors in the circuit to get the zero tempco. Record the two resistor values.

Let me know when you have those three numbers so I can check your model is working OK.