# Heat Dissipation and Max Frequency Question

1. Apr 1, 2014

### sailmike

I have two questions, one is how to calculate the heat dissipation from M1 and M2, and the second is how to calculate the maximum frequency M2 can handle. The LED is a Cree XLamp XQ-D running at 3.1V and 350mA. There is a 3.1 volt drop across the LED leaving 1.9V across M1, M2, and R2. I did find the heat information on the data sheets, I just don't know how to use them. My guess is 1.9 volts times 350mA times 65 degrees C/W giving 43.225 watts. I don't know if this is right. I've attached a picture of the circuit and the data sheets for M1/M2 and the LED.

Thanks a lot,

Mike

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2. Apr 1, 2014

### meBigGuy

Since M2 just switches on and off, its power dissipation will be (Id^2)Rds while it is conducting, and 0 while it is off. The power while switching is the integral of Id*Vds (where both Id and Vds are changing as it turns on)

M1's power will be Id*Vds while it is conducting, where Vds is 5V-Vdiode-Vr. Vdiode and Vr are determined by Id.

So, the steady state power in M1 is 350ma * (1.9V - .5) = 0.49 watts. (ignoring the drop across M2)

The M1 temp rise above ambient is 0.49W*65C/W ~= 31.85 C.

3. Apr 2, 2014

### sailmike

Is that a double integral? One is from 0 to 350mA and the other is from 0 to (1.9 - .5)? Ah, it's plus ambient. That I didn't think of.

And how do I calculate the maximum frequency M2 can handle?

Thanks a lot,
Mike

4. Apr 2, 2014

### meBigGuy

Not sure what the max frequency would be. Probably dominated by the 25K resistor charging the M1 gate (30 to 40 pf). Time constant is 1uS.

When the circuit is off for a bit, the voltage on the gate of M1 will raise to 5V. When M2 turns on, there will be a HUGE current spike (D1 in series with 1.4 ohms driven by 5 volts) for the time it takes Q1 to turn on and limit M1's current. When M2 turns off, M1's gate voltage rises again. So there will always be a turnon surge current.

5. Apr 2, 2014

### meBigGuy

You posted while I was writing. LOL

M2 switching will be determined by how fast you can change the gate voltage. Power fets (these are 7.9A fets) have high gate capacitance. That is what limits their speed, generally. Put some resistance in series with V2 and you will see what I mean.

But, I'm not entirely sure. You have the simulator. You will probably want to decrease R1 (but I'm guessing).

6. Apr 2, 2014

### sailmike

For this simulation I'm using the PHT6NQ10T, which has a Vth of about 3.8V and the MOSFET I want to use has a Vth of 1.6V so I'll be raising R1 to 50k for that. This FET has an input capacitance of 360pF. Is this high?

Where's V2? Do you mean the input to M2? I'm not seeing the current spike you mentioned. I moved the current probe to various spots and don't see any current spike. Also, when I put the voltage probe between D1 and M2 I don't see a clean square wave as seen in the attached screenshot. How do I clean this up clean this up?

Thanks,
Mike

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7. Apr 2, 2014

### meBigGuy

I can only guess what you are seeing. It would help to also see the voltage on gate of M1, M2 and the current through the diode. Then I could explain.

I *think* what I am seeing is that the you turn on M2 at the large negative drop. This will be the max current time. Then the thransistor turns on and M1 starts to current limit, so the voltage rises. Then you turn off M2 and the voltage shoots up. Set the frequency to 250Khz and you will get a better idea of what is really happening. But, show all the waveforms.

360pf is not the gate capacitance. Look in the graphs for gate to source capacitance.

The 25K has nothing to do with the threshold voltage of M1. Increasing to 50K will only make things worse, BTW.

Why do you want to run this at 1Mhz?

8. Apr 2, 2014

### sailmike

My internet went down last night, so I couldn't reply sooner.

I'm experimenting with flashing the LED's one at a time at high speed so it looks like all the LED's are on at the same time and still putting out a lot of light. This will go into a 25 LED flashlight. I know that the faster the LED's flash, the brighter and more intense they will be. Learned that in one of the salt water aquarium forums.

Ok, I got some screenshots. Hope it's enough to paint a better picture of what's going on. This circuit will have a 5 volt regulator, so just a resistor to limit the voltage and current may be enough rather than using a constant current source. I'd like to better understand this circuit though. I wonder if there is another kind of switch I can use that will work better than a MOSFET?

Oh, the gate voltage of M2 is just a 0-5 volt square wave.

Thanks a lot,
Mike

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9. Apr 2, 2014

### mjhilger

I think I'm misunderstanding your intent. I believe an LED puts out its max light at a particular current specified by the manufacturer. The ability for the light to continue to operate at that current level is dependent on the substrate keeping below a particular temp to prevent damage to the structure of the LED. You can pulse the LED with a very intense current well above the max continuous level specified by the manufacturer for short periods, provided that you do not exceed the max voltage and the max operating temp. Pulsing allows the designer to control the brightness by coming down from the max output. But I don't think you can exceed the light intensity output by pulsing it as opposed to straight level DC driven. The only difference is the ability to provide control easier and through several at the same time in a more efficient switching manner - so switching just makes it easier to provide control. But not overall any brighter.
If I'm wrong I hope someone else chimes in and corrects me.
Most switching for LED's is in the KHz range so as to not take the switching time power loss (and heating of that transistor to the point that it would need a heat sink) that MeBigGuy pointed out earlier. So put a big heat sink on the LED's that you are driving and drive them to their max at a reasonable KHz switching speed and I think you will be more than happy. MeBigGuy has given you great information on the circuit and its behavior.

10. Apr 2, 2014

### sailmike

Actually, the maximum current for these LED's is 700mA. I'm just running them at the current they will naturally take for a given voltage. My goal is to flash 25 LED's one at a time so that only one is on at any time rather than all of them at the same time. It seems logical to me. The flashing will be controlled from 3 cascaded decade counters and a couple of AND gates. It's basically a high speed chaser. If you don't think this will work, please explain.

Thank you,
Mike

11. Apr 2, 2014

### meBigGuy

Can't you do multiple traces at one time? if you want to understand what is really happening I need a waveform that shows what I asked for and a switching rate of 100KHz or so. If you really want to understand what is happening, display the startup cycles as the gate voltage ramps down to an average value.

I think you are switching rapidly after the circuit has somewhat stabilized and not seeing all effects.

As for flashing, The eye integrates the light so an LED switching 50% at 20ma is the same brightness as an LED running at 100% 10ma (assuming similar LED efficiency). The output curve for the LED seems pretty linear to me.

The main reason people switch (besides multiplexing applications) is to produce variable brightness without high power dissipation in the switch. The switch is either on or off. There is NO advantage to switching faster.

Are you going to have "many M2's" each driving an LED (shared M1)? Or are you going to build 25 M1-M2 pairs?

The brightest output will be 25 LED's, each running at X ma continuously. (where X is below the 700ma limit).

You do realize that reducing the 1.37 ohm resistor will increase the current limit?

12. Apr 3, 2014

### sailmike

I'm planning to use a single M2 transistor for each LED. It seems to me that, since the supply voltage will be held constant via the voltage regulator, I won't need the constant current source and could just use a resistor. I'm not sure what you mean by multiple traces at one time.

I got the screenshots I think you asked for.

You've helped me a lot meBigGuy.

Thanks a lot,
Mike

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13. Apr 3, 2014

### meBigGuy

something is not right.

M1 is supposed to be a current limiter. When the current across the 1.4 ohm resistor is enough to turn on Q1, it decreases the voltage on the gate of M1, which reduces the current. It finds a balance and limits the current.

But, with 350ma, there is not enough voltage to turn on Q1 350ma* 1.4 = .49 volts.

So, the capacitive coupling from M2 gate shoot-through must be causing the M1 gate to rise., and vice versa.

To see the current limiter actually work increase R2 until the on current drops in half. Then you will have real waveforms of what the circuit is supposed to do.

One side effect of FETs is that there are good sized capacitors from the gate to source and drain. When the gate switches, this signal "shoots through" the capacitor. For example, the spike on M2 drain is caused by the 5V signal on its gate. If you put a resistor is series with M2 gate to slow it down, you will see what I mean.

As far as the waveforms go, I thought you could probe 3 or 4 points on a single display. If you can't, so be it.

14. Apr 3, 2014

### sailmike

I can put multiple probes at the same time. I just forgot I could do that! lol

There's a voltage drop across M1 and M2, so the resistor has to overcome those drops to put enough voltage on the base of Q1. If I built this circuit as seen in my previous post, the current could rise considerably before Q1 turns on and brings it back down to 350mA? With R2 at 3 ohms, the waveforms appear the same as before. I just looked up 2N2222 and it has a Vbe minimum of 0.6 V. The formula I used to calculate R2 = 0.5/I is therefore, wrong and should be 0.6/I or 0.7/I. The transistors M1 and M2 have a fairly high Vth, around 3.5 to 4 V. With this in mind, I played around with the values of R1 and R2. With R2 at 1.71429 and R1 at 1k, got a current of 325 mA and a voltage at the base of about 558 mV.

Thanks,
Mike