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One of the key elements of electronics design beyond the elementary workings of the circuits is the thermal management and the EMC (Electro Magnetic Compatibility). On a 3D-printer controller board, the MOSFETs tend to be one of the major contributors to both the EMC part as well as the thermal management part.

For Replicape Rev B, I’ve opted to cut out the gate drivers on the MOSFETs in favour of having the gates driven directly form the PWM driver (PCA9685). The main reason for this was to cut down cost in order to reach the target price of $99 MSRP, but this was only after careful consideration and calculation of the expected heat dissipation from the MOSFETs driving the extruder heaters and heated build plate. I do not recommend doing this if you are going to drive the MOSFETs directly from a BeagleBone Black. It has a mere 6 mA source current and 4 mA sink current on the output pins as well as only 3.3 V output causing the MOSFETs to spend more time in transitions and (most likely) have a higher Rdson value due to the lower gate voltage.

The first prototype board is now back, thank you Elmatica for the quick service and supberb PCBs! I can compare the calculations to actual measurements to better calculate the heat dissipation from each of the MOSFETs. The strange thing is that the rise time across the load did not change for the worse as expected, instead it improved! Considerably!

Here is a picture of the layout of the PCB. The MOSFETs are those three in line with the fuse. The fuse is the yellow thing.

Here are some of the important parameters:

PWM frequency: 1 KHz

Replicape Rev A4A:

Gate voltage: [latex]12 V[/latex]

Rise time measured across load: [latex]50 ns[/latex]

Fall time measured across load: [latex]200 ns[/latex]

[latex]R_{DS(ON)}[/latex] at [latex]12V(GS)[/latex] is [latex]3 m \Omega[/latex]

Replicape Rev B1:

Gate voltage: [latex]5 V[/latex]

Fall time measured across load: [latex]50 ns[/latex]

Rise time measured across load: [latex]50 ns[/latex]

[latex]R_{DS(ON)}[/latex] at [latex]5 V(GS)[/latex] is [latex]4 m \Omega[/latex]

**Why has it improved?**

Looking at the schematic for the Rev A4A and the Rev B1 board, the difference is the lack of a gate driver in the latter. The gate drivers from the Rev A4 board has a drive capacity of 500 mA both source and sink. This is great if the load that is being switched is in close proximity to the MOSFET, but on a 3D-printer, the load is usually located about a meter away from the board, causing the heater cable to act as an antenna. Therefore, a series resistor of [latex]100 \Omega[/latex] was introduced in series on the gate pin order to limit the rise and fall times and thus limit the bandwidth of the “antenna”. Apparently, the guy doing the calculations on the series resistor was a little eager so the resulting slope became too flat. By removing the gate driver and the series resistor, the slope is instead limited by the maximum source and sink currents of the PWM driver, 10 mA and 25 mA respectively.

The rest of this post will show the calculations on where power is dissipated in the MOSFET and compare the two designs.

**Power dissipation during on-time.**

To get a base-line on what the worst case heat dissipation is, calculations are done for a MOSFET being fully-on (in saturation) 100% of the period. Looking at the data sheet for AON6758 (figure 3), the [latex]R_{DS(ON)} = 4 m\Omega[/latex] for[latex] V_{GS} = 4.5 V[/latex]. All measurements have been done with a 40W @ 12 V heater element giving a current of

[latex]I_D = \frac{P}{U} = \frac{40 W}{12 V} = 3.33 A [/latex]

and having a resistance of

[latex]R_{HEATER} = \frac{U}{I_D} = \frac{12 V}{3.33 A} = 3.6 \Omega[/latex]

Knowing the current through the heater and the on-resistance of the MOSFET, we can calculate the power dissipated in the MOSFET:

[latex]P_{MOSFET} = U_{DS} I_{D} = R_{DS(ON)}xI_{D}^2 = 0.004 \Omega x 3.33^2 A = 0.044 W[/latex]

On Replicape Rev A4A, the gate driver gave a [latex]V_{GS}[/latex] of 12 V which in turn gave an [latex]R_{DS(ON)} = 3 m\Omega[/latex], giving a heat dissipation of 0.033 W.

**Power dissipation during off-time**

For completeness, I’m including the power dissipation during off-time as well. Intuitively it is negligible, but let’s grab the boob and do the math to be sure.

Looking at the data sheet, the leakage current when the MOSFET is fully off is

[latex]I_{DSS} = 0.5 mA[/latex] for [latex]V_{GS} = 0 V[/latex]

In this case all the voltage lies across the drain-source, so we have

[latex]P_{DSS} = U_{DS} \times I_{DSS} = 12V \times 0.5 mA = 0.006 W[/latex]

This is an order of magnitude lower than on-time, but this increases as a function of temperature up to a reported 100 mA for a junction temperature of 125 degrees.

**Power dissipation during transitions.**

So what about during the transitions? To calculate that, it’s necessary to look at the power as a function of time. The average power is the integral of the pulse divided by the period. Look:

[latex] P_\mathrm{avg} = \frac{1}{T} \int_{0}^{T}p(t) \mathrm{d}t [/latex]

Looking at the rise and fall transitions, we can approximate the curve by a second order polynomial curve. With a little trial and error,

I landed on the following values:

[latex] V(t) = 4.4 \times 10^{8} t + 4.0 \times 10^{15} t^2 [/latex]

From this, the drain current was calculated and also the voltage across the drain and source of the MOSFET giving the necessary

parameters for modelling the power dissipated in the MOSFET during the transitions. Here is a plot of the [latex]V_{DS}[/latex], the drain current and the resulting power dissipated. Note that even though the peak power reaches 10 W, this is only for a short time. The picture below shows the plot of the voltage across the MOSFET, the current going through it and the resulting power being dissipated during a rise transition.

Some quick calculations reveal that

[latex]P_{AVG} = \frac{1}{T}\int_0^T At – Bt^2 dt = \frac{1}{T}(\frac{1}{2}At^2 – \frac{1}{3}Bt^3) |_0^T[/latex]

[latex]= \frac{1}{T}(\frac{1}{2}AT^2 – \frac{1}{3}BT^3) = \frac{1}{2}AT – \frac{1}{3}BT^2[/latex]

Using the values from before, reveals an effect loss of 7.66 W during the rise transition. a Similar value can be expected during the fall transition.

**Steady state heat dissipation**

For a 50% duty cycle at 1 KHz the average dissipation will then be **0.78 W**. Now please note that this is a fairly crude first order

approach to the heat dissipation calculations. Still, with the right layout with at least 1 in^2 copper, the packages should be able to dissipate between 2.6 and 4.1 W according to the data sheet which is far below what these calculations show. Comparing this to some of the other boards that has had this calculation done for them, it looks pretty good!

Comments are closed.

A few questions in random order. 🙂

Did you try, for curiosity’s sake, to just reduce the series resistor?

As you say, it would likely alleviate the problem, but it would be interesting to explore the difference in this application, with or without a gate driver, on equal terms.

What kind of load did you use for the measurements?

I’m guessing it is a purely resistive load, but how would the two circuits compare with the heaters as load? (with a more suitable series resistor, of course)

Of course, I may be a bad guesser..

1) I’ve not tried reducing the series resistance on the A4A revision with the gate driver, but that would be interesting in order to verify the theory. I’ll see if I cannot get that done!

2) I used 40 W heater elements from China: http://www.aliexpress.com/item/Single-end-cartridge-heater-dia-6-x-L20mm-12V-40W-140cm-lead-cable-mini-cartridge-heater/738354287.html They are wire wound, so they are probably pretty inductive.

3) Yes, you are a bad guesser. This setup is totally realistic 🙂

Then I must say, those are excellent results. 😀

Sir,

For Vgs = 12V, you have written R(ds on)= 3m Ohm

and for Vgs = 5V, you have written R(ds on)= 4m Ohm,

can you please elaborate on how you computed the values of R(ds on) for different Vgs,

Is there a equation relating the two. I have come across only this equation:

R(ds on) = Vds/Id when Vgs=constant,

how to obtain R(ds on) values for different Vgs values?

Thanks

If i want to make an RF amplifier of 1KW, Power Mosfet of what rating should be chosen for this purpose ? futher given i have to use a power supply of 300VDC. What is the difference between power dissipation and power handling of a MOSFET ?