Electronic Load MOSFET balancing

[set]

I am building an electronic load with a similar block diagram as the one described here (EEVblog 102), though mine is a little more complicated. I have a single 100 ohm resistor coming out of the 2nd op amp into the gates of 3 MOSFETs that are in parallel. When I apply a 5A load one of the MOSFETs gets significantly hotter than the other two, say 80 C vs 35 C. I understand that this is because the gate threshold voltage is slightly different across the 3 FETs and some are turning on more than others. The common solutions for this problem is either to have separate control circuitry for each MOSFET or to put a resistor between the source of the FETs and the common node above the current sense resistor. My problem is that my electronic load has a wide current and voltage range (10mA - 25A / .5V - 32V). To allow the wide range I have multiple current sense resistors that I switch in and switch out. Only a single current sense resistor path is active at a time. If I duplicate the control loop for each FET I also need to duplicate multiple resistors/switches which makes that solution less than ideal. I also don't want to add additional resistance to the load because some of my loads will need an equivalent resistance of 0.150 ohms while others would need around 15 ohms.

Would there be any way that I can balance out the power dissipation between the multiple FETs without significantly adding to my parts list? As a note this load will not be for transients and will be used for set CC loads.

[diyaudio]

The common solutions for this problem is either to have separate control circuitry for each MOSFET

You answered your question.

You need to main a separate control loop for each active cell so each current sink is working independently from each other, with this approach you now have a module based active load, where you can chain more "cells" and increase current capability.

If you study the tear-down of the BK 8500, there are 2 quad opamps TL084 (I think), 8 * 2 IRFP 250 mosfets and x8 0.05 ohm 8 watt current balancer resistors,  just by glazing over the tear-down its obvious they using the same approach. 

Have you calculated SOA ratings for your fets ?

[Jay_Diddy_B]

Hi,

In this thread I showed how to use two MOSFETs. This technique can be expanded to as many MOSFETs that are needed to dissipate the power.

https://www.eevblog.com/forum/projects/dynamic-electronic-load-project/msg288313/#msg288313

This technique is used by HP and BK and many other manufacturers' electronic loads.

Regards,

Jay_Diddy_B

[dannyf]

one of the MOSFETs gets significantly hotter than the other two

Use a temperature sensor (ntc or ptc, for example) on each of the mosfet. The opamp drives the mosfets via voltage dividers, one of the "resistors" in the dividers is the temperature sensor. Set up the dividers so that when the temperature goes up, the divider generates a lower Vgs on the corresponding mosfet.

In the end, all the mosfets will see the same temperature -> assuming identical divider / sensor parameters.

One opamp is needed.

[electros6]

In my electronic load I dsigned it for 20mA to 20A  and almost 0 to 80V 150W . I used two two mosfet in parallel total of four mosfet and drive it from 1k resistor from the opamp and its works fine. I used two 0.01 ohm resistor to sense current and use two  OP07 to amplifiy the drop across the sense resistor for each branch and add the two result. 

[set]

The common solutions for this problem is either to have separate control circuitry for each MOSFET

You answered your question.

You need to main a separate control loop for each active cell so each current sink is working independently from each other, with this approach you now have a module based active load, where you can chain more "cells" and increase current capability.

If you study the tear-down of the BK 8500, there are 2 quad opamps TL084 (I think), 8 * 2 IRFP 250 mosfets and x8 0.05 ohm 8 watt current balancer resistors,  just by glazing over the tear-down its obvious they using the same approach. 

Have you calculated SOA ratings for your fets ?

It seems like I may need to go with separate control loops but I was trying to find another solution before going down that path.

I'm using the irfp3306pbf. Figure 8 has the SOA. I am in the SOA if I were to only use a single FET however I don't believe I can keep the FETs cool enough with a single FET. I could be using 2 FETs but I decided to use 3 FETs for an additional safety margin. If I do need to use separate control loops for each FET I will likely go back to using 2 FETs again.

one of the MOSFETs gets significantly hotter than the other two

Use a temperature sensor (ntc or ptc, for example) on each of the mosfet. The opamp drives the mosfets via voltage dividers, one of the "resistors" in the dividers is the temperature sensor. Set up the dividers so that when the temperature goes up, the divider generates a lower Vgs on the corresponding mosfet.

In the end, all the mosfets will see the same temperature -> assuming identical divider / sensor parameters.

One opamp is needed.

Good idea! I will be giving this a try.

In my electronic load I dsigned it for 20mA to 20A  and almost 0 to 80V 150W . I used two two mosfet in parallel total of four mosfet and drive it from 1k resistor from the opamp and its works fine. I used two 0.01 ohm resistor to sense current and use two  OP07 to amplifiy the drop across the sense resistor for each branch and add the two result. 

What were the ranges of your OP07 output? How accurate was the current sense? Could you post a schematic? It seems like you didn't have the problems I thought I was going to have and I may have over designed my solution. I am using a 4 channel ADC to measure the drop across each current sense resistor which have been sized so that I am always at least measuring >20mV as I didn't want the ADC to measure down too low.

[retrolefty]

Seems to me the just selecting a single MOSFET suitable for the load ranges would be a lot simpler (cheaper? less PCB room, etc). Or would that be cheating? 

[set]

Seems to me the just selecting a single MOSFET suitable for the load ranges would be a lot simpler (cheaper? less PCB room, etc). Or would that be cheating? 

If I could find a suitable FET that was cost effective I would go with it. I didn't find any that would let me dissipate the heat I needed to. I ran some basic thermal dissipation calculations and I would have a hard time getting all of the heat out of a single TO-247. It may be possible to use a TO-264 but I didn't see any FETs with a low enough Rdson in that package.

[dannyf]

Another way to get it to work, without balancing resistors, is to use lateral mosfets.

They are expensive (10x or so than the typical vertical mosfets), and of limited sourcing alternatives.

[DanielS]

I would go with load-balancing resistors between source and sense resistors: simply match your FETs a little better and use the highest value load-balancing shunts your lowest voltage can accommodate at its highest current. With three FETs, the load-balancing should prevent the "worst" one from drawing more than 10A and if you use 2W balancing resistors (possibly in the form of a bunch of SMD parts or a piece of wire), that allows you to use 0.02 ohm shunts and you only need to match FETs within 150mV of Vgs. FETs also usually have a positive thermal coefficient, so they should self-balance themselves reasonably well as long as you do not give them a chance to self-destruct first.

[Jay_Diddy_B]

Hi,
If you study these graphs on the IRFP3306 MOSFET datasheet:



This curve shows that the RDSon increases as a function of junction temperature. It is normalized, to get the actual RDSon you multiple the RDSon at 25C by the coefficient on the vertical axis.
Under these conditions the RDSon has a positive temperature coefficient. This only helps sharing if the MOSFETs are being used as a switch and the gate source voltage is high enough to fully enhance the device.



In this curve, the transfer characteristics, the Drain current versus VGS is shown for two temperatures. As the temperature increases more current will flow for a given VGS. This is a positive temperature coefficient, but of transconductance. This is the curve that is applicable to an electronic load when the MOSFETs are being operated in the linear region. In the linear region there is the possibility of thermal runaway, because higher junction temperature, leads to more current, leads to a higher temperature etc.


The op-amps are cheap compared to cost of the MOSFETs. I would strongly recommend 1 op-amp per MOSFET.

I suspect that using an op-amp to servo the case temperature of the MOSFETs, as proposed by Dannyf, may work, it might also be too slow. It would be harder to implement than controlling the current in each MOSFET.

Regards,

Jay_Diddy_B

[Pjotr]

Good point Jay. I don't know if this is already mentioned, but the same mechanism can lead to hot spots on the chip that leads to premature break down of some cell's. This causes an avalanche of broken cells, and finally the whole chip is broken. Therefore at DC, de-rate the maximum dissipation given for the device, by say a factor of 4 to be on the safe side.

[dannyf]

A quick simulation.

Here is basically a circuit as we discussed earlier, no temperature sensor is employeed: R1's value is not influenced by M1's dissipation (power, a proxy for its temperature).

As I used purposedly two vastly different mosfet, the current imbalance is great: at 10v, M1 has 8amp going through it and M2 has 2amp going through it.

[dannyf]

Now, the same circuit, the same mosfets, with a ntc in R1: R1's value goes down as power dissipation (again, a proxy for M1's temperature) goes up.

The current imbalance still exists but it is much smaller: at 10v, M1 flows 5.4amp and M2 flows 4.5amp, approximately.

[dannyf]

Obviously, you will need to do some analysis in picking the values of the dividers. Ideally, you want to set the dividers so that at full power, they flow roughly the same amount of current -> lower power imbalance doesn't matter as much.

The beauty of lateral mosfets is that they have much lower knee point for their tempco so that such an imbalance is taken care of automagically. For vertical mosfets, the transition points are so high that the mosfets are practically destroyed long before.

Hope it helps.

[c4757p]

Surely you don't think that when investigating thermal runaway (which happens quite fast when it gets going) and thermistors attached to the package (rather slow), a simple simulation with no time delay whatsoever has any relation to the real-life performance of the circuit...

[diyaudio]

Question. with the selection of the irfp3306pbf...

why use a mosfet primarily designed for switching applications? its rated at 4.2mill ohms RDS(on), @ 60V VDSS... The eload mosfets are functioning in its linear region.(most of the time)  this would impose on cost as manufacturing processes with low rds(on) increases in cost. Its better to use something in the ohm region of 0.02 there about, older power mosfet technology is suitable for applications like this.

I paid $1 (locally) for an IRFP260 VDSS = 200V, RDS(on) = 0.04 ohm, ID = 50A,  where the irfp3306pbf costs $3. 

[dannyf]

I was asked that as the current going through the two devices seems to be so un-linear, will the circuit generate a linear output on the load (R3 in my circuit).

Here is the simulation, with the load current on R3 plotted in the red trace.

Hope it helps.

[set]

Question. with the selection of the irfp3306pbf...

why use a mosfet primarily designed for switching applications? its rated at 4.2mill ohms RDS(on)... The eload mosfets are functioning in its linear region.(most of the time)  this would impose on cost as manufacturing processes with low rds(on) increases in cost. Its better to use something in the region of 0.02 there about old power mosfets are great for this I paid $1 for an IRFP250 where the irfp3306pbf costs $3.

My application requires some of the total equivalent resistance to be around 150 mohms. I am want to minimize the potential on resistance of the FETs in case my system approaches 150 mohms on its own. The IRFP250 looks like it has an on resistance of 85 mohms and even is parallel that is around 1/3 of my total resistance when fully on.

[dannyf]

why use a mosfet primarily designed for switching applications?

Practically all of today's mosfets are designed for switching applications, your IRFP260 included.

The only known exceptions are the lateral mosfets and power jfets (which are no longer produced).

As to why one switching mosfet vs. another: it could be a gazillion (valid) reasons, like availability, .... What cost you $3 apiece could cost someone nothing, and vice versa.

[dannyf]

My application requires some of the total equivalent resistance to be around 150 mohms.

Rds(on) has no meaning for linear applications.

[diyaudio]

The IRFP250 looks like it has an on resistance of 85 mohms and even is parallel that is around 1/3 of my total resistance when fully on.

"fully on in saturation mode"  the issue im trying to understand is this, the mosfets are in linear operation, so why the focus on a design parameter like "fully on in saturation mode" and its not used ?

[diyaudio]

My application requires some of the total equivalent resistance to be around 150 mohms.

Rds(on) has no meaning for linear applications.

you beat me to that one.

[Jay_Diddy_B]

Hi,

While you guys were busy typing away, I was running simulations of a sharing circuit. I used two very different MOSFETs in the simulation.

Here is the LTspice model that I used:




And the results from a DC sweep showing that there is perfect matching between the two MOSFETs. In practice there will be a few percent difference resulting from the tolerance of the 0.1 current sense resistors:



I also tested the circuit under dynamic conditions. You can see that even with the different MOSFETs there is very good tracking between them:




This is several orders of magnitude faster than trying to force sharing by measuring the case temperature of the devices. If you look at the Safe operating for the iRFP3306, under DC conditions the MOSFET is capable of 40W dissipation and 80W for 10ms. This SOA graph shows the thermal instability region, indicated by the breakpoints in the SOA graphs. The heat will not transfer fast enough to force the sharing.




You will find this method of forcing the sharing in HP 6060A, 6060B and 6050x series of electronic loads.

Regards,

Jay_Diddy_B

[Pjotr]

This is several orders of magnitude faster than trying to force sharing by measuring the case temperature of the devices. If you look at the Safe operating for the iRFP3306, under DC conditions the MOSFET is capable of 40W dissipation and 80W for 10ms. This SOA graph shows the thermal instability region, indicated by the breakpoints in the SOA graphs. The heat will not transfer fast enough to force the sharing.

Regarding this, the old style large chip mosfets are in favor. Those have also a lower channel to case R_Th which helps to avoid thermal runaway. Found some good reading about it: http://www.infineon.com/dgdl/Infineon+-+Application+Note+-+PowerMOSFETs+-+OptiMOS%E2%84%A2+-+Linear+Mode+Operation+and+SOA+Power+MOSFETs.pdf?fileId=db3a30433e30e4bf013e3646e9381200

And IXYS seems to make special powerfets for linear applications: http://www.ixys.com/Documents/Articles/Article_Linear_Power_MOSFETs.pdf