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Ideal Transistor for battery discharger

rooster92

Dec 8, 2011
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I'm have a couple questions about using transistors. I have very little experience with them. I'm working on building an automatic discharger for some LIPOs that I use for RC. I have most of the circuit built. Basically, I'm using an op-amp to compare the voltage of the pack to a reference voltage. If the pack is above 15v or 22v (depending on the mode controlled by a switch), a transistor will be activated which will turn on a bank of bulbs. Attached is the schematic. I picked up a NPN 3055T (10A rating, 75W dissipation) and mounted a heat sink (see attached image). During my initial testing, I limited my power supply to 2A @ 15.5 volts (appr 31 watts) flowing through the heat sink and after only 8-10 seconds, the heatsink was much too hot to touch.

My questions:

1. How hot should or can a transistor get? I'm only running it at less than half of it's power rating so I'm a bit surprised it is getting so hot.

And because I know that is probably a very hard question to answer, perhaps I should be asking this:

2. What would be the best transistor to use for this application? Ideally, I'd like to discharge packs at ~10 amps.

Cheers!
 

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Harald Kapp

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Welcome!

There is no ideal transistor, you always have to make compromises.

As for the temperature: You'll have to look up the limits in the transistor's datasheet. The 2N3055, for example, has a max. operating junction temperature of 200 °C. If you want to operate this transistor at room temperature, say max. 30 °C, your thermal budget is 200 ° - 30 ° = 170 °. Add some safety margin, that leaves you with 150 ° temperature difference between junction and ambient temperature. The 2N3055 has a thermal resistance of 1.5 °/W junction-case. This means that at 100 W the temperature of the junction will rise to 100 W * 1.5 °/W + case temperature which results in 150° + case temperature. If you manage to cool the case to 30 ° C, the junction then would be at 150 ° + 30 ° = 180 °. This is at the limit. In order to keep the case at 30 °, however, you would need a huge heat sink and possibly active cooling (fan).

Another option would be a MOSFET, e.g. NTD4856N. It has a max. Drain-source resistance of ~5 mOhm (o.k., the datasheet says 4.7mOhm). 5mOhm at 10A generate 0.5W of power dissipation (I²*R). This is easily manageable with a small heat sink. If mounted properly (see datasheet), the thermal resistance junction-ambient is 113 °/W which at 0.5 W translates into 113*0.5 ° = 57 °C over ambient. AT 30 °C ambient temperature the junction heats up to 30 ° C + 57 ° = 87 ° which is much less than the allowed 175 °C junction temperature for this transistor.
Drawback: Its SMD and less easily handled than the 2N3055.


Regards,
Harald
 
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rooster92

Dec 8, 2011
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Thanks for the information. It sounds like I should definitely be looking at a MOSFET. I found a NTD4856N (http://goo.gl/iQzEx) that is through hole as I'm no where near being able to deal with surface mount. However, it is only rated at 25V and some of the packs I'll be discharging are right around 25-26 volts. I'm guessing that I'll want a little safety margin?

I found a hundreds of MOSFETs on mouser. How do you think one of these will work? The second one looks as though it would give me a ton of margin.

FQPF20N06 (http://goo.gl/4fo0P) 60V, 15A, 30W, .06 OHM
FDP070AN06A0 (http://goo.gl/HcKgB) 60V, 15A, 175W, .007 OHM

Last question. It is fairly common that MOSFETs switch on at ~5v?
 
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jackorocko

Apr 4, 2010
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Thanks for the information. It sounds like I should definitely be looking at a MOSFET. I found a NTD4856N (http://goo.gl/iQzEx) that is through hole as I'm no where near being able to deal with surface mount. However, it is only rated at 25V and some of the packs I'll be discharging are right around 25-26 volts. I'm guessing that I'll want a little safety margin?

I found a hundreds of MOSFETs on mouser. How do you think one of these will work? The second one looks as though it would give me a ton of margin.

FQPF20N06 (http://goo.gl/4fo0P) 60V, 15A, 30W, .06 OHM
FDP070AN06A0 (http://goo.gl/HcKgB) 60V, 15A, 175W, .007 OHM

Last question. It is fairly common that MOSFETs switch on at ~5v?

I think you have misinterpreted what harald has said.

In the first link you posted, the drain - source on resistance is 10x that of the one harald linked to. sqr(10) = 20 * .06 = 1.2W of dissipated power. Not all mosfets are created equally. Note, it is not that you can't use the mosfet linked, but it will not run as cool as the one harald linked to which means you will need a bigger heatsink
 
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rooster92

Dec 8, 2011
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I think you have misinterpreted what harald has said.

In the first link you posted, the drain - source on resistance is 10x that of the one harald linked to. sqr(10) = 20 * .06 = 1.2W of dissipated power. Not all mosfets are created equally. Note, it is not that you can't use the mosfet linked, but it will not run as cool as the one harald linked to which means you will need a bigger heatsink

I did notice the first one I linked to was 10x more on the resistance but thought it would fall in an acceptable range of dissipated heat in this application. Is the voltage rating going to be a problem with the MOSFET harold suggested being that I'll be running it right around 25-26V? That's what worried me and prompted me to look around a bit.
 

Harald Kapp

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A good voltage margin of at least 10% is always a good idea - unless you can ensure really tight specifications on the voltage supply and the temperature. Note also, that at higher temperatures some derating (worsening) of transistor parameters can occur.
On the other hand, the higher the drain-source voltage, the higher the drain-source resistance usually is (unless you go for "exotic" and expensive transistors).
Your second transistor looks good and leaves you with a comfortable safety margin.

Question 2: no, it is not common that MOSFETs turn on at Vgs=5V. These types are sometimes called "logic MOSFET" or "logic level MOSFET".
The FDP070AN06A0 (datasheet: http://www.fairchildsemi.com/ds/FD/FDB070AN06A0.pdf) will give you only 10A @Ugs=5V under worst case conditions (datasheet, figure 7). With Ugs=5.5 V it will happily conduct 20 A regardless of temperature. So for a reliable circuit you will need to step up the gate drive voltage to at least 5.5 V.
The NTD4856N (http://www.onsemi.com/pub_link/Collateral/NTD4856N.PDF) on the other hand will deliver >20 A at Ugs >= 3 V.


Thirdly, there is the matter of price and availability. If you start working with MOSFETs, my tip is to buy some more than required. Chances are that the first experiments will send some of your transistors to silicon heaven.



As I said, there are many options, not a single optimum solution.

I hope this helps.

Regards,
Harald
 

BobK

Jan 5, 2010
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During my initial testing, I limited my power supply to 2A @ 15.5 volts (appr 31 watts) flowing through the heat sink and after only 8-10 seconds, the heatsink was much too hot to touch.
You are computing the power dissapated by the transistor wrong. Only the voltage across C to E of the transistor should be used. This better not be 15V!

The Vce at saturation for an 2N3055 is listed as 1.1V at 4 amps, and 400ma Ibe. So at 2 amps, the power dissapated should be about 2W, which is nothing for this transistor and it should stay cool with even a moderate heat sink.

Which brings me to the real point. I think the problem is that you are not saturating the transistor. In fact, I know you are not from your circuit diagram.

The 560R resistor in the base limits the base current to 20ma if the comparator is outputting 12V. (Why you mentioned 5V puzzles me since the circuit shows 12V going into the comparator, and it's output should be at least 11V). The gain of the 2N2055 is 20, so to get 2 amp output you should need a base current of 100ma. The comparator probably cannot provide this, so I would use a darlington configuration. Just connect an 2N2222 or 2N3904 up with collect to collector, emiter to base of the 2N3055 and the base where the base of the 2N2055 is now connected. I think you will see a big improvement in the power dissapated.

Bob
 
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TedA

Sep 26, 2011
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Rooster92,

I concur with the previous posts pointing out the lack of base drive for the power transistor in your original circuit. There are quite a few more potential problems.

I'll try to add to what the previous posts have said.

Can you reveal the identity of your op-amp? Looks as if it may be the DIP-14 device on the breadboard in the picture, but I can't quite read the P/N on it.

You should know that not only may this device not be able to fully turn-on the power transistor, it may also be unable to turn it fully off! Some devices can pull the output pin very close to the negative power supply rail, others only within a few volts.

The heatsink in the picture is only good for a few watts, nowhere near 75W. You would need something much larger, perhaps with a fan.

Transistor current and power ratings must be derated in use. The data sheet values are invariably given for 25C at the silicon. The silicon runs much hotter than the heatsink temperature. The datasheets for the transistor and the heatsink will help calculate what to expect, but a reasonable expectation is to run the device at 1/4 to 1/2 the datasheet values.

3055T is an incomplete part number. The 2N3055 is a different part, in a metal can, rated at 15A and 117 W. The "T" suffix usually means a TO-220 style plastic package as seen in the photograph. The 10A & 75W ratings reflect this difference.

None of these parts is much good beyond 4-5A. The gain and saturation voltages degrade as you approach the maximum rated current.

To get a low saturation voltage at several amps collector current, you need to drive the base of most bipolar power transistors at about 10% of the collector current. So you should try to provide about 1A base current for a 10A load.

If you already have a bag full of the plastic 3055T parts, you might parallel 2-4 transistors to drive the lamps, then use one more of these, plus a small signal transistor, to boost the base drive to around 1A, total. You would need to either add small emitter resistors for the output transistors, or divide the lightbulb strings among the transistors' collectors. Separate base resistors will be required, too.

Using multiple power devices helps spread the heat around, and helps keep the devices away from their maximum ratings.

Starting from scratch, I think N-channel power FETs might be easier to use.

If your op-amp output does approach the supply rails, with the 12V supply, you should be able to drive an ordinary power FET directly. No logic-level gate device would be required.

Any of the 60V power FETs mentioned might be made to do your job, particularly if you parallel several devices.

None of the posts so far address what happens as the battery voltage reaches the voltage at which the discharger is to remove the load from the battery ( "threshold voltage" ). You don't say exactly what should happen, here.

The present circuit will attempt to throttle back the transistor to keep the battery voltage at the threshold setting. More than likely, you will get oscillation, instead of a nice steady decrease in current as the battery discharges further. In either case, the transistor will see a lot more power than it would sitting there turned on fully.

Perhaps you want the discharge to end altogether when the threshold voltage is reached under the full load. That is, as soon as the threshold voltage is reached, the load is switched off and stays off. This makes the transistor's situation easier, as it never has to dissipate much power.

Conversely, you might want the discharge current to taper until the open circuit voltage of the battery is below the threshold voltage.

Depending on how you want the circuit to behave, parts can be added around the op-amp ether to calm any oscillations while throttling-back the load current, or to make the current to go to zero and stay there, once the threshold voltage is reached.

If the transistors are to throttle the discharge current, you will need to arrange for them to dissipate lots of power. Multiple transistors in parallel is a good idea. A resistor bank to share the dissipation is, as well.

Lightbulbs are not the best answer for your load bank if the current is to be throttled. Their resistance rises as they see more voltage. One would prefer something with a resistance that falls with rising voltage, in order to reduce the power that the transistors have to dissipate.

Depending on the battery's capacity, and on how long the set-up may go unattended, you might want to add a circuit to turn the discharger completely off, once its job is done. The control circuit will continue to discharge the battery slowly, even with the power transistors fully off.

One more thing: the 7812 may oscillate without an input capacitor. A 10uF 35V electrolytic should calm it down.

Ted
 

(*steve*)

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A couple of things:

1) Mosfets are easier to parallel in this type of circuit
2) Does the battery voltage ever get below 16V? If so your 12V regulator might drop out. In the worst case your discharge may never end.
3) Check the specs for Vgs(th). In logic level mosfets it is typically much lower than 5V. However more important are the graphs showing saturation current at various Vgs.
4) Darlington bipolar transistors can reduce base current, but at the cost of Vce(sat)
5) I'd consider some hysteresis so that when the voltage drops to turn the discharger off, it has to rise significantly to turn it back on. An additional RC delay could also be employed to limit the rate at which it turns on and off.
 

MZEROKWR

Dec 9, 2011
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Hi Rooster92
I am new on here to and I find it a very interesting forum, I enjoyed reading all the replies you received but one thing i can't understand is why you would want to discharge your lipos.
Is this just a project you are trying?
I have been fly electric Helicopters for the past 4 years + and I invested over £800 in my cells so I am paranoid about looking after them and to make sure i get the max out of them. I don’t recall having needed on any occasion the need to run them down.
I would be interested in the reasoning behind the need to run them down maybe I am missing some important information here.

My helis are on http:// rotorgraph.co.uk
please take a look.
 

rooster92

Dec 8, 2011
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Wow the last few posts have highlighted how little I know. As they say, "in over my head." I don't mind though as it is my style of learning. Up until a few weeks ago, I'd never used any components other than LEDs and resistors. haha The learning process has been great and I really appreciate the time people take to answer questions and pose questions back!

In regards to my question about 5 volts applied to a gate of a mosfet, I was watching a video on YouTube that talked about using MOSFETs and applying a "logic signal" to turn them off or on. For whatever reason, that bit stuck in my head. I know so little about transistors that I didn't even know until TedA mentioned above that I might not be turning it on all the way.

It also never dawned on me that when the pack discharges to ~3.7 volts and turns the transistor off that the pack will gain voltage back and just keep cycling. So, there is that problem I need to figure out. Right now, I have a $12 USD device (http://goo.gl/vTEcA) that monitors the voltage of each cell while I discharge into a dummy bank of bulbs. If any cell drops to a preset voltage, an alarm goes off. It works great and I don't even really need to be attempting any of this. I'm just trying to build this circuit to learn something. :)

By the way, the two different types of packs that I discharge are 4 cell and 6 cell LIPO.

I'm going to experiment some more tonight with the random transistors I have on hand but also, I have ordered some MOSFETS too. My plan is to remove/replace the resistor on the output of the op-amp to see how much more current will flow across the transistor. I also have some small transistors picked up from radioshack and will attempt to use one of those to drive the larger power resistors.

By the way, I checked and the part number on the power resistor is "MJE3055T" and the op-amp is "LM324N". Next time I ask a question, I'll try and be more complete and clear with what I am working with. Going forward, I'll also label my schematics better. Part of the problem with knowing so little is not being able to ask intelligible questions. I appreciate everyone's patience. :)

I also need to read much much more on transistors! Much of what is listed on the datasheets is way over my head. Lots to learn but I'm having a blast!

cheers
 

jackorocko

Apr 4, 2010
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My plan is to remove/replace the resistor on the output of the op-amp to see how much more current will flow across the transistor.

Nope, not gonna work. You need that resistor there to set the current through the base which controls the C-E current. The transistor is an amplifying device,C-E current is proportional to B-E current.

example: If you need 1A current going through the C-E junction and the transistor has a gain (Hfe) of 100 then you will need 1/100 = 0.01A base current. The resistor connected to the base of the transistor is what provides this. (5V - 0.7)/0.01A = 430Ohms. If you remove the resistor, there will be nothing to provide the current to the base of the transistor. The resistor is vital to how the transistor works, it is the load that provides the current through the circuit.

Maybe you should look up kirchoff's law.

edit: most people will double the base current required (to make sure the transistor is fully on), simple because not all transistors are created equal. The gain can drift quite a bit even with two seemingly identical transistors. Normally you calculate using worse case scenario.
 
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jackorocko

Apr 4, 2010
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As bob has stated, your best bet is to probably wire a set of transistors up in a darlington configuration. This will allow you to have as little base current as possible with the biggest gain.

Bob already made it clear that the op-amp can not properly source enough current with the transistor you are using. You fix this by using a darlington setup which has more gain and requires less base current.
 

rooster92

Dec 8, 2011
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A couple of things:

1) Mosfets are easier to parallel in this type of circuit
2) Does the battery voltage ever get below 16V? If so your 12V regulator might drop out. In the worst case your discharge may never end.
3) Check the specs for Vgs(th). In logic level mosfets it is typically much lower than 5V. However more important are the graphs showing saturation current at various Vgs.
4) Darlington bipolar transistors can reduce base current, but at the cost of Vce(sat)
5) I'd consider some hysteresis so that when the voltage drops to turn the discharger off, it has to rise significantly to turn it back on. An additional RC delay could also be employed to limit the rate at which it turns on and off.


2. Yes, my plan is to discharge them to apprx 3.7v per cell so ~14.8v total. I'm using a simulation program on my mac that seems to show the 7812 will work down to 14.2v. I'll check the datasheet for it be sure. Of course, when discharging 6s packs, it shouldn't be a problem
 

rooster92

Dec 8, 2011
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Hi Rooster92
I am new on here to and I find it a very interesting forum, I enjoyed reading all the replies you received but one thing i can't understand is why you would want to discharge your lipos.
Is this just a project you are trying?
I have been fly electric Helicopters for the past 4 years + and I invested over £800 in my cells so I am paranoid about looking after them and to make sure i get the max out of them. I don’t recall having needed on any occasion the need to run them down.
I would be interested in the reasoning behind the need to run them down maybe I am missing some important information here.

My helis are on http:// rotorgraph.co.uk
please take a look.

Hi MZEROKWR. I've read a lot of threads on RCtech and Helifreak (in particular) that talk about not leaving your packs charged for more than a few days --> one week. Seems like the general consensus is to store your LIPOs at a "storage voltage". In fact many chargers have a "storage" mode. There is quite a discussion about what the ideal voltage is but most seem to agree storing them ~3.8v. As the weather is getting much colder here in Minnesota, I want to cycle my batteries occasionally, even when I am not flying. Some nights, I'll charge my packs planning on flying the next day and for whatever reason, I don't get out. Rather than leaving them charged, I discharge them.
 

rooster92

Dec 8, 2011
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Update: I found some MOSFETs in an old UPS. They are IRFZ46N (http://goo.gl/ntBKB). I swapped out the NPN I was using. When the voltage is above the cut-off voltage, the IRFZ46N hardly gets warm and passes all the current so no issues at all. However, as the voltage decreases near to the cutoff threshold, the MOSFET gets very hot, very fast. This will be a problem as the battery nears the cutoff.

I still have the other MOSFETs on order along with some larger heat sinks and will play around with various configurations. I also need to try a darlington configuration as well, per suggestions above.
 
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TedA

Sep 26, 2011
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rooster92,

Re the hot FETs.

You need to decide how the circuit should function as the cutoff voltage is approached, then modify the LM324 circuit to either shut down quickly and completely, or to control the load current as it tapers down, without oscillating.

In the second case, you will need to provide a big heatsink for the power transistors.

Ted
 

BobK

Jan 5, 2010
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It is probably oscillating as someone suggested earlier when the voltage gets near the cutoff. This would cause the MOSFET to heat because it would be switching rapidly, faster than it could be fully turned on or off and therefore operating in the linear region.

A possible solution is to add a lot of hysteresis to the comparator so that once it crosses the threshold and turns off, it does not turn back on when the voltage rises a bit.

Bob
 
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