Power FET instead of SSR? - Page 1
Forum section: electronics / beginners

author: anvoice
author_rank: Frequent Contributor
timestamp: April 18, 2015, 02:21:06 am
content: Hello,I'm trying to control a 24V DC resistive load (1.4 ohm 3d printer heated bed) drawing about 20A (continuously while heating up, then pulsed). Since I have to power it from a separate PSU, I was thinking initially of using an SSR. The already existing power circuitry powered by the first PSU would then activate the relay instead of directly feeding the bed, and the relay would transmit power from the second PSU to the heated bed. While this works, the DC SSRs I got from ebay (only place I could find them) turned out to be massively under specs (FET rated at 2A in a 40A relay), and even if I order from a different place from China I'll be waiting weeks just to see if it works. The only reasonable solution seems to build my own relay, but I haven't found any good guides on building DC SSRs online.So I was wondering: if it's reasonable to just use a heat-sinked power FET instead of the relay? Seems like if I get one rated at 40A or so, equip it with a good heat sink and cooling, it should do the job just fine, so why do I need the SSR? Am I missing something? Thanks!

author: Psi
author_rank: Super Contributor
timestamp: April 18, 2015, 03:01:53 am
content: FET will work fine.Yeah, just put a good heatsink on it.Maybe the SSR was spec'ed because it is isolated?Is your +24V referenced to your control GND ?

author: anvoice
author_rank: Frequent Contributor
timestamp: April 18, 2015, 03:22:51 am
content: By isolated, you mean optically? I'm not sure how that allows a 2A FET to carry 40A...PSU #2 (which drives the load) is connected to the load via its 24V channel and its own ground, so I guess that means no? (Very new at electronics, so apologies for any inaccuracy in my reasoning). PSU #1 (powering the circuitry which sends the switching signal) is also a 24V one.

author: Psi
author_rank: Super Contributor
timestamp: April 18, 2015, 04:13:58 am
content: An SSR is usually optically isolated.I just meant that if you swap to a FET you need to be aware that you have lost that isolation.In your example PSU2 is totally isolated from PSU1. If you use a mosfet you need the same ground point. (unless you use a relay or something)

author: anvoice
author_rank: Frequent Contributor
timestamp: April 18, 2015, 04:30:33 am
content: If I understand correctly, optical isolation is more relevant for switching with a low voltage DC signal a high voltage AC load, to protect the control circuit. I'm also guessing I can add an optical isolator pretty easily to the setup, it's just that I don't have a good circuit diagram for a full-fledged DC SSR I could make, but I need something that works.What about one of these setups?To be honest I have no idea what happens when you connect 2 different grounds together (current flow if different potential = bad?) or connect the negative side of the load to a different PSU's ground. If there's a good way to do this, I'd appreciate hearing it.

author: Psi
author_rank: Super Contributor
timestamp: April 18, 2015, 04:58:04 am
content: To control the mosfet you need two connections,-a signal connection to the gate-a ground connection to the source.Since you also need 2 connections to "switch" something you must have a "common" connection on the control side and the switching side. (Ground)

author: anvoice
author_rank: Frequent Contributor
timestamp: April 18, 2015, 05:29:12 am
content: Ah, right, the FET needs a lower voltage at the gate. So I'd essentially need a regulator that takes the 24V down below 15V. Thanks for the info!Edit: further research reveals that a DC voltage regulator such as a L7812 may not work well when pulsed (when bed is at temperature and the signal turns on and off to keep it there). Is there some way around that? The pulse frequency is probably around 4Hz if I remember correctly.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: April 18, 2015, 07:05:32 am
content: QuotePerfect application for a Zener-diode. Just put a 1k resistor between the gate and the control-output and the 15V Zener straigt across Gate-Source.Make sure to use a MOSFET with lowest possible on-resistance. The IRFP7530PBF or IRFP4368 would be well suited and they should be able to carry 20A without any haeatsinking.Quote4Hz is basically DC - no problems there with the Zener. This simple solution would work just fine even in the lower kHz-range. Above 10kHz you would probably need a dedicated MOSFET-driver.

author: anvoice
author_rank: Frequent Contributor
timestamp: April 18, 2015, 08:10:09 am
content: Quote from: PSR B1257 on April 18, 2015, 07:05:32 amWill the Zener be ok with the pulsed DC signal? I realized that I can just put the regulator on the upstream side of the controller where the pulsing is not a problem.I'm guessing the turn-on resistance is what determines how much current the FET will carry before overheating? Problem is I'm having some trouble finding those FETs here in the US, and they're quite expensive. Could I use something like a IRF540N (33A, 100V)? Probably wouldn't be quite as efficient overall, but I can get them faster and cheaper here.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: April 18, 2015, 08:29:13 am
content: QuoteYes, I edited my last post regarding this issueQuoteYes, they are - for a good reason.QuoteYes.QuoteWell, this type has a on-resistance of 44mOhm, meaning it dissipates about 18W at 20A drain current. Not a good choice.The IRF3205 with 8mOhm would dissipate about 3W and therefor requires only a little heatsink.

author: anvoice
author_rank: Frequent Contributor
timestamp: April 18, 2015, 08:46:57 am
content: Quote from: PSR B1257 on April 18, 2015, 08:29:13 amDon't doubt it, just thinking cost-benefit.Quote from: PSR B1257 on April 18, 2015, 08:29:13 amThe heat sinks I got for the SSRs can disspiate about 50W I believe, + I'll have a pretty hefty fan pointing at the electronics, so I don't think it'll overheat. The efficiency (possibly aesthetics) is a different matter. Perhaps I'll get a few of the IRF540N since they'll get here in days, and also order a couple of the IRFP4368 from China for cheap. If I get conned again, I'll just get a refund and try a local seller.Thanks, that was quite informative!

author: PSR B1257
author_rank: Frequent Contributor
timestamp: April 18, 2015, 10:42:47 am
content: QuoteGood. You can connect them in parallel to share the load current.It will not be a perfect 1/n ratio, but good enough.

author: anvoice
author_rank: Frequent Contributor
timestamp: April 25, 2015, 10:12:04 am
content: Fun fact: my controller actually switches the negative output pin instead of the positive pin. That is, the positive pin will always be at 24V, while the "negative" pin is at 0V when the heat signal is on, and 24V when the signal is off... Naturally I'll have to drive it from that negative pin, so I'll basically need an NPN transistor added to the circuit to trigger the power FET.Could anyone suggest a good P-channel power transistor, perhaps an analogue of the IRFP4368? Fewer components to fail/short would be far more pleasant.A problem I see with this setup though is that if PSU1 stops working, a P-channel FET (or a NPN transistor) would basically remain on forever as long as PSU2 works fine...

author: Psi
author_rank: Super Contributor
timestamp: April 25, 2015, 10:22:09 am
content: Your zener is on the wrong side of the resistor. Look at what will happen when your transistor switches on. You will have 24V directly on the 15V zener.Also you cant feed 24V into a transistor like that, you need a resistor in series.Check your transistor polarity as well.You could change the code in the arduino so the logic is reversed. 24V=on etc.You could also use a P channel mosfet, that way you don't need the extra transistor to invert the logic. (be sure you understand how a mosfet gate works and where to put the zener to limit the gate voltage to 15V on a p channel fet)To make sure the heater stays off when PSU1 is off or disconnected you need to have a resistor on the mosfet gate that pulls it to the "off" state. Then the controller has to do some work to overcome this, so a loss of PSU1 power will prevent this from happening.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: April 25, 2015, 10:45:37 am
content: Quote from: anvoice on April 25, 2015, 10:12:04 amThe gate-resistor is at the wrong location. It has to bein frontof the Zener, otherwise the current through the Zener will not be limited.The NPN transistor is also connet incorrect. If the emitter is at the positive supply, you need a higher voltage on the base to turn it on.QuoteP-channel MOSFETs are always worse compared to N-channel types. But you don't need a P-channel MOSFET, you just need to reconfigure the circuit a bit.That's the most simple way to achieve it.But the MOSFET in this case will be always on, if there is no control signal. And it is not suited for any high frequency PWM switching by the way, but since you only switch at a couple Hz it's fine.

author: Psi
author_rank: Super Contributor
timestamp: April 25, 2015, 10:58:13 am
content: Quote from: PSR B1257 on April 25, 2015, 10:45:37 amP channel aren't as good, but they still work fine.I have a 60V 60A 16mR P-channel mosfet doing 12V 10A 100hz PWM with no problems.I have no doubt it would handle 24V 20A fine if heatsinked well.Its dpak SMT though, so probably not any use to OP.Just saying there's plenty of p channel fets that can do the job.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: April 25, 2015, 11:09:55 am
content: QuoteYes, they work. I just wont to point ot, that the N-cannel type in this case will also work.

author: SeanB
author_rank: Super Contributor
timestamp: April 25, 2015, 11:35:17 am
content: I will also add that you need a discharge resistor across the zener diode as well, otherwise the MOSFET will turn on, and stay turned on, until the charge stored at the gate has discharged through the zener diode leakage. If the transistor driving it is only slightly leaky, it might never turn off at all.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: April 25, 2015, 12:14:00 pm
content: Unfortunately it's not that simple. If you choose a resistor <1k it will make the Zener useless, since it cannot work as a shunt regulator anymore. If you choose a resistor >1k you will always have a potential >VSupply/2 at the gate and therefore the MOSFET will stay on anyway.

author: anvoice
author_rank: Frequent Contributor
timestamp: April 25, 2015, 07:24:09 pm
content: Thanks for pointing out the resistor problems, I'll make sure to put those in properly.I actually realized that the controller's output to the heater is derived from a single dedicated input, thus I might be able to hook up PSU2 to that and avoid the issue of the FET staying on, basically excluding PSU1 from the circuit diagram.Reprogramming the arduino does sound like by far the easiest solution, as it would work fine with the original circuit. I'll have to look for a way to do that (pretty new to programming too).

author: Zero999
author_rank: Super Contributor
timestamp: April 25, 2015, 08:19:15 pm
content: Why not drop the zener and use two resistors to make a potential divider?If the heater is going to be left unattended, then there should be a thermal fuse in series with the element to provide protection against fire, in case the temperature controller fails to turn off.

author: anvoice
author_rank: Frequent Contributor
timestamp: April 25, 2015, 08:47:53 pm
content: Quote from: Hero999 on April 25, 2015, 08:19:15 pmAny advantage/disadvantage over Zener? More reliable?Quote from: Hero999 on April 25, 2015, 08:19:15 pmThe heater shouldn't be able to go above about 200C without a short-circuit though (power supply can't provide enough power) at which point the PSU's overload protection would kick in and shut off power. 200C isn't enough to ignite anything but flammables (possibly paper), of which I don't have any in the printer's vicinity. I'm worried adding the thermal fuse itself would produce more risk by introducing a possibly high-resistance connection at 17A.

author: Zero999
author_rank: Super Contributor
timestamp: April 26, 2015, 03:52:42 pm
content: The advantage of a zener is the output will be regulated, so it's the same, regardless of the power supply voltage but this shouldn't be necessary if you're planning to use around 24V all the time.A thermal fuse will have negligible resistance, providing it's rated to the appropriate current.If you need isolation, you can use an optically isolated MOSFET driver such VOM1271 but the switching speed will be slow.http://www.vishay.com/docs/83469/vom1271t.pdf

author: anvoice
author_rank: Frequent Contributor
timestamp: April 26, 2015, 10:29:10 pm
content: Thanks, I'll look into the thermal fuses.