Regulated Supply with OpAmp and Transistor works, MOSFET doesn't - Page 1
Forum section: electronics / beginners

author: rentner
author_rank: Regular Contributor
timestamp: March 28, 2013, 08:31:02 pm
content: In the attachment I show you a standart voltage regulator for supplies with variable output voltage. This works just fine. But I have a big problem with this circuit: If I want to replace the NPN transistor with a Mosfet (N-Channel type), the circuit just does not work. Why? Well, somehow the Gate won't charge up enough. Referenced to ground it has a voltage of maybe 10, but referenced to the source it is just about 3 Volts - this won't work of course.Any way to fix that? I mean, a Transistor needs current, to let current flow, a Mosfet just needs a voltage applied. Since I want to regulate ~160VDC to 130V clean DC at a current of about 5 Amps, I would have quite some losses. To get rid a bit of this, I want to use a Mosfet, since the losses can be quite low with low RDS and paralleled Devices for even lower resistance. But the gate won't charge up enougth. I get about 7V into 1k resistive load.Opamp is, by the, the LM358. Negative voltage supply to circuit ground and positive supply to extra 12V regulated supply. It works well with a transistor, so the OpAmp is definitely working quite well...Please help me... If I put source of the mosfet to ground, the supply (of course, seems logical), won't work correct, since one side is connected to the 160V side and the other to the unfixed drain of the Mosfet. Would not work at all...

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: March 28, 2013, 08:52:30 pm
content: Hi,The circuit will not start unless there is voltage on the Zener. With the BJT there is enough conduction to get the circuit started.With the MOSFET connect a resistor, say 100K between pins 7 & 3 of the op-amp and the circuit will probably start.You may have some oscillations because of the gate capacitance of the MOSFET. There are no real advantages of using a MOSFET in this circuit.Jay_Diddy_B

author: rentner
author_rank: Regular Contributor
timestamp: March 28, 2013, 08:59:17 pm
content: Forgot to mention: Zener diode is connected to the 12V aswell. I also DO get some voltage. But not a "high" current like over 50mA and just 7 Volts. They are clearly regulated, but don't change, no matter, what I do... Changing the feedback still changes nothing... the Voltage won't rise, no matter, what the feedback Voltage is...Edit:I mean... I need quite some current to trigger the Transister... I want to switch 1kW with the Transistor and came to just below 8 Amps. If I just use 1 simple 15 Amp NPN (MJL 3281A ISC) and a darlington Stage, to give it 130mA, it will conduct at least 8 Amps (Rising temperature, more gain, more current possible), which is not that much power lost. Let's pretend I am using 3 Volts at 130mA, it loose just 0.39W at max. Maybe that Circuit would be better.Since I don't want to change the gain and just use it for on and off, will a 200W power dissipating Transistor be enought? How much losses should I expect, when switching 160VDC unregulated to 130VDC regulated?If I just loose a few watts, it won't be that horrible actually.

author: rentner
author_rank: Regular Contributor
timestamp: March 28, 2013, 09:24:42 pm
content: Last Edit: I will simply use a sperate, completely isolated supply and let it float on the Output, use an Optocoupler, to isolate the OpAmp from the Switching stage and that worked very, very well. It does not need much current, so It won't take much place at all on the PCB. Works perfectly fine!Still thanks for help!!!

author: w2aew
author_rank: Super Contributor
timestamp: March 29, 2013, 03:47:56 am
content: Some basic comments:dropping 160V to 130V, and supplying 5A - that's 150W (30V * 5A) dissipation in your pass device!Your circuit certainly won't work at those high voltages, but I'm sure you know that.The LM358 output can swing near ground, but can't get near the positive supply. The MOSFET will need a higher gate voltage for a given regulated output than the BJT will need.

author: amspire
author_rank: Super Contributor
timestamp: March 29, 2013, 05:27:15 am
content: Quote from: w2aew on March 29, 2013, 03:47:56 amYes. That means at least a 300W mosfet on a very good heatsink.It is not easy to use mosfets in parallel in linear mode. When you see it, it is usually special high channel resistance mosfets designed for parallel use.Transistors are much easier to connect in parallel with a small emitter resistor on each transistor. Works great!QuoteThere are two options - connect the LM358 regulator to a +/- floating supply with the opamp's 0V rail connected to the +ve supply output, or add a high voltage driver transistor at the opamp output.I always prefer power transistors to mosfets for linear amplifiers, as mosfets driven from a low current driver have the characteristics of allowing massive current spikes though to the load if you short the output - even if you have a current limit loop in the circuit. Transistors tend to be slower and behave more like current sources, and it is easy to prevent the big current spike.I would definitely add a current limit, even if it is for an internal supply that is always meant to operate at 130V. At very least, you have to add fuses. With 160V at 5A = 900W of power, there is plenty of power here to melt the wire in transformers or burn big holes through circuit boards. Even with a current limit, if the output is overloaded, the power transistors/mosfets will overheat and fail. When they fail, they will short, so the 160V will go straight to the load and you have to consider the effects of that. Adding thermal protection as well as current limiting is a good idea.The circuit may be easy to stabilize with a resistive load. If the load may have some big electrolytic capacitors across the supply output, it will probably hard to stabilize, particularly if you have a huge mosfet with several thousand nF of gate capacitance in the circuit.Quote from: rentner on March 28, 2013, 08:31:02 pmThe RDS has nothing at all to do with the efficiency when using mosfets as linear regulators. The mosfet power = current x voltage drop regardless of the RDS, and in general, the best linear mosfets will have a high RDS. If you want it to work from 140V input or higher, then for a 130V output, the ideal mosfet could have a RDS as high as 2 ohms. The big mosfet with the 0.002 ohm RDS will probably be useless for high power linear mode. Check the Safe Operating Area graph in the data sheet.Richard.

author: rentner
author_rank: Regular Contributor
timestamp: March 29, 2013, 11:03:03 am
content: That much power cannot be dissipated. That is impossible! Just remember: There are only two states: On and Off. On charges the caps and Off does not charge any further. The only losses are through RdsOn and switching losses. Impossible, to dissipate that much power, because it is switching. It only CAN dissipate that much power, if I use the Zener - NPN Voltage regulator, because that is a resistive Regulator.

author: rentner
author_rank: Regular Contributor
timestamp: March 29, 2013, 12:12:02 pm
content: Well, I edited the Circuit quite a lot, now using an OptoIsolator and I will also use NPN high power switching Transistors. That works fine and the good thing is: It realy changes between on and off. That means, I will not dissipate that much power at all. I will just need a Power Darlington, the OptoIsolator connected to that with 50k in series and than the switching Transistor. It will all the time (at least almost all the time) be in Saturation or Off. I simulated that very succesfully in LTspice, but with different Transistors, because there are no high current Transistors for high voltages like that in LTspice.I hope to have less than 5% Switching losses - that would be great and not totally impossible. And with a bit less luck, it will not be more than 10%.

author: amspire
author_rank: Super Contributor
timestamp: March 29, 2013, 01:06:32 pm
content: Quote from: rentner on March 29, 2013, 11:03:03 amIf you make a switching regulator, that is different. But a switching regulator must have an inductor and a different circuit. Without the inductor, the power dissipated in the mosfet or some other component will be exactly the same as a linear regulator.Mathematics are straight forward. In linear mode, the power dissipation is (160-130) * 5A = 150WIf the mosfet has a 0.1 ohm RDS and you switch it, then when it switches on, the voltage across the mosfet will be 30V at a current of 30V / 0.1 Ohm = 300A (if you assume the capacitors are perfect). With a current of 300A, the mosfet only has to be on for 100% * 5A/300A = 1.6666% of the time but for that time, it dissipates 30V * 300A = 9000W.So the average power in the mosfet is 9000W * 1.66666% = 150W. Same as before.If there is less then 30V across the mosfet when it switches on, that only means there are resistive voltage drops elsewhere in the circuit in places like the wiring, the input filter capacitor, the transformer, the rectifier diodes and the output capacitor. None of these parts can possibly cope with 150W of resistive losses. The mosfet is the only device that has a chance of handling the power.To put it another way, there is 160V *5A = 800 W of power going in. There is 130V * 5A - 650W of power going out.The difference is 150W. The laws of physics require that energy is conserved - the regulator must dissipate 150W, no matter what you try and do.In the circuit you showed, the mosfet will be operating in linear mode and not switching mode. The mosfet is not a device that magically always switches. It is the circuit driving the mosfet that makes it behave like a switch.If you used a switching circuit with an inductor, that is different as the input current will be less then 5A.An inductor is a device that can absorb energy from a constant voltage and dump the energy out at a different voltage with (ideally) zero energy loss. I will include transformers in the class of devices called Inductors.A capacitor is a device that can absorb energy from a constant current, and later dump it out at a different current with (ideally) zero energy loss.An ideal Mosfet is a resistive device, and it can only dissipate energy - it can do nothing else.Richard.

author: rentner
author_rank: Regular Contributor
timestamp: March 29, 2013, 01:36:35 pm
content: Hm... That's bad. Would I get more efficiency with a DC/DC Converter? Man, now I will have find proper inductors for that. Well, I cannot live with losses of 150W which is way to much... Lowering the Voltage of 160V to 135 is not a solution in my opinion, since the Voltage could drop to low, if the current rises too much. For a supply of over 1kW, it is not a good idea. It's propably better with a DC/DC converter.Only question I need about that: How efficient will that possibly be?

author: eevblogfan
author_rank: Frequent Contributor
timestamp: March 29, 2013, 02:19:24 pm
content: 1+ for amspier for thatI agree that mosfets are not needed , transistors is the linear approch !now for the topic , I am not pro to that field But I strongly reccomand figure if buck converter will do . because buch converter is MUCH MUCH better to that circuit considering it's DC and not AC ,that as far as I can helpCheers !

author: amspire
author_rank: Super Contributor
timestamp: March 29, 2013, 02:32:21 pm
content: Quote from: rentner on March 29, 2013, 01:36:35 pmFirst back to the linear supply. you need to allow for mains variations - lets say +/-10%. That means you need at least 135V + 20% peak on the input = 162V.Then if you have a 10,000uF filter cap after a full wave rectifier, that will have a ripple at 5A of 5V at 50Hz, so you need at least 162+5 = 167V. Then for regulation, you need at least a couple of volts across the power device, so that makes 169V peak. Then you will get variations from the transformer since as it heats, the resistance of the windings increases, so lets make that 175V peak.Then unless you find a transformer with exactly the right output voltage to give exactly 175V peak on the rectified output at 5A, you have to go for the next nearest available transformer, so that might be 190V. Now your power dissipation is not 150W but closer to 300W realistically.With the switching circuits, you have done very well if you can get 90% efficiency. That would mean a 65W dissipation at full load.Designing high power, high voltage switching circuits is not easy if you are not experienced - particularly if they have to directly connect to the mains.Your best bet might be something that is much simpler - a SCR preregulator to limit the voltage to the linear regulator to an average of 5V above the output voltage to the linear regulator. This reduces the linear regulator power to 5 *5 = 25W, and it also means you can include overload limiting without excessive dissipation.Look at this Application Note for an explanation of a SCR preregulator:http://cds.linear.com/docs/en/application-note/an32f.pdfThis is much simpler as it only switches at the mains frequency, and it means you still have a transformer that makes the debugging far easier and safer. (I hope you plan on having a mains transformer).Richard.

author: amspire
author_rank: Super Contributor
timestamp: March 29, 2013, 02:54:13 pm
content: By the way, just be aware that working at 150V plus DC is really dangerous - much more dangerous that AC. It kills you far more efficiently.And with the energy available in the circuit, when you get a failure in development, you will get power devices exploding - often with case fragments flying everywhere and sometimes an impressive burst of flame.My recommended way of working at such voltages, is you only ever touch the test circuit or connect test leads when it is switched off and the capacitors have been discharged, and you either put the test circuit behind a protective shield when it is on ( eg put a cardboard box over it) or wear safety glasses.Don't wear metal banded watches, metal bracelets or rings when you are testing.Have a kill switch to quickly shut off the power.Richard.

author: rentner
author_rank: Regular Contributor
timestamp: March 29, 2013, 03:45:50 pm
content: That's not a problem. I was working on a running Amplifier whith +70V/-70VDC. I already know, how dangerous that is. Wanted to probe the Voltage across 70VDC and left the Probes in the 10 Amps jack accidentally... The board was a bit... just a bit... toasted.Will never do that again. From now on I always check first, which mode on the Multimeter and what jack...Well, just 90% and not more is possible? Why is this technology so unefficient? I mean, there are ATX Supplies with 300W at 12 Volts or even more and they can reach 95% efficiency... How is that even possible?

author: Christopher
author_rank: Frequent Contributor
timestamp: March 29, 2013, 05:08:46 pm
content: Quote from: rentner on March 29, 2013, 03:45:50 pmwith both linear/smps there is always power lost as heat.For Linear supplies: Pwasted = (Vin - Vout) * I You'll be lucky to get 60% efficiency with a linear reg (Efficiency is inver. proportional to I).SMPS Buck the power is stored in an inductor, so less current is drawn from the input therefore less power wasted. SMPS are much much harder to implement right, though. Look for a simple chip without external fets, TI do some good ones.

author: amspire
author_rank: Super Contributor
timestamp: March 29, 2013, 10:57:34 pm
content: Quote from: rentner on March 29, 2013, 03:45:50 pmMost ATX supplies are under 90% efficiency when properly bench tested, but I guess if you are a company that for the last three decades has done nothing but make millions of ATX power supplies, you gradually learn how to save 0.3% here and 0.1% there and it eventually adds up. Every time your engineers finish a design, they start again design exactly the same supply, but just that tiny bit better based on newer parts, and on ideas for improvement they got from from their last design effort. By about the 20th time round this loop, you start to get pretty good.For a home constructor, you are doing very well if your one of prototype supply struggles to 80% efficiency. For a start, you will not be able to build the same transformer that the power supply manufacturer managed to build - you will not have access to the same core, the custom gap they ordered from the manufacturer, perhaps you cannot get the same wire, you don't have the machines they used to wind the very precisely designed windings. You will not rewind the transformer 50 times until is is absolutely an optimal design. You will not have big stocks of every possible wire type you might need - you will have a few wire sizes available and you will make do with those even though they are far from optimal. You will wind it, and if it is good enough for 80% efficiency, you will use it. Your transformer will probably generate much more heat in the windings and magnetic core then you were expecting, and will have much more leakage inductance then you were hoping for. You have to absorb the energy stored in the primary leakage inductance, but you find that trying to add a snubber to absorb that energy (to stop the switching mosfets from blowing up), you loose 10% of your efficiency just in one single design problem.After you have finished your first supply, you will have learned heaps, and I am sure you could make a better supply next time. But after all the work, you will realize you can live with the 70% to 80% efficiency - as long as it works. You will be happy if it is reliable, if the regulation is stable, and if it doesn't ruin the TV and radio reception for the neighborhood.

author: AlfBaz
author_rank: Super Contributor
timestamp: March 30, 2013, 12:11:51 am
content: Quote from: amspire on March 29, 2013, 02:54:13 pmFriend of mine showed up to work with a big red burn mark around his wrist. He was working on his car and somehow managed to short the +ve to ground via his steel watch. It welded itself to the post and metal and struggled to get it off as the current heated the wrist band. Ended up with burns and cuts when he yanked the watch free

author: BobbyK
author_rank: Regular Contributor
timestamp: March 30, 2013, 02:19:56 am
content: Quote from: rentner on March 29, 2013, 03:45:50 pmActually, a switching supply that can certifiably work under full load at over 80% efficiency in the PC industry is rather rare, even though the PC is a rather well-behaved and well-defined load. Only the top PC supply manufacturers are able to make supplies that are better than 85% and they reserve that for the highest power rated supplies only (e.g. 1200 Watt PC supplies) due to overhead costs.see:http://en.wikipedia.org/wiki/80_Plus

author: rentner
author_rank: Regular Contributor
timestamp: March 30, 2013, 09:07:02 am
content: Wow thanks guis. Just woke up and? BAM! Tons of informations.You guys rock! I love this forum. Nobody tells me: "Ahw, you will die, you won't make it" or something like that. No, you just continue giving informations to me and I love that. This is realy appreciated!Thanks to ALL of you. I learned a lot in this topic.