How to generate +15V and -15V  pulses from +5V pulses ? Simplest way ! - Page 1
Forum section: electronics / projects

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 23, 2016, 04:29:21 pm
content: Hi All,A friend of mine wanted to generate +15V and -15V pulses wtih his 5V arduino(Say with 100mA, 10khz).To get +15V was easy with a transistor (2n3904) and mosfet. But what is the simplest way to getThe -15V pulses, without an opamp ?He has a power supply with 5V, +15V, -15V.Thanks

author: Benta
author_rank: Super Contributor
timestamp: December 23, 2016, 07:29:52 pm
content: Why "without an opamp"?That IS the easiest way.Although 100 mA is quite a number.

author: ebclr
author_rank: Super Contributor
timestamp: December 23, 2016, 07:50:00 pm
content: Optocoupler is a very simple and easy solution

author: MK14
author_rank: Super Contributor
timestamp: December 23, 2016, 08:14:54 pm
content: Quote from: JacquesBBB on December 23, 2016, 04:29:21 pm2N3906 (using the +5V (Emitter) and a signal into its base via a resistor, from the MCU) feeding through a resistor (connected to its collector), a few milliamps current into the base of a small/medium PNP power transistor, on/for the -15 Volt rail. Part of your output.If I've got it a bit wrong, I really needed to see it as a schematic. Words are often problematic.Solid State Relays, would be even easier. But could be too/very expensive, and are difficult to find at 10 KHz, with a quick look on my part.

author: bktemp
author_rank: Super Contributor
timestamp: December 23, 2016, 08:43:51 pm
content: Easy solution: TLP251 optocoupler/mosfet driver followed by a buffer stage (complementary emitter follower).Cheap solution using only more common parts: Level shifter using multiple stages for shifting the voltage first from 0V/5V to 0V/15V and then to +/-15V followed by the buffer stage.

author: Zero999
author_rank: Super Contributor
timestamp: December 23, 2016, 09:31:17 pm
content: Capacitively coupled, complementary, common emitter amplifiers.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 23, 2016, 10:27:12 pm
content: Thanks, this is brillant, and indeed what I was looking for. I was stuck with the problem of DC level, and the capacitive coupling is the trick.I made a small simulation on everycircuit herehttp://everycircuit.com/circuit/6588028428746752

author: MK14
author_rank: Super Contributor
timestamp: December 24, 2016, 12:06:48 am
content: Quote from: Hero999 on December 23, 2016, 09:31:17 pmWhen the circuit first switches on, and C1/C2 are both discharged = 0V across them. What is going to stop both the top and bottom transistors switching on at the same time (R1 and R2 will limit the max base currents, but they seem low enough to be problematic), and short circuiting the +15 and -15V rails (shoot through), potentially causing a very high current to flow, which may exceed (depending on the PSUs, output capacitors and the transistors real life Hfe's etc) the transistors max ratings and/or cause other undesirable effects, elsewhere ?Similarly (during initial switch on, power up), real life circuits will be likely to have a partial non-symmetric, differences, between the plus and minus sides (because of C1 & C2's tolerance and other factors). Which can make the MCU output pin, temporarily negative (or too high a positive voltage), via C1/C2. Which may latch up the MCU or even break it ?

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 24, 2016, 01:04:46 am
content: I was mostly interested on how to make the negative pulse. Indeed, the positive and negative pulses can be generated by two different CPU pins.In this case, the load can be put across the circuit. As well the current from the CPU is limited or can be limited by a resistor. Something like this

author: MK14
author_rank: Super Contributor
timestamp: December 24, 2016, 02:31:08 am
content: I've drawn out the circuit I was trying to describe in words, and simulated it.I'veNOTspent the time to optimize the values, specify the transistors (should be real easy) or check it out properly.But it seems to simulate as intended.Edit: I've noticed that I mis-connected the output load R5. I've corrected it, and the simulation seems unchanged. I've NOT spent the time re-pasting and updating the files for the forum.Edit2: One mistake I've just noticed. If the output pin (which supplies the PWM signal), floats, initially while the MCU boots itself up and starts running the application software. It could attempt to turn on both the top and bottom transistors. There is probably a moral to that somewhere.One solution is to just put a pull down resistor, between the floating output and ground, of say 1K (may need a different value). That should be high enough for the MCU to over-ride it when it turns the pin into an output and yet low enough to force it into a defined state.Edit3: Being critical of my own circuit. Now I've seen it in the flesh (schematic). It seems to be too reliant on the MCU pin, providing output voltages very close to either exactly 5V or 0V. That WON'T necessarily happen in practice, especially if there are other things it connects to (I DON'T know what you plan for this). So it would probably be better to put in a logic "buffer" stage, to replace the early parts of this circuit. So it can cope with a much wider range of output voltages. Not easy or so straight forward, because of the positive and negative supply needs.SORRY!.In the following plot. The blue trace is the 0 to 5 volts, PWM signal (fed into the circuit). The Green trace is the resultant -15v to +15V output.

author: MK14
author_rank: Super Contributor
timestamp: December 24, 2016, 03:41:01 am
content: On reflection. I now think the earlier opinions were best.I.e. Something on the lines of a simple opto-isolator, on the 5V logic side. Followed by a simple, driver stage, driven by that opto-isolator.Hence just two small and low cost integrated circuits, and a lot less hassle.The driver stage (+ and - 15 volts) could be a motor driver (H-bridge), op-amp, and many, many other options. In a single IC package (plus the opto-isolator).In some cases, you may not even need the opto-isolator (e.g. op-amp, which could be a high current output voltage one).I suppose "Simplest solution", actually means different things to different people.To me it means one or two ICs and not much else. Job done. Rather than fiddling with a dozen or so small passive and active components, for ages, to get it working. There are many ways such a circuit could misbehave.

author: basinstreetdesign
author_rank: Frequent Contributor
timestamp: December 24, 2016, 03:57:31 am
content: The easiest way to make negative pulses to a power rail below ground is with a PNP BJT with base connected to ground through a resistor (say about 5K), the emitter goes to the uC output and the collector connected to the negative rail, -Vrail, through a load resistor(say 10K). With the uC o/p low the Xsistor is off and the collector is near -Vrail. When the uC pulls the emitter high. the grounded base turns the Xsistor on and yanks the collector high to the uC output voltage. If you want the pulse to be limited to ground then clamp the collector with a diode to same. 4 parts. Easy.Follow this with the power stage of your choice to get more current.Tim

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 24, 2016, 05:30:56 am
content: Thanks, this is a good one.With the diode, the output is at 0.6V and not 0, but if a NPN BJT is used with base at the collector of the first stage, everything is perfect.

author: Zero999
author_rank: Super Contributor
timestamp: December 24, 2016, 10:49:40 am
content: Quote from: MK14 on December 24, 2016, 12:06:48 amYou're right about shoot through but it won't be an issue as long as the power supply's current is limited to a safe level. Failing that, it's fairly easy to limit the current with emitter resistors and diodes to clamp the base voltages.The transients on the MCU pins will be clamped by the ESD diodes. If it's an issue, add Schottky diodes from the output to the 5V and 0V power supply rails.

author: MK14
author_rank: Super Contributor
timestamp: December 24, 2016, 12:58:46 pm
content: Quote from: JacquesBBB on December 24, 2016, 05:30:56 amIt does seem to be working nicely!Quote from: Hero999 on December 24, 2016, 10:49:40 amYour modifications seem to fix the various issues.I find that in H-bridge (or similar) configurations. Shoot-through can easily break the output transistors. If you are NOT careful.I guess part of the problem is that such circuits are often used with DC motor drivers (H-bridge), which can need quite large currents (at start up). Giving the potentially too high shoot-through currents, enough current to potentially break the output transistors. Depending on the transistors specifications.It is neat doing it with the capacitors.

author: Zero999
author_rank: Super Contributor
timestamp: December 24, 2016, 01:49:59 pm
content: Here's another solution. It's similar to the one you posted, except it won't turn on both transistors if the MCU's output floats. There will also be some shoot through during switching, which is solved by current limiting.By the way, in LTSpice, you can rotate components by pressing ctrl+R and mirror them with Crtl+E; hopefully that'll make producing more readable schematics easier.

author: MK14
author_rank: Super Contributor
timestamp: December 24, 2016, 03:11:29 pm
content: Quote from: Hero999 on December 24, 2016, 01:49:59 pmThat's looking even better than your previous circuit. Without doing a huge analysis of it. It looks like it will work well.Although I really, really like LTspice. I try and use the on screen buttons, to do the components/mirroring and the schematic tends to just scroll like crazy, when I do that. So I must be doing it wrong!Using the keyboard short-cuts (like you suggested), is a good solution to that/my problem with that.Sorry that the schematic was a bit sloppy and badly drawn (I fully agree). I was trying to do it quickly, for the forum post. Clearly, doing it that way (in a big rush, to minimize time spent on doing post), has produced poor results.I'm finding the time gaps in my usage of LTspice, are big enough, to somewhat forget how to use it. Sorry again, about the poor quality schematic.Also those output transistors you keep on using, are amazingly well specified. The guaranteed, minimum Hfe of 250 at 100 mA's (for both NPN and PNP, On-Semi version), is very, very good. It is all too easy to lose too much Hfe (via the datasheet), at higher currents and it is often not specified (at such a generous amount), for small signal transistors (or very low end/small power transistors).

author: basinstreetdesign
author_rank: Frequent Contributor
timestamp: December 24, 2016, 06:11:09 pm
content: Quote from: JacquesBBB on December 24, 2016, 05:30:56 amNot quite. I should have added that with the diode clamp a resistor of a few hundred Ohms to 1K would be needed in my circuit in the emitter leg to limit the clamp current.But with your variation, you need a resistor in the base of the NPN to limit NPN base current. Otherwise when the uC o/p goes high and turns on the PNP, the forward biased eb junction of the NPN will not limit current putting both transistors and the uC in jeopardy.

author: Zero999
author_rank: Super Contributor
timestamp: December 24, 2016, 09:16:31 pm
content: Quote from: JacquesBBB on December 24, 2016, 01:04:46 amThat doesn't sound like a very good idea. If something goes wrong with the MCU/software, then both transistors could turn on, causing smoke.Here's another variation of the circuit I previously posted, which avoids shoot-through by introducing some dead time. The turn-on is delayed slightly by adding an additional diode-RC network to the input stage.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 24, 2016, 10:28:51 pm
content: Thanks to All for their input. This was very enlightening.In fact, It happened that the real need of my friend was quite different from what I originally thoughtfrom his (not so well formulated ) question.But it made me think to this theoretical problem of how to generate a negative 15V pulse with significant current from a positive 5 V pulse.I was surprised to see that it was not as obvious as generating a positive 15V pulse, and it was nice to see all the solutions that came along,based on very different principles.The optocoupler is a solution, but certainly not an answer to the question that I was thinking of which was to provide an answer,with only very basic components, as for the positive pulse : essentially transistors and resistors with minimal number.My question was really : what is so different with a negative pulse ?I am impressed by the capacitive coupling solution of Hero999 and the clever use of the PNP BJT by basinstreetdesign,but all the other inputs were really amazing in such a small time.So thanks again to everybody.Jacques

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 24, 2016, 10:43:42 pm
content: Quote from: Hero999 on December 24, 2016, 09:16:31 pmYou are right, but as I said above, I am not sure that this was the real requirement. But although I do not have the precise question, your solution is very nice. Thanks.

author: bktemp
author_rank: Super Contributor
timestamp: December 24, 2016, 10:50:19 pm
content: Quote from: JacquesBBB on December 24, 2016, 10:28:51 pmNothing, but positive+negative requires level shifting in both directions.And both level shifters must not generate overlapping pulses, otherwise both output transistors are switched on at the same time.

author: BrianHG
author_rank: Super Contributor
timestamp: December 24, 2016, 11:37:44 pm
content: A 10Khz square wave, you can always use a line transformer... Well, LOL,,, ok, I guess today that's more complex and pricey than a few transistors & resistors & caps. Or use the output of that transformer to drive a class AB emitter-follower. That should give a really good current on the +/- rails.Other IC Alternative: instead of an OPamp, a RS-232 line driver would do the trick & many have a built in switched cap power-supply which will convert your source 5v to +/-15 volts for you! They will also digitally clean your source signal and operate from DC up to the MHz range. Unlike the OP-Amp, the output with these will hit the +/- rails with balanced current each way & it will be switched super fast & some transmitter ICs have a tristate option on the output and they draw much less current than most OP-amps which can go up to +/-15v.Question JacquesBBB, do you have anything against using small signal mosfets like 2N7000, or VP2106 instead of BJT?Would you accept a constant current source single ended class A output? This means your power supply will draw between 100ma and 200ma during a low output depending on your load, and between 0ma and 100ma for a high output depending on your load. Your transistors will get toasty as well...

author: MK14
author_rank: Super Contributor
timestamp: December 25, 2016, 12:38:11 am
content: Quote from: BrianHG on December 24, 2016, 11:37:44 pmThat's a very clever and neat idea.They are usually robustly built as well. Since they are designed to cope with a significant number of feet of cabling and probably a degree of abuse as well. Especially in noisy environments.It will cope with the TTL (or similar) logic levels as well.I think there are many different, one or two IC solutions to this kind of problem. Hence my change of tact, earlier in this thread.Since in my opinion, a single cheap IC (such as you describe) is quick, easy, cheap andsimple. Which seemed to be the original objective.Rather than spending ages, messing around with and fine tuning a ten or twenty passive/small component solution. When a $1 common bread and butter chip or two. Will do it, with little design effort or messing around (hopefully!).

author: Zero999
author_rank: Super Contributor
timestamp: December 25, 2016, 12:57:21 am
content: The trouble with an RS-232 driver is it won't be able to output the required 100mA. The current is typically limited so something small such as 10mA. The voltage drop will also be higher than plain old transistors configured as common emitter amplifiers.Source, data sheet for the old 75188.http://www.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=mc1488&fileType=pdfThere may be better ICs around now with a higher current output. If anyone knows of any, please say.

author: MK14
author_rank: Super Contributor
timestamp: December 25, 2016, 01:29:44 am
content: Quote from: Hero999 on December 25, 2016, 12:57:21 amThere are many examples, such as ST's L6362A.ST's selection guide for more options:http://www.st.com/content/ccc/resource/sales_and_marketing/promotional_material/selection_guide/group0/ac/24/ac/22/0d/da/4e/40/sgips/files/sgips.pdf/jcr:content/translations/en.sgips.pdfWhich can do something like 0.3 Amps output (Adjustable limit).http://www.st.com/content/ccc/resource/technical/document/datasheet/12/f7/23/8b/fa/07/48/98/DM00170703.pdf/files/DM00170703.pdf/jcr:content/translations/en.DM00170703.pdfAbout £2.12 for one off's. Currently available and in stock.http://www.digikey.co.uk/product-detail/en/stmicroelectronics/L6362A/497-16410-ND/5962628

author: BrianHG
author_rank: Super Contributor
timestamp: December 25, 2016, 01:32:48 am
content: Quote from: Hero999 on December 25, 2016, 12:57:21 amUse a transmitter IC with 4 outputs, wired in parallel with one MC1488 at 0.50$ each.http://www.ti.com/lit/ds/symlink/mc1488.pdfhmmm, ok, with current limited devices, you'll get +/- 40ma this way and we wont sit 2 or 3 on top of each other to get that 100ma.Taking the output of 1 transmitter and driving a 2N3904 and 2N3906 in an AB emitter / follower, that's a 50cent IC + 2 x 5cent transistors, with an optional 2x 1 ohm resistors in series between the emitters as a little protection, you will get +/- 14.5v out, with a clean kick of way over 100ma. If you used beefier transistors, you will get over +/- 1amp output. This circuit gets rid of all those diodes/caps/resistors in the previous circuits + no worries of both transistors switching on at the same time. If you need the full +/-15v, I would just increase the power supply by the 0.5v.

author: MK14
author_rank: Super Contributor
timestamp: December 25, 2016, 02:08:04 am
content: Quote from: BrianHG on December 25, 2016, 01:32:48 amAnalogy:My common sense argument (agreeing with what you just said, and what I said on similar lines, earlier in this thread), is that if a project/design/requirements, basically needed a few logic gates, flip-flops and a simple amplifier to perform it.Ignoring the fact that you may just elect to use an MCU or similar, in practice.You might use a few IC's, such as a 74HC quad Nand, binary counter and an op-amp IC.Rather than mess around with 25 discrete transistors and mosfets. Along with dozens of diodes, resistors and small capacitors.So I don't necessarily see why this threads solution should be any different.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 25, 2016, 06:00:48 am
content: Quote from: BrianHG on December 24, 2016, 11:37:44 pmYes, this will qualify for my initial question. Show us your solution.

author: MK14
author_rank: Super Contributor
timestamp: December 25, 2016, 10:20:54 am
content: Another way of doing it would be using (single IC power Op-Amp, there are many different ones available, this is ONLY one example)...On-semi's LA6500-E, which can supply up to about an Amp, at up to +/- 15V (or 30V DC).It is in a TO-220-5H case, so it can easily be heatsinked if necessary.In stock, £1.45 each.http://www.digikey.co.uk/product-detail/en/on-semiconductor/LA6500-E/LA6500-EOS-ND/2748211Datasheet:http://www.onsemi.com/pub_link/Collateral/LA6500-D.PDFYes, it could be designed out of discrete transistors, and many passive component. But why bother (unless you want to do it that way for fun, or to create a retro/vintage item, etc). But opinions can vary (as to when to use ICs and when to use discretes), which is fine.Or you can also do it by using a normal op-amp and power transistors or mosfets on its output, with feedback, to make your own, power op-amps.(Edited from my original post, to make a lot shorter and made more general purpose).

author: MK14
author_rank: Super Contributor
timestamp: December 25, 2016, 10:48:41 am
content: Quote from: JacquesBBB on December 25, 2016, 06:00:48 amIn all fairness to your original question.It does look (at a quick glance), like something which can be solved with a few discrete components and bit of skill. For $1 in total jelly bean parts.But in practice, it looks like (as others have already said) for a number of reasons, that might not be the case. Such as the problems of the top and bottom transistors, conducting at the same time, and the level shifting, etc.I've come across the phenomenon, with a number of other things in electronics. Where the component costs, say £5 each. Yet one is convinced that a £0.03 component should be able to do it.It is only when you fully understand all the technical problems and issues, with what you are trying to do, when you realize that the £5 component, was worth every penny.E.g. £299 Fluke multimeter vs £1 bargain bin multimeter.

author: Zero999
author_rank: Super Contributor
timestamp: December 25, 2016, 04:36:38 pm
content: Quote from: MK14 on December 25, 2016, 01:29:44 amThere are other factors than cost. For example, L6362A would require a reflow oven, which many people don't have.Quote from: MK14 on December 25, 2016, 10:20:54 amI wouldn't recommend an op-amp, which will be very slow. Generating a 10kHz squarewave with a decent rise/fall time requires a high slew rate and many op-amps also take awhile to recover when the output is saturated to either supply rail. The LA6500 certainly won't do. It has a slew rate of just 0.15V/?s and would take 200µs to go from -15V to +15V, not including the time required for it to come out of saturation, which isn't specified on the data sheet. The period of a 10kHz waveform is 100µs.A comparator would be better but it could be difficult to find one which will operate from a +/-15 power supply and has a +/-15V push-pull output.An h-bridge IC, such as the SN754410 or L293D, is another option but it would still need level shifting, since most h-bridge ICs only level shift upwards, not downwards.http://www.ti.com/lit/ds/symlink/sn754410.pdfhttp://www.ti.com/lit/ds/symlink/l293.pdf

author: MK14
author_rank: Super Contributor
timestamp: December 25, 2016, 05:42:29 pm
content: You're right, sorry (packaging, slow slew rate, etc).But it depends on exactly what they want to do, which is not clear to me, from the OP.You can get considerably better ones, but I don't know enough about, what they are trying to do, to choose for them the right one.Also I agree with you. In some use cases, a power op-amp would be the wrong choice.I messed up the post, by shortening it too much.Unfortunately I cut out the section, which explained that you can get a very large variety of different ones.So go ahead and choose the right one, to suit your own requirements.Since I'm NOT 100% sure what those requirements are, I will leave it to the user, to choose which one to go for.You can get them in different case styles, to your preferences.This example is through hole...It takes something like 20 ns to reach the final value. (or 200 ns, depending on exactly what you are measuring and/or configuring it). ns figures via a quick look at some of the test graphs. I'd need to know more, to calculate better figures.But it could be done, from the details supplied. But I prefer to do things more thoroughly.At a little over £4 each.http://www.digikey.co.uk/product-detail/en/linear-technology/LT1010CT-PBF/LT1010CT-PBF-ND/891417Datasheet:http://cds.linear.com/docs/en/datasheet/1010fe.pdfPower op-amps are just one variety of IC which can do this.There are many other types which will do it as well. I was just trying to show one of those ways.

author: BrianHG
author_rank: Super Contributor
timestamp: December 26, 2016, 12:08:11 am
content: Here is my super simple 5 component solution. I hope I didn't break the original question's rules.Now, in the attached schematic, there are only 5 parts.Though, since Q1 radiates up to 3 watts max, I had to use something stronger than a 2N3906 with the ability to add a heatsink.Resistor R1 requires the +15v to be +15v, plus, (R1 may need tuning!) as the transistor heats up, the 100ma constant current source will increase, but, this doesn't break the rules and it wont affect functionality.Mosfet Q2's RDS on at 4.5v is so low, that the output will be -15v with both the CC pull-up of Q1 and the 100ma output load combined. Under this circumstance, the -15v power supply will have a 200ma load while the +15v power supply will always have a 100ma load.The output is obviously inverted.You can add a 100k resistor across the Zener is you want the Mosfet to default to off without a switching output.This is not the neatest solution, just the fewest parts and no power-up worries.

author: Zero999
author_rank: Super Contributor
timestamp: December 26, 2016, 12:20:41 am
content: Quote from: BrianHG on December 26, 2016, 12:08:11 amI think it might affect functionality quite severely. There's a huge risk of thermal runaway. The current increase as the transistor warms up, causes the it to dissipate more power and the transistor to heat further. This positive feedback loop can result in destruction of the transistor. You need an emitter resistor, plus a couple of diodes to clamp the base voltage to limit the current to a safe level.

author: BrianHG
author_rank: Super Contributor
timestamp: December 26, 2016, 12:38:09 am
content: Quote from: Hero999 on December 26, 2016, 12:20:41 amYes, in class AB amps where the emitters of the PNP and NPN are tied together with only a slight increase of beta, the amps, yes, going from 0.01 to 0.1, then .5, then 1, then 2... ramp up like crazy and thermal run-away kills such designs in a puff of smoke. Remember, the change in temp also changes the Vbe characteristic of both transistors which have been tuned to get rid of that voltage gap between the positive and negative!Remember, this is not an emitter follower amp with 1 emitter tied to the next with sitting on a bias margin tuned to remove that voltage gap between the positive and negative. The thermal change wont affect the current by more than 100ma in my circuit, still operating within a safe margin.This is a common emitter current amp. The current set across the base by the 9k at 15v is approximately 1.7ma regardless of transistor temp. The transistor at 25 degrees is rated with a current gain of 60. The result is the transistor should cc at 100ma and radiate 1.5w with a 50/50 duty cycle with this circuit. Now when the transistor heats up, the maximum current gain of my selected transistor at 150 degrees Celsius is 100. This will only increase the CC to 167ma, or, 2.5 watt with a 50/50 duty cycle. Since the current isn't sky-rocketing like when you have a PNP and NPN emitter-follower configuration, this circuit wont go completely out of control.

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 26, 2016, 01:07:13 am
content: Quote from: BrianHG on December 26, 2016, 12:08:11 amChange the bipolar to a NPN with the emitter to the output and the collector to the supply. Change the base resistor to one going to the supply, the value of which depends on what droop and power dissipation you're OK with. Move the MOSFET drain to the base of the NPN, then finish the design by adding a diode from the output so the MOSFET can pull down the output.A variant on that would use a second N channel MOSFET instead of a bipolar along with a charge pump to get enough gate drive voltage. And another step is to add a small NPN and diode to accelerate turn on of the gate. That's way overkill for this use, but I have seen that design in cheap power inverters. No shoot through and a short dead time are its main advantages. The main disadvantage is that it doesn't tristate.

author: BrianHG
author_rank: Super Contributor
timestamp: December 26, 2016, 01:47:11 am
content: Quote from: NiHaoMike on December 26, 2016, 01:07:13 amI see what you have done. With 3 mosfets total, a few caps and 2 additional diodes, it is possible to make a +/-4 amp, near zero power wasting +/-15v switch using just the same mosfet I have in my circuit. You would also should be able to use a 2N7000 in such a scheme, but, the switch on for the 2N7000 at 4.5v for the gate will only be slightly more than 300ma.I was trying to get that +15v output right at 15v and the -15v right at -15, with the fewest parts possible. The constant current transistor amp trick with the ultra-low Rds on mosfet was the quick cheap root. Making a bootstrapped output switch eliminates all the wasted current, but, requires the additional switch for the high-side gate with it's own protection diode + charge cap.

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 26, 2016, 02:02:43 am
content: Quote from: BrianHG on December 26, 2016, 01:47:11 amThe clever part of the design is that the bottom MOSFET acts as the driver for the top MOSFET or bipolar. The only other part that has to handle the full output current is a diode.

author: BrianHG
author_rank: Super Contributor
timestamp: December 26, 2016, 02:19:40 am
content: Quote from: NiHaoMike on December 26, 2016, 02:02:43 amOoops, I took a things 1 step further, well, the original question was to be as simple as possible with 1 transistor and 1 mosfet.I'll post my bootstrap circuit in a few hours. In my circuit, there is no current being driven across a diode, but, it does need the addition of a 2N7000 used as an inverter gate.

author: Zero999
author_rank: Super Contributor
timestamp: December 26, 2016, 09:59:04 am
content: Quote from: BrianHG on December 26, 2016, 12:38:09 amI see your logic but still don't think it's good design practise to rely on the Hfe of a BJT to be a specific value. The gain on the data sheet may be specified as 60@25oC which is just the typical value. In reality, it could be over double that, so you'd need to take that into account when sizing the heat sink. On the other hand, at low temperatures, the gain might be lower, so it will need to warm up before it starts working properly.

author: bktemp
author_rank: Super Contributor
timestamp: December 26, 2016, 11:59:16 am
content: If ac coupling is allowed, the circuit postet by Hero999 is a good start.If you remove all resistors and use only a single resistor at the input, there is virtually no cross-conduction during switching, because the voltage can only turn on one transistor if the base of the other transistor has reached 0V base-emitter voltage.The only problem is turn on, but if the power supply is current limited and turned on slowly this shouldn't be a big problem.I often use similiar circuits for driving 12V latching relays using +/-5V.

author: Zero999
author_rank: Super Contributor
timestamp: December 26, 2016, 06:09:38 pm
content: Quote from: bktemp on December 26, 2016, 11:59:16 amWell that does simplify it a bit. The idea with two base resistors was to limit the turn on transient. The other resistors in series with the diodes didn't do much though. I'd still include the emitter resistors and clamping diodes, unless you're certain the power supply is adequately current limited. To simplify things D1 & D2 could be 2.7V zener diodes, then 2R2 emitter resistors would limit the switch on transient to about 1A.

author: BrianHG
author_rank: Super Contributor
timestamp: December 27, 2016, 02:29:26 am
content: Quote from: bktemp on December 26, 2016, 11:59:16 amJust placing 2 resistors at the base inputs in your design, inbetween C1 and C2, connecting the input to the center of the 2 resistors & using a value of, say around 2.2k when using a 3904/3906 combo would limit the power-up current to a short 400-600ma spike as the caps charge and also limit the output operation current around 100ma as well. This would be my choice for the optimum solution for few parts and safe power-up and an elegant circuit solution. Now, this solution relies that the input hit the 0 and 5 volt rails equally with a 2.2kohm load.Note that with all these solutions, a ripple noise on the +/-PSP at around of 1v with reference to your digital IO GND could cause the transistors to switch on sporadically. Your should have decoupling caps at the transistor emitters to the GND of your IO's output connector. (This problem includes my mosfet circuit as well....)