How to generate +15V and -15V pulses from +5V pulses ? Simplest way !

[MK14]

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=pdf

There may be better ICs around now with a higher current output. If anyone knows of any, please say.

There 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.pdf

Which 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.pdf

About £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

[BrianHG]

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.

Use a transmitter IC with 4 outputs, wired in parallel with one MC1488 at 0.50$ each.  http://www.ti.com/lit/ds/symlink/mc1488.pdf
hmmm, 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.

[MK14]

Use a transmitter IC with 4 outputs, wired in parallel with one MC1488 at 0.50$ each.  http://www.ti.com/lit/ds/symlink/mc1488.pdf
hmmm, 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.

Analogy:
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.

[JacquesBBB]

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...

Yes, this will qualify for my initial question.  Show us your solution.

[MK14]

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/2748211

Datasheet:
http://www.onsemi.com/pub_link/Collateral/LA6500-D.PDF

Yes, 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).

[MK14]

Yes, this will qualify for my initial question.  Show us your solution.

In 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.

[Zero999]

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=pdf

There may be better ICs around now with a higher current output. If anyone knows of any, please say.

There 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.pdf

Which 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.pdf

About £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

There are other factors than cost. For example, L6362A would require a reflow oven, which many people don't have.

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/2748211

Datasheet:
http://www.onsemi.com/pub_link/Collateral/LA6500-D.PDF

Yes, 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).

I 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.pdf
http://www.ti.com/lit/ds/symlink/l293.pdf

[MK14]

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/891417
Datasheet:
http://cds.linear.com/docs/en/datasheet/1010fe.pdf

Power 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.

[BrianHG]

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.

[Zero999]

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.

I 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.

[BrianHG]

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.

I 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.

Yes, 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.

[NiHaoMike]

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.

Change 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.

[BrianHG]

Change 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.

I 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.

[NiHaoMike]

Change 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.

I 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.

The 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.

[BrianHG]

The 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.

Ooops, 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.

[Zero999]

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.

I 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.

Yes, 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.

I 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@25^oC 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.

[bktemp]

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.

[Zero999]

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.

Well 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.

[BrianHG]

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.

Just 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....)