Common emitter push-pull?  - Page 1
Forum section: electronics / beginners

author: TheBaconWizard
author_rank: Regular Contributor
timestamp: June 22, 2018, 06:47:15 pm
content: Hey guys ‘n gals,Been looking at push-pull circuits and therefore amplifiers.What I want to do is actually power a voltage multiplier, it has nothing to do with audio.I am looking for amplified (or in fact switched, since the transistors can and will be in saturation mode when on) voltage gain, not only current. To this end I found about a class A amplifier ie common-emitter.Now, I definitely want a push-pull arrangement, but it seems to me that preserving the common-emitter topology will help me. Does that sound right?Here is the usual arrangement minus biasing etc:And here is the arrangement I think I want, taken from the web but the values aren’t mine, just an example. I found so few such examples that I wonder I it’s actually an error:It’s first principles I need here rather than specifics, i’d Want to use midgets eventually anyhow. My question is, does the 2nd example work and does it give me what I want? If not, what will?I’m feeding the gates/bases with a clock signal of between 0 and 12v, and Vcc is 140v

author: rstofer
author_rank: Super Contributor
timestamp: June 22, 2018, 06:59:52 pm
content: Given that your control voltage is 12V and your V+ is 140V, how do you propose to shut off the top transistor? You need to get the base to within 0.7V of the rail. You might be able to do that with another NPN controlling the base. Use a pull-up resistor from the base of the PNP to V+ and let the NPN pull it down to turn on the PNP.How fast is the rise time? More to the point, if both transistors are conducting even a wee little bit of current, the 140V is going to make for a spectacular failure mode. If the bottom transistor has 0.7V on the base, it will be conducting. Without the inverter described above, that same 0.7V on the base of the PNP will have it full on. Should be fun to watch - from a distance.I haven't thought about how to guarantee there is no overlap but given the lack of current limiting in V+ (like a resistor or something), overlap will be a huge problem. Even at lower voltages...

author: Buriedcode
author_rank: Super Contributor
timestamp: June 22, 2018, 07:03:04 pm
content: In the second figure, what happens when you're driving signal (to the bases) is greater than ~0.7V and less than 19.3V ? Hint: Both transistors will be onFor very fast transitions this cross conduction can be brief, but your driving signal would have to hit both rails to fully turn off either transistor and not "linger" inbetween. I've used common emitter push pull to drive medium power charge pumps, as they are essentially just amplifying current.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 22, 2018, 07:08:37 pm
content: What are you actually after -- what is the load, how much voltage and current? What is the supply? How fast are you driving this, how fast does it need to go?Common emitter is tricky because the base voltage is relative to the emitter. That's fine for the NPN, but you need to drive the PNP at +140V.Common collector saves you the drive current, but has the problem that you need to generate the 140V swing. Needing less current does open up the option of a class-A gain stage (you'd probably use some circuitry up at +140V, but it's passive, you don't need to pass any signals up to it), which makes it a popular solution to this kind of problem.Tim

author: JS
author_rank: Frequent Contributor
timestamp: June 22, 2018, 07:17:10 pm
content: Vcc 140V?That looks like a lot and you are going to need some serious tranaistors. Also, you are probably better off with a voltage gain stage and then a classic class B push pull config.To get to 140V you can't work directly with an opamp, so you will need some more circuit arround it. Also, for the common emitter oush pull output stage you need quite some circuit to keep it stable and the bias under control, so the complexity of an extra stage might be lower than the single stage with all you need around. There are reasons to go this route but you haven't mention any of them, like low drop, close to rail output, as you could get only a saturation voltage to the rail with common emmiter config while you could loose over a volt from rails with common collector.JS

author: TheBaconWizard
author_rank: Regular Contributor
timestamp: June 22, 2018, 08:01:47 pm
content: Sounds like that’s the plan, then: go for a common-emitter stage or stages and finally a standard push-pull arrangement

author: tpowell1830
author_rank: Frequent Contributor
timestamp: June 22, 2018, 09:33:47 pm
content: Since you want full ON when your signal ticks, have you considered using a MOSFET H Bridge driver?Hope this helps...EDIT: I might add that the high 140V lends itself better to MOSFETs as well.

author: Zero999
author_rank: Super Contributor
timestamp: June 22, 2018, 10:17:13 pm
content: Add a common base stage, before the common emitter push-pull output.By the way, I can't remember whether the 150V BJTs used in the .asc file are included in the default LTSpice install. If not, it should simulate fine with the generic models, but I'll post them if you want. Oh and I know 150V maximum rated parts, is a bit close on a 140V circuit, but it's what I had in the library: use >160V parts, in real life.

author: JS
author_rank: Frequent Contributor
timestamp: June 22, 2018, 11:01:22 pm
content: FWIW I have some 200V BJT so voltage it's well within common parts and no need to go for mosfets, but you could.How did that sim run?JS

author: Zero999
author_rank: Super Contributor
timestamp: June 22, 2018, 11:53:41 pm
content: Just thinking Q3 will be dissipating a fair amount of power. I'm not sure if it's a good idea to drop that kind of voltage and current, even at 50% duty cycle, so a collector resistor is a good idea: I'd go with 22k.Another option is to add 100V zener diode and biasing resistor, so Q3 no longer needs to drop such a high voltage.

author: JS
author_rank: Frequent Contributor
timestamp: June 23, 2018, 12:22:31 am
content: Quote from: Hero999 on June 22, 2018, 11:53:41 pmMeh.. You need to drop that energy somewhere, expend your money in a zenner or a bigger tranny, I'd go for the single part.JS

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 23, 2018, 01:16:06 am
content: OP hasn't mentioned how much current or capacitance drive is required, so it's impossible to say how much drive the transistors also need.Tim

author: JS
author_rank: Frequent Contributor
timestamp: June 23, 2018, 03:07:01 am
content: Quote from: T3sl4co1l on June 23, 2018, 01:16:06 amHe used 1k load in the schematics, looks like a lot, 20W to the load but who knows...JS

author: TheBaconWizard
author_rank: Regular Contributor
timestamp: June 23, 2018, 11:35:01 am
content: Some seriously fantastic and detailed help here, MANY thanks everyone. I have work to do!

author: Zero999
author_rank: Super Contributor
timestamp: June 23, 2018, 05:17:06 pm
content: Quote from: JS on June 23, 2018, 12:22:31 amI suppose the zener could be replaced with a cheap resistor, but it would still require a high voltage transistor.Quote from: JS on June 23, 2018, 03:07:01 amAre you referring to my schematics? I guessed and a 1k load would not dissipate 20W, but 4.9W, since it's capacitively coupled it would get 70V, not 140V.

author: JS
author_rank: Frequent Contributor
timestamp: June 23, 2018, 07:10:08 pm
content: Quote from: Hero999 on June 23, 2018, 05:17:06 pmDoh!The only part seeing almost the entire 140 volts at full load would be that tranny, but at the vase you would only need a few mA so will only be dissipating a watt or so, no worries, can be done and you need a big tranny anyway to stand the 140V, it could probably be the same as the output one.JS