| Electronics > Beginners |
| Common emitter push-pull? |
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| TheBaconWizard:
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 |
| rstofer:
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... |
| Buriedcode:
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 on :) For 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. |
| T3sl4co1l:
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 |
| JS:
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 |
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