High Voltage Op-Amps

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It won't work, when switched on two tiny black holes will
be created and the rest of the board will be sucked in.
The Vceo of the output transistors is only 140 and 150 volt.
This is a perfect example of what stupidities you can do
with a simulator.

on1aag.
 
Hero999 said:
You're right and Q5 is the constant current load for it.

I've just simulated it in LTSpice and it works perfectly.
Click to expand...
I'm guessing that his transducer needs to be driven at ≈40kHz. The amplifier's response as drawn is about 1.3dB down at 40kHz. It also slews at about 12V/usec, meaning you can only get about 100V p-p at 40kHz out of it, even with no load. Sinewave simulation shows lots of distortion at 40kHz with 80V p-p output. The output of the LT1001 is a triangle wave under these conditions.
If 40kHz is not the target frequency, then never mind!
There are probably things that can be done to improve this performance.
 
It can be fixed by doing the following:
  • Use a faster op-amp.
  • Reduce R12 to increase the voltage gain of the buffer and put less strain on the op-amp, be careful too higher gain will cause oscillation.
  • Reduce the compensation cacitor values, again watch out for oscillation.
 
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Hi there,


Hmmmm, did somebody say 120 volts at 1.2 amps? This is got PWM written
all over it. Anything linear is going to eat up lots of power reproducing
sine waves, unless square waves are acceptable of course, but then rise
and fall times have to be carefully considered also. MOSFET's anyone?
They are quite efficient for rectangular waves.

In any case, calculate the power lost in the output transistors
before deciding on a particular design and go from there.
 
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The problem is to build a decent class D amplifier, the switching frequency needs to be much higher the the bandwidth, for 40kHz the switching frequency would need to be 320kHz minimum.

on1aag,
Regarding the transistors: I just used the higest Vce transistors on LTSpice, I agree they need to be rated to at least 300V.
 
on1aag said:
Maybe it's a symmetrical transistor ?

on1aag.
Click to expand...
It turns into a low-beta transistor when you swap emitter and collector.
The simulation oscillates when you place the transistor correctly.
 
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Sounds good. I will do that along with a minor change to the emitter - follower circuit. I would like to use Darlington pair. Doing so I need to add 2 more diodes. The signal frequency will be varied from 6Hz - 60Hz so I probably won't have any issues with gain roll-off. I never dealt with switching amplifiers or components before so I doubt I can design a Class D amplifier at this point. So I will just stick to linear for now.

Thank you for your help.
 
Random

Oh and if anyone is getting tired of my old picture, I have a new one I can upload. But before I do that, I would like some comments on it...

Here it is:

**broken link removed**
 
Roff said:
Q3 is upside down.
Click to expand...
My (poor) excuse is that it's the default orintation for PNP transistors with LTSpice and all other simulators I've used have the emitter pointing upward as the default for PNP transistors.

on1aag said:
Hi Ron,

I noticed that too, at 1 MHz and above it looks a bit "playfull".

on1aag.
Click to expand...
Yes, as it now has more gain, it ocillates around 500mV at about 1MHz.

I've experimented a bit with the compensation capaictors and for some reason increasing their values makes it worse not better.

I came to the conclusion that they got a bit carried away with the compensation capacitors, removing C2 solves the problem, it only adds another phase shift which makes it more prone to oscillation.

It looks like you're building an inverter.

This isn't a very good way of doing it, PWM or modified sinewave is better.

Are you intending to use this to control the speed of an AC motor?

If so it might not work with a single phase motor. The motor cerainly won't start at 6Hz as the starting mechanism is frequency sensitive. You could have a go at running it at full speed, then slowing it down which might work.

Another thing is that at low speeds the voltage must be reduced to avoid core saturation. With a PWM or modified sinewave inverter this is easy, the duty cycle is just reduced.
 

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Hero999 said:
Another thing is that at low speeds the voltage must be reduced to avoid core saturation. With a PWM or modified sinewave inverter this is easy, the duty cycle is just reduced.
Click to expand...

So the voltage has to be reduced If I were to reduce the speed of AC motor by lowering the frequency? But that would reduce the torque. And how does a PWM work with AC motors? Will a duty cycle other than 50% not burn the winding of AC motor? Oh, and can you please explain core saturation?

Thanks once again.
 
Core saturation occurs when stong magnetic field applied to the iron core causing it to loose it's magnetic properties. There's a limit to the amount of magnetic energy the core can store, once the field exceeds a certain strength the core might as well not be there. This means that when the current flowing through the wire surrounding the core, the inductance will fall, if there's no inductance then a large current will flow, causing the motor to burn out.
**broken link removed**)

A PWM inverter uses class D amplification of a sinewave (the motor does the filtering itself), as the frequency is reduced the ampltude of the sinewave is also reduced.

A modified sinewave inverter uses a steped sinewave, the lengh of the on time is just reduced as the speed is reduced.

The chances are that if you applies a 60Hz 170V 50% duty cycle squarewave to the motor it would burn out. To work at 60Hz the RMS value of the voltage going into the motor needs to be the the same as a 120V sinewave, a 170VAC squarewave has an RMS voltage of 170V - far too high for the motor.

PWM is probably your safest option, for 60Hz the frequency doesn't need to be too high 1kHz would do but it would whine so you're better off keeping it to >20KHz.
 
Frosty_47 said:
So the voltage has to be reduced If I were to reduce the speed of AC motor by lowering the frequency? But that would reduce the torque. Thanks once again.
Click to expand...

The torque deppends on the applied current, not the voltage.

As you reduce the frecuency the reactance of the different inductances is also reduced, so with less voltage you have the same current.

Remember that XL = ω x L = 2 x Pi x f x L

Where XL = inductive reactance
ω = Angular frecuency (radians/second)
f = frecuency (Hertz)
 

Man I hate learning. Everytime I learn something, I discover that there is always something new I don't know about...

Anyways thank you for educatting me on this matter. Is there a way to calculate V/Hz ratio? And will LM25037 be a suitable choice for this application ?

Thanks
 
Hero999 said:
It's linear, for example for to run a 120VAC motor at 6Hz the voltage needs to be reduced to 12VAC.

That would do it.
Click to expand...


Thank you. I apriciate the time you spent educating noobs like me. I hope someday, I will be close to your lvl of knowledge.
 
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Frosty_47 said:
Oh and if anyone is getting tired of my old picture, I have a new one I can upload. But before I do that, I would like some comments on it...
Click to expand...

I've always thought, as a canadian, that our flag should be an eagle riding a beaver.
 
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