200V 10MHz squarewave

Discussion in 'General Electronics Chat' started by TheComet, Mar 15, 2013.

  1. TheComet

    Thread Starter Member

    Mar 11, 2013
    88
    12
    *sigh* I'm stuck.

    Requirements
    Wave from MUST be symmetrical. This means the same two MOSFETs need to be used, where one of them has to be bootstrapped.

    Output Current: 1mA
    Output Voltage: +-100V (bipolar)
    Output frequency: Anything between 50kHz - 10MHz
    Output waveform: Squarewave (at most 20ns rise/fall between 100V and -100V)

    The driver has 3 states: Either it's outputting 100V, -100V, or GND.

    NOTE: It's not a constant frequency, but 3 successive pulses followed by a long pause of 10ms.

    Setup
    Available Voltage Sources: +100V, -100V, +5V, GND

    The signals for driving the low side and high side transistors are already available at 10 MHz, 5V with deadtime:

    [​IMG]

    As mentioned, they aren't continuous. The 3 pulses only occur every 10ms (I set it to a lower period for simulation purposes):

    [​IMG]

    Here's the idea I'm currently working on. A manual bootstrap:

    [​IMG]

    No matter what I try though, I just can't get it to look right. The output wave doesn't go negative, the capacitor C1 doesn't charge enough to power the upper nand gate for long enough, and there are probably other things completely wrong with this schematic.

    Any help or ideas are greatly appreciated!

    TheComet
     
  2. shortbus

    AAC Fanatic!

    Sep 30, 2009
    4,010
    1,530
    Don't think your going to be able to do this with surface mount mosfets. You also need to pick mosfets with "head room", rated at higher voltage and amperage than what your switching. Around two or more times higher is a good place to start.

    You should be looking at using mosfet drivers to drive the mosfet gates. To turn them on and off fast enough. Logic gates can't supply enough current to the gates. Since the mosfet gate is basically a capacitor. C2 makes that gate inoperable, since a capacitor can't pass DC.

    Does your "out" voltage power something? If so your mosfets are "high side switches".

    This should give you some things to research for your project. Some one else may have other ideas for you.
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    200V in 20ns is 10^10 v/sec. I=C*dv/dt. How much capacitance (including stray) are you driving?

    EDIT: Also, P=F*C*V^2. So, if F=10MHz, and V=200V, then P/C=1e7*4e4
    P=4e11 Watts/Farad, or
    P=400mW/pF.

    A 10pF load will require 4 Watts just to charge and discharge it, dissipated by the transistors. This is in addition to shoot-through losses (during rise and fall times), which will be hard to avoid, IMHO, at 10MHz.
    I'm not trying to scare you off. You just need to be aware of the power requirements.

    And what's up with R8? That will require 2A p-p!
    There are so many problems with your circuit, I don't know where to begin.:( Shortbus gave you good advice.
     
    Last edited: Mar 16, 2013
    shortbus likes this.
  4. timescope

    Member

    Dec 14, 2011
    298
    44
    Hi,
    BSS123 is a 100v transistor and your peak to peak voltage is 200v. Do a google search for 400v logic level mosfets.

    Design a 5v supply at the -100v rail to power a mosfet driver for the lower output transistor. This could be a simple resistor (> 1W) in series with a zener diode connected between GND and -100v with a capacitor in parallel with the zener diode.

    I have attached a schematic of some ideas to drive the lower transistor. I have used current switching in an attempt to obtain the required speed.
    I have not simulated it. You can replace the complementary transistors that drive the lower transistor with a mosfet driver.

    Timescope
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    The delay of the level translator required to drive the low side MOSFET will totally change the timing required to maintain the dead zones.
    At 10MHz, you should be able AC couple the load, and just use one high voltage (200V) supply, unless the duty cycle changes.
     
  6. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    Also, what will the load be?
     
  7. timescope

    Member

    Dec 14, 2011
    298
    44
    Agreed. We could also move the IR2110 down to the -100v line and drive the inputs with two identical level translators.

    The supplies are specified : +/- 100v, +5v.

    Timescope
     
  8. TheComet

    Thread Starter Member

    Mar 11, 2013
    88
    12
    Sorry for the very late reply.

    Forgot this important fact.

    It's supposed to drive a piezo-crystal, which is designed to have a reactance of 100 Ohms @ 10 MHz. This crystal is in series with a 100 Ohm resistor, so the total impedance is 144 Ohms.

    Believe me, I looked for a solution with an IC, but they don't seem to exist. Either they can withstand 200V and can't switch within 20ns, or they can switch within 20ns but can't withstand 200V. I couldn't find any that meet the requirements, that's why I'm forced to design my own high side driver.

    Your calculations are neglecting the fact that the driving pulses only occur every 10ms. You're calculating the power dissipation during switching times - and you're right, a lot of power is being consumed - but there's a long pause of 10ms in between every burst, that's why I can get away with it.

    This is what I ended up doing for the high-side driver, thanks for the tip!

    @all

    Here's an update on the circuit, and it's kind of working the way I want it to now (I know some of the components wouldn't be able to handle the voltages, I'm working on that part now).

    What do you all think? I've attached the simulation files to this post.

    Total power dissipation is ~200mW with a load of 150 Ohms.

    [​IMG]

    blue: high-side signal
    red: low-side signal
    light-blue: output signal
    VCC=5V
    VDD=100V
    VSS=-100V
    HS=high side signal (0..5V)
    LS=load side signal (0..5V)
    [​IMG]

    TheComet
     
    Last edited: Mar 18, 2013
  9. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    I'm just throwing this out there as it is in an area that I don't yet know much about, but don't RF circuits commonly play a game where they generate a signal at one voltage/current level and impedance and then use a transformer to match the impedance to the load which changes the voltage/current levels at the same time?

    So would it be possible to perhaps work at a lower voltage and higher current and then transform them to the higher voltage and lower current just prior to the load?
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Comet, I zoned out on the 10ms burst rate.
    Do you have a datasheet for the piezo?
     
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