BJT amplifier/buffer with offset?

Discussion in 'General Electronics Chat' started by tom66, Sep 20, 2010.

  1. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    I want to be able to set the 'offset' of an a buffer. That is, given a signal of 5V and an offset of -4V, the output would be 1V (the offset could be positive as well, in which case the output would increase in voltage.) I could do this with op-amps, but I'm looking for high speed operation, and high-speed op-amps are expensive. I'm only doing this to teach myself about practical discrete amplifiers (as it's something don't know much about.) I've searched the AAC book but found little info.
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    What about the negative side of the pulse, is it -4V?

    If so, you'll need a negative power supply, if not, then it is clipping.
     
  3. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    Negative power supply is available. Assume maybe +/- 15V rails.

    The offset could be from -5V to +5V.

    The input is already buffered and with relatively high impedence (FET buffer.) But the signal has a 3.2V dc component that needs to be removed, along with any other dc component that may be on the input.
     
  4. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    OK, it is a given you will have a 0.6BE drop, this can not be corrected. I'm going to assume you aren't after an summing circuit, though that may be what your after.

    [​IMG]

    Just pick a bias point that creates the voltage you want.
     
    Last edited: Sep 20, 2010
  5. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    A summing circuit could well be what I need. I don't mind having to apply a negative signal to increase the offset or positive signal to decrease it, as long as I can control it with say a microcontroller or a slow op-amp (the slow op-amp inverting a DAC output perhaps.)
     
  6. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    Thanks for the circuit Bill, but wouldn't that remove all of the dc component? I want to be able to offset it under the control from some external voltage; I'm unsure on how to do this.

    Say I have a 1Vp-p sine wave with an offset of 3V, (going from 2.5V to 3.5V.) If I offset it by -3V, then I get effectively a 1Vp-p sine wave with no dc, but if I offset it by -2V I still have 1V dc. It would be easy with op-amps but challenging to find one capable of 100MHz+.
     
  7. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Replace C1 with another resistor and you have what passes for a summing circuit. It is drastically inferior to a op amp version, as there is no isolation between the two inputs. With an op amp both inputs go to a virtual ground, but no such animal exists here. I am also merging R2 with the second input that would be summed with the input. There may be better out there, but I'm not aware of them. Want a drawing? I'm warning you in advance though, it sucks.

    ***************

    BTW, this is covered here.

    http://www.allaboutcircuits.com/vol_3/chpt_8/8.html
     
    Last edited: Sep 20, 2010
  8. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    I think I know what you mean. Something like what's attached. I will have to think about adding a BJT to it. The signal is being attenuated unfortunately by this summer circuit, will have to look into that.

    Code ( (Unknown Language)):
    1.  
    2. $ 1 5.0E-6 15.472767971186109 50 5.0 50
    3. j 272 192 304 192 0 -4.0
    4. R 304 176 304 144 0 0 40.0 15.0 0.0 0.0 0.5
    5. r 304 208 304 272 0 10000.0
    6. g 304 272 304 288 0
    7. r 304 208 384 208 0 1000000.0
    8. r 384 208 384 272 0 1000000.0
    9. R 384 272 384 304 0 0 40.0 -3.25 0.0 0.0 0.5
    10. O 384 208 432 208 1
    11. R 272 192 240 192 0 1 40.0 1.0 0.0 0.0 0.5
    12. o 8 64 0 291 2.5 9.765625E-5 0 -1
    13. o 7 64 0 290 1.25 9.765625E-5 1 -1
    14.  
    Using Windows on the college network here, use falstad.com/circuit to simulate the circuit.
     
  9. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    [​IMG]

    Figure 1 is the basic model. It assumes you will always have the inputs connected to something, and gives the best results. However, the two signals are coupled with 200Ω between them.

    Figure 2 is when you may not have guaranteed inputs, and gives the circuit a guaranteed input when the other two are open. This resistor does interact, and adversely affects the performance. I figure 150KΩ or more.

    Figure 3 is for when you don't have a ground, it makes a pseudo ground. Figure 300KΩ for both bias resistors, it has all the same problems as Figure 2.

    If the beta of the transistor is 150 then you have an input impedance on the transistor of around 150KΩ, and this does interact with the adder, which is the reason you have such small resistors for the adder inputs.
     
    tom66 likes this.
  10. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    Thanks Bill!

    I'm still thinking about how amplification would happen, and I'll post another thread when I'm ready.

    Turns out, in AS Electronics (studying now) only MOSFETs are discussed, and AFAIK BJTs are very useful for analog circuits and MOSFETs are really only used as switches, except in rare cases. But maybe we will learn some analog stuff.
     
  11. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Tom, MOSFET's are very popular in analog IC design which is why they are taught so much.
    Discrete signal MOSFET's aren't common which is where that impression comes from.
     
    Last edited: Sep 20, 2010
  12. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    Yeah, but in my course description it says "MOSFET as a switch."

    But maybe more advanced topics will be discussed in A2 Electronics.

    This is only college, so I don't expect it to be too advanced.

    We were testing 4069 inverters to see at which point the output transitions. Turns out, with the inverters available (CD4069 IIRC) the transition point is 2.35V, with a 5V supply.
     
  13. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    JFETs and BJTs are generally used for analog, MOSFETs almost never. Part of the reason is they are not very linear. Linearity is more than a convenience if you are going to use it as such. Their smaller cousins, CMOS, are also used almost exclusively for digital. Wishful thinking doesn't change basic physics.

    If you look up the op amps using FETs you will see JFETs.

    I have seen some designs using digital CMOS inverters for analog designs, but they have never really taken off because they simply don't work well in that mode.
     
  14. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    Microchip's op-amps use CMOS technology pretty much entirely I believe. That is, MOSFETs all around, not a BJT or JFET in sight.
     
  15. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Care to show an example? I looked on their parts list, they don't show schematics of their op amps.


    **************

    OK, found the application notes AN722.

    http://ww1.microchip.com/downloads/en/AppNotes/00722a.pdf

    Interesting reading, using CMOS comes with a price...

     
    Last edited: Sep 20, 2010
  16. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    http://www.microchip.com/wwwproducts/Devices.aspx?dDocName=en528727

    No schematic, but "The MCP6031 uses Microchip's advanced CMOS technology, which provides low bias current, high-speed operation, high open-loop gain and rail-to-rail input and output swing."

    This is a low end chip having only a 10kHz GBWP. But they also make ones up to about 60MHz GBWP.
     
  17. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    CMOS is made of MOSFETs by definition. A MOSFET does not mean power MOSFET.


    CMOS offers lower power dissipation with higher density due to a smaller size with a simpler manufacturing process. Their drawback is lower gain (and hence less possibility for linearity from lots of feedback) and more capacitance, making for smaller bandwidths.
    Switched filters, ADCs, DACs, sample & hold, mixed-signal FPGAs... they're very very often CMOS.

    Pick up any recent textbook on analog circuit design and they will without a doubt focus on CMOS design.

    Some National CMOS products I found in just a few minutes:
    http://www.national.com/pf/DA/DAC121S101.html#Overview
    http://www.national.com/pf/LM/LMP8358.html#Overview
    http://www.national.com/pf/DC/ADC16V130.html#Overview
    http://www.national.com/pf/LM/LMP7709.html#Overview
    http://www.national.com/mpf/LM/LMC6484.html#Overview
     
  18. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Funny thing though, very few op amps have found use at this and similar sites. It may change, but not yet.
     
  19. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    So, what, are you implying that semiconductor manufacturers and electronics designers don't know what they're doing and design with CMOS analog devices just because they want to?

    One of the key advantages of CMOS is low power. Hobbyists usually don't care about microwatts.
    CMOS is a modern design process, hence it's used in modern devices. Modern devices are generally SMD. Most hobbyists don't like SMD.

    People in general like to work off of what they're used to. The majority of projects online and in magazines etc. are done by guys who started before CMOS was popular, why would they change devices when they don't need to?

    There is a million reasons why they're not popularly used on publicly visible projects.
     
  20. tom66

    Thread Starter Senior Member

    May 9, 2009
    2,613
    214
    Agreed. Just like that 10KHz op-amp. Pretty much useless for audio. But maybe it would be used in a multimeter, or a temperature probe, due to its low power consumption and low input current.
     
Loading...