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H-Bridge

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by kal.a, Feb 6, 2016.

  1. spec

    spec Well-Known Member Most Helpful Member

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    What are barrels Keep? Is that a type of watch in the States?
    spec
     
  2. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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  3. spec

    spec Well-Known Member Most Helpful Member

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    Barrels just the same, but nothing in the context of watches- or am I missing one of your subtle jokes. :wideyed:
     
  4. dave

    Dave New Member

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  5. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    Related to post 85. 86. 87. 88 I remember the insignificant stuff. So, I guess, it's a KISS/Keep thing.
     
  6. spec

    spec Well-Known Member Most Helpful Member

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    Sorry Keep- got it. Yes, barrels. Then there are rods, pecks, bushels, carats, hands, acres, hectares, knots ... :hilarious:
     
  7. gophert

    gophert Active Member

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    And fortnights and barleycorns for those in the kingdom.
     
  8. MaxHeadRoom78

    MaxHeadRoom78 Active Member

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    Odd that UK was always CC for engine size, just as N.A. is metric now with Litre.
    I really prefer the metric thread/tap drill size, the auto industry has already gone over.
    No more BSF, BSW, BA, AF, US Machine screw etc.
    Max.
     
  9. spec

    spec Well-Known Member Most Helpful Member

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    Yeah, engine sizes in the UK have always been in cc, but this may because many of the early engines, both car and aeroplane, came from the continent, mainly France.
    Threads used to be a nightmare: BSW, BSF, BA, UNC.... Now, apart from specialist threads: gas etc, you only have metric coarse or fine- job done.
     
  10. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    All it means is more money hat i have to spend for tools.
    let's see now:

    We have the metric adjustible wrench.
    The metric Vise Grips
    The metric vise
    The metric flat blade
    The metric pipe wrench.
    The metric band wrench.
    The metric hack saw.

    and the list goes on.
     
  11. Mikebits

    Mikebits Well-Known Member

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    Just be glad your right handed. Have you priced out a left handed metric flat blade? :eek:
     
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  12. kal.a

    kal.a Member

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    Hi spec,

    I'm back from my trip and into my lab dissecting and reviewing this thread to understand even more of it.
    Why was it better to increase R10 and R11 on this to 10K?
    Q6 with base resistor of 5K6 will have 4.3mA and with a 10K there will be 2.4mA and the transistor will be fully saturated in either case but you mentioned speed at some point and that that was the reason for making the value of R24 on post #88 , for example, a 470Ω which will supply the base with a whopping 17.8mA.
    And R10 on the the same schematic is well within the Ic rating.

    I'm still struggling with base resistors and how to determine their values. I'm using them as switches and they will be either fully on or off. I keep in mind the following when considering their values:
    Saturation: the minimum current required to saturate the transistor.
    Speed: I try to keep the value as small as possible though I still haven't wrapped my head around "parasitic capacitance" and how that affects the discharging of the transistor. It seems to me that if I make the resistor's value small it will turn on the transistor fast but then the high current will slow its turning off.
    I've asked those same questions a few times and have read the thread multiple times but the concept is not getting into my thick skull. Not yet anyways.

    Thanks
    Kal
     
  13. KeepItSimpleStupid

    KeepItSimpleStupid Well-Known Member Most Helpful Member

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    The one little concept will help:
    The voltage across a capacitor can;t change instantaneously.

    That being said, the initial capacitance acts like a "short" for a very short amount of time. The way to overcome that is to use a "bigger hammer" or current source.
    The values of Hfe in saturation is generally different for an amplifier.

    So, you do need the values of Hfe(min @ saturation), Vbe and any temperature variations as well as Ic. Use worst case values. Don;t forget to look at the driving source too.

    Ib <= Ic / (Hfe min)

    Rb =< (Vdrive-Vbe)/Ib

    There is the concept of a "Speed up capacitor": http://www.engr.usask.ca/classes/EE/392/Lab-Guides/9-Electron Devices'12.pdf

    This effectively "shorts" the base resistor momentarily so it initially sees more current.
     
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  14. spec

    spec Well-Known Member Most Helpful Member

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    Nice to see you back on line kal,

    Hope you had a good trip.

    Do you mean R4 and R5?

    If you remember, the 470 Ohm base resistors are not desirable. They are only included so there is no direct path for current to flow from the 24V line and through the ORT into the base of the transistor and to the 0V line. In this instance the optocoupler receiving transistor (ORT) is the current limiting factor. The opto coupler current transfer ratio (CTR), worst case is 20% and as 10mA is being sunk through the opto transmitting LED (OTL), only 2 mA will be available from the ORT, worst case. By contrast, you may get an optocoupler which has a CTR of 178%, the data sheet does not preclude it, in which case yo would have a base current of 17.8 mA. If you had a CTR of anything higher than 178%, say a million percent, the current would still be limited to 17.8mA. Because the current would be limited to this value, no damage would be done though.

    There are many resistor values that would be suitable in this design. I have chosen certain values for certain functions and it is best to keep the same design for the same circuit function.

    Two components ultimately limit the speed of the H bridge circuit:
    (1) The MOSFETs
    (2) The opto couplers

    Paradoxically MOSFETs are searingly fast. It is just that MOSFETs have massive parasitic capacitances from the gate to source (cGS) but, much more importantly, from the gate to drain (cGD). The gate drain capacitance can be as high as 1nF, so if you are driving the gate with a voltage signal with a 1K Ohm source resistance the time constant would be 1K * 1nF = 1 micro second. This is an age in electronic terms. For this reason I would guess to fully turn the MOSFETs on would probably take 10 micro seconds (voltage drive source 1K) and 66 microseconds to turn them off (voltage drive source 5K6 + 1K). This description of speed effects is greatly simplified and for that reason is not accurate.

    In fact the drain gate capacitor appears as a virtual capacitance at the gate of value, MOSFET voltage gain* cGD . So, if you assume that the voltage gain of the MOSFETs is 10, a virtual capacitor of 10 * 1nf = 10nF would appear at the gate. This means that the MOSFETs would switch on and off ten times slower than stated above.

    One of my bridge circuits had high speed gate driver opto couplers. These have a very low output impedance, both turning the MOSFETs on and off. They also have massive source and sink current capability to charge and discharge any capacitances.

    Parasitics and propagation delay and other speed affecting aspects may sound complicated but in principle they are not. If you tried to describe to someone who had never seen a car (auto) before, especially a manual (stick-shift) type, they would be overwhelmed and think that they would never be able to drive. Yet millions of people drive a car with no problems whatsoever.

    You are on the right track thinking about speed, saturation etc, but don't let it cloud the issue. The fact that you are even considering these effects is a good sign. The other thing is that what you see is my design approach. Other designers would probably arrive at different values. Also what you have from me are paper designs. In a development environment, a paper design would be built and developed. This would inevitably show that certain adjustments could be made here and there to improve the performance. For example the four MOSFETSs may hoot like hell, in spite of the gate stoppers and decoupling capacitors. Similarly the opto receiving transistors may hoot, especially with that open circuit base just hanging in the air.

    There is generally a lot of confusion about transistor collector voltage saturation. In general saturation is the point where any increase in base current produces no further drop in VCE. And data sheets assume this point to be when IB is IC/10. But in circuit design, a transistor with a low VCE -some voltage less than VCB- is often described as being saturated. I tend to use Ib= IC/20 for light saturation with high hFE transistors.

    The more heavily a transistor is saturated the longer it takes to recover from saturation and, of course, the base current is wasted current. Incidentally, the 74 series TTL logic with S in the part number are non saturating- that is why they are faster. The lightning fast ECL is non saturating too, but that also uses another technique to increase speed.

    As I have implied before, your skull can't be that thick when you consider the wide range of techniques you have covered in the last few weeks.

    What you say is true about turning a transistor on fast by forcing a lot of current into its base. The other side of the story is that you can turn a transistor off fast by sucking out a load of current from the base. If you had an NBJT that was turned on hard the fastest that you could turn it off would be to instantly make the base negative with respect to the emitter (on most transistors you can't go much more negative than 5V without blowing the base emitter junction). Similarly to turn an NMOSFET off fast, it is common to make the gate negative with respect to the source. With MOSFETs you can make the gate much more negative without blowing anything, 20V being common. Normally 5V to 7V would do the job though.

    The other aspect of speed is what is termed propagation delay. Taking a BJT this may be the time that a signal on the base takes to have an effect on the collector. Some transistors (microwave transistors) are greyhounds and some are slugs (audio power BJTs). MOSFETs are greyhounds but, hell, are they difficult to drive.

    spec
     
    Last edited: Mar 23, 2016
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  15. kal.a

    kal.a Member

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    Nop. R10 and R11. If you look at post #33 you commented on my modification of your circuit by recommending changing R10 and R11 to 10K which I've actually done and that's what is on the board that I completed and is working quite well.

    And bingo. That was the missing link that I needed. I completely missed that part about the ORT being current limiter. I've read your explanation in an earlier post a few times and didn't get it till now.
    Thanks a lot spec. I'm quite relived to have finally understood it.:woot:
     
  16. kal.a

    kal.a Member

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    Hi spec,

    I did a few more experiments now that I understand how the opto works and to see things as they happen and what I found is the following:

    The opto's If (forward current) has proportional effect on the opto's Ic current as described by the CTR graph in the datasheet.
    R35 has zero effect on the Opto's Ic . I changed the value of the resistor and the current remained the same.
    R35 has an effect on the current going through R24 and consequently through the base of Q27 . The higher the value of R35 the higher the voltage and current applied to R24 so if I reduce R35 to let's say 2K2 I would get something like 6.7mA (1mA less than with a 5K6) at the base of Q27. No real reason for doing that just playing with the circuit.

    Two issues I'm still working on :
    1- The voltage regulator will generate heat and will need a heat sink for sure but I'm not sure if the heat is too much to dissipate by a heat sink. I'm expecting to run a motor on about 38 V and let's say I go for the maximum current of 1A then:
    38-9= 29 x 1A = 29Watt. Is that too much?

    Another issue is that the output of the PLC is 24V+ for pulse and the logic gates input have a maximum of 7V+. So I will have to modify the circuit with a transistor circuit similar to what you had in the original design on post #25. But then it would defeat the purpose of the gates and I may as well just use transistors. (not to mention that I still can't get the gates to work :D)

    I'm sure I said this before but this stuff is surprisingly fun :)


    Cheers
    Kal
     
    Last edited: Mar 23, 2016
  17. spec

    spec Well-Known Member Most Helpful Member

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    Hy kal,

    Am on social things for the next few hours- will hopefully answer your post around 7pm GMT.

    spec

    (For some reason I don't get notifications for new posts on threads I am watching)
     
  18. kal.a

    kal.a Member

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    Thanks spec, no rush at all as I'm working on it at my own pace......which is pretty slow.
    I greatly appreciate you helping me out at your convenience. :)
     
  19. spec

    spec Well-Known Member Most Helpful Member

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    Hy kal,

    Like you, I can't understand what that spec chap was on about.:banghead: Your circuit is correct.
     
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  20. spec

    spec Well-Known Member Most Helpful Member

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    You are starting to sound like a seasoned design engineer now.:)
     
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  21. spec

    spec Well-Known Member Most Helpful Member

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    It is good that you are considering the dissipation of the voltage regulator, as you should for every component in your circuit. But you are not right in thinking that the motor current passes through the voltage regulator though. So it makes no difference to the voltage regulator what current any motor that you drive with the H bridge. The voltage regulator even has the same dissipation if no motor is fitted.

    The dissipation in a voltage regulator, Ptot= (Vin- Vout) * Iout in Watts.
    Vin = 38V
    Vout= 9V
    Thus Vin- Vout = 29V
    Iout = (Voltage defining resistor chain/Vout) + (Imax opto receiver 1) + (Imax opto receiver 2)

    (Vout/Voltage defining resistor chain) = 9V/(180R + 1K1) = 7mA
    Imax opto receiver transistor 1= 17.8 mA
    Imax opto receiver transistor 2= 17.8 mA
    Thus the total Iout is (7mA + 17.8 mA + 17.8 mA) = 42.6mA

    Thus Ptot= 29V*42.6mA= 1.24W

    However it is highly unlikely that both optocouplers would have a CTR of 178% or more. As the worst case data sheet figure gives an CTR of 20% a good practical best case CTR would be 100%.

    This would mean that the voltage regulator dissipation would drop to 29V * (7mA+10mA+10mA)= 783mW.

    From the data sheet, the LM317 in a TO-220 case has a junction to ambient air resistance of 19 deg C/W.
    Thus the temperature difference between the junction and the ambient air will be (783mW * 19 deg C/W) = 14.88 deg C.

    Assume an ambient air temperature of 70 deg C inside your equipment under worst case conditions. Thus, the voltage regulator junction temperature will be, ambient air temperature + temperature difference between between junction and ambient= 70 deg C + 14.88 deg C= 84.88 deg C.

    As the maximum operating junction temperature for the LM317 from the data sheet is 125 deg C there is a safety margin of 40.12 deg C without the use of a heat sink.
     
    Last edited: Mar 23, 2016
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