Interfacing cmos to ttl

Discussion in 'The Projects Forum' started by simeonz11, Sep 8, 2009.

  1. simeonz11

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    Apr 29, 2008
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  2. simeonz11

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    Apr 29, 2008
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    Bump , this is urgent , I am waiting for a solution so I can buy some better transistors , or I may try to scavange some lol :)
     
    Last edited: Sep 8, 2009
  3. SgtWookie

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    Change the 2.2k resistor on the base to 10k.
    Change the 10k resistor on the collector to 1k.
    Ensure that you have identified the transistor's leads correctly.
     
  4. THE_RB

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    Feb 11, 2008
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    Just use a resistive voltage divider. You will save parts and get rid of the transistor saturation turn off delay and the annoying inversion.

    Just remember that TTL input will draw a couple of mA so you need to use a smaller resistor as the top resistor than the 12v:5v ratio would suggest.
     
  5. simeonz11

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  6. SgtWookie

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    I'm afraid that a resistive divider approach wouldn't work very well in this application. 4000 series CMOS has a pretty low current source ability to begin with, unless you don't mind stressing the outputs quite a bit.

    In the schematic he linked to, the base resistor being 2.2k was too low of a value for being driven by 4000-series CMOS; it would result in about 5.45mA current. Increasing it to 10k decreases the base current to 1.2mA. This will also keep the transistor from becoming deeply saturated.

    The 10k collector resistor was much too high of a value to drive standard TTL inputs. Even 1k is a bit high, but should work if the transistor is connected properly. 470 Ohms would be the best value to use.
     
  7. SgtWookie

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    OK.
    Try increasing the base resistor to reduce saturation depth. 2N4401's can have an hFE up to 300 or so. Try values from 20k to 50k or so. Don't use a pot.

    Keep lead lengths short. While 1.2MHz is pretty slow, square waves are the sum of all of the odd harmonics, so digital signals need quite a bit of bandwidth. Long leads = inductance = whacky waveforms.
     
  8. SgtWookie

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    Threw some quick simulations together. 2N4401 is definitely too slow.

    The best results I got with existing library components on short notice was using a 2N2369 with a 150k resistor on the base; the output was more like a triangle wave than a square wave.

    Do you have a spare CMOS gate lying around? If so, go to schematic #3. [eta] However, use a 10k resistor between the two CMOS IC's so they don't get stressed.
     
    Last edited: Sep 8, 2009
  9. simeonz11

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    OK

    I am gonna spend the gas to go to that place thats 45 minutes away , its either that or shipping and getting it in 3 days .
     
  10. Wendy

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    There is always a dirty method, use signal 12 diodes in series to drop around 7V from the CMOS high.

    I didn't see a schematic of the transistor circuit, I suspect it was there and went away. I'd be looking for a higher frequency transistor somewhere, then use a common collector with a voltage divider on the output.
     
  11. Damo666

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    Aug 25, 2009
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    Use a 2n2219 high speed switching transitor or similar.
     
  12. Wendy

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    A 2N2219 is virtually identical to a 2N2222 except for power dissapation. The case is larger, so it tends to dissipate more heat, which is the biggest difference. All in all, they are both similar to a 2N4401 in speed and gain, but the 2N4401 has the lowest wattage dissipation.

    2N2222 Datasheet

    2N2219 Datasheet

    The PN2222 is also the same transistor, except for it has a plastic case.
     
    Last edited: Sep 9, 2009
  13. Ron H

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    A brute force, high speed saturating BJT generally requires much lower collector resistance than 10k. Below is a simulation with lower Rc. The two base resistors are chosen to cause switching to occur at ≈6V. Note that the duty cycle would be easier to maintain if the rise and fall times were insignificant relative to the period, which generally isn't the case with 4000 series logic running at 1.5MHz (50ns typical transition times @ Vcc=10V).

    You might have to juggle the base resistor values a little to maintain duty cycle, mainly due to CMOS output drive characteristics, which I ignored.

    2N2369 is an excellent high speed saturating switch, due in part to the fact that it has low storage time.
     
    Last edited: Sep 9, 2009
  14. SgtWookie

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    Interesting, Ron_H ... but unfortunately, the 4000 series CMOS' output impedance is a good deal higher than the signal generators', and you coupled that zero impedance through via a 1F cap, giving a mighty high base current.

    But I swiped your idea and tinkered with it a bit. Even works with a 2N4401 now, and keeps within the load limits of 4000 series CMOS. R2/R3 will probably need some tweaking to work with real-world components, but it's not too bad looking at the moment.
     
  15. Ron H

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    The 1f cap is 1 femptofarad. I was playing with the cap value, and set it to 1f to make it ≈0 without removing it. I have occasionally had sim problems when making C=0. I should have removed it before posting, but I forgot.

    I believe the CD4000 series has plenty of drive capability for this circuit. See the attached pullup curves of TI's CD4001.

    The results you posted don't preserve the duty cycle, which was an implied requirement of the OP.
     
  16. THE_RB

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    The CD4047 should drive a few mA output with no problems.

    You only need to source enough output current to pull up the TTL input (1.5mA ??) and maybe another 0.5mA extra source current for the lower resistor current.

    I don't think running the 4047 at 2mA source and <1mA sink would cause any problems.
     
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