CMOS verses TTL
- Thread starter Marcwolf
- Start date
One of the main differences between TTL and CMOS from a user perspective are the signaling levels.
Common TTL circuits operate with a 5V power supply. A TTL signal is defined as "low" between 0V and 0.8V, and "high" when between 2.0V and 5V.
In contrast, CMOS usually defines low as <Vcc/2 and high as >Vcc/2 (Vcc being the supply voltage, commonly 5V but for newer devices - LVCMOS - also 3.3V, 2.5V or 1.8V).
In reality, low for CMOS outputs is close to 0V and high is close to Vcc, so CMOS devices usually have no problem driving a TTL input. The opposite is not necessarily the case (a 2V TTL high is not enough to exceed the 2.5V CMOS threshold), but very often works just as well.
There is also a class of "TTL-compatible" CMOS devices out there that is guaranteed to work with TTL.
Wolfgang
Common TTL circuits operate with a 5V power supply. A TTL signal is defined as "low" between 0V and 0.8V, and "high" when between 2.0V and 5V.
In contrast, CMOS usually defines low as <Vcc/2 and high as >Vcc/2 (Vcc being the supply voltage, commonly 5V but for newer devices - LVCMOS - also 3.3V, 2.5V or 1.8V).
In reality, low for CMOS outputs is close to 0V and high is close to Vcc, so CMOS devices usually have no problem driving a TTL input. The opposite is not necessarily the case (a 2V TTL high is not enough to exceed the 2.5V CMOS threshold), but very often works just as well.
There is also a class of "TTL-compatible" CMOS devices out there that is guaranteed to work with TTL.
Wolfgang
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Quite true - the 74HC logic family is speed, function, and pin-out compatible with the once common 74LS types, & offers attractive advantages of wide supply voltage (2V-6V) & lower power consumption. They work a real treat when driven with 3 x AA ( 4½V) PICAXEs.
Because not every manufacturer supports every device, several logic families continue to be used. With the industry move to lower & varying supply voltages (espec. 3.3V), "HC" (high speed CMOS) has become a versatile champ compared with picky "LS"(Low power Schottky) logic. For logic donkey work "HC" (in the style 74HCxx) is HIGHLY recommended. See a family overview => http://en.wikipedia.org/wiki/Logic_family
Because not every manufacturer supports every device, several logic families continue to be used. With the industry move to lower & varying supply voltages (espec. 3.3V), "HC" (high speed CMOS) has become a versatile champ compared with picky "LS"(Low power Schottky) logic. For logic donkey work "HC" (in the style 74HCxx) is HIGHLY recommended. See a family overview => http://en.wikipedia.org/wiki/Logic_family
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Many many thanks for the info.
You have also solved another problem. I am making a circuit that will be run from batteries and it has worried me re what happens when the VCC drops below 5V. Using 74HC or pure CMOS this will solve that problem also.
You folks might answer another Q along the line.
The battery pack I am using for the project will supply approx 7.5V. I was thinking of using a 7805 to regulae the voltage but I am unsure of what happens when the input voltage gets to around 5V. Does it just passs through what it can or does it shut the votage off completly
Second option is a Zener diode which could handle the 5V and below since if it is below the zener cascade voltage the zenor will not act.
Many thanks
Dave
You have also solved another problem. I am making a circuit that will be run from batteries and it has worried me re what happens when the VCC drops below 5V. Using 74HC or pure CMOS this will solve that problem also.
You folks might answer another Q along the line.
The battery pack I am using for the project will supply approx 7.5V. I was thinking of using a 7805 to regulae the voltage but I am unsure of what happens when the input voltage gets to around 5V. Does it just passs through what it can or does it shut the votage off completly
Second option is a Zener diode which could handle the 5V and below since if it is below the zener cascade voltage the zenor will not act.
Many thanks
Dave
hippy
Ex-Staff (retired)
I've never known a 7805 to actually turn off its output when its input is too low. There may be some regulator types which do, and some have an "enable" which can be controlled by additional circuitry.
How a regulator behaves as voltage drops depends on the regulator. Some don't work well without a few volts headroom above 5V on input, other LDO (low-dropout) require much less. It also likely depends upon what current you are drawing through the regulator.
I've run a PICAXE from a 7805 driven from a Lab PSU where I've reduced the voltage and found that the output is usually Vin-Vheadroom approx once the input gets below Vout+Vheadroom. I didn't notice any problems with PICAXE operation sinking just a couple of mA but it's is almost certainly the case that the regulator is working out of specification so its characteristics will be largely undefined. I didn't do exhaustive testing and a recent post on "brownouts" suggests that running with less than the headroom required for a 7805 can cause problems, at least in some circumstances.
hippy
Ex-Staff (retired)
An alternative solution is to use a "less than 5V" regulator and use components which can run at that voltage. If 3V3 is too low, that can probably be jacked-up to 3V9 by putting a diode between regulator Common and 0V, pointy-end to 0V. Multiple diodes can be used. Different diode types affect the voltage adjusted. I suspect zeners can be used as well; Google may be a friend there.
This trick also works well in my experience with 7805's to create a 5V6 or 6V2 supply. Very handy if something wants 5V after being powered through a diode ( such as in battery backup situations ) where adding a diode to the 7805 Common compensates for the subsequent diode drop.
This trick also works well in my experience with 7805's to create a 5V6 or 6V2 supply. Very handy if something wants 5V after being powered through a diode ( such as in battery backup situations ) where adding a diode to the 7805 Common compensates for the subsequent diode drop.
Marc, this really is the case of reading Data Sheets, sorry to mention it.
Different regulators behave differently. For micropower regulators the usual thing that happens as the supply reaches the reg voltage is that the quiescent power consumption goes up which kills the battery off quicker. Then you will get a non-linear voltage drop acroos regulator.
What happens with the dinosaur 7805 I don't know but it does require a relatively large bit of voltage headroom. It obviously depends on the circuit requirements but for low current <50mA battery powered stuff I would always use a low-dropout regulator eg. LP2950 or LM2936. Dropout also varied with current load. Read up.
Bottom line; your Vsupply shouldn't drop below Vreg+Vdropout or else funny things will happen. Sometimes a bit of a droop doesn't matter but if you're doing things like ADCing then it does.
This is where Data Sheets, reading and looking at graphs is very handy.
Different regulators behave differently. For micropower regulators the usual thing that happens as the supply reaches the reg voltage is that the quiescent power consumption goes up which kills the battery off quicker. Then you will get a non-linear voltage drop acroos regulator.
What happens with the dinosaur 7805 I don't know but it does require a relatively large bit of voltage headroom. It obviously depends on the circuit requirements but for low current <50mA battery powered stuff I would always use a low-dropout regulator eg. LP2950 or LM2936. Dropout also varied with current load. Read up.
Bottom line; your Vsupply shouldn't drop below Vreg+Vdropout or else funny things will happen. Sometimes a bit of a droop doesn't matter but if you're doing things like ADCing then it does.
This is where Data Sheets, reading and looking at graphs is very handy.
Both PICAXEs & "HC" run perfectly from ~3-5½V. Why hence are you using a 7.5V battery pack? In an age when electronic devices are increasingly being designed for "smell of an oily rag" supplies, this seems unwarranted.
Suggest you tell us more about the circuit. Mmm- let me guess - a grunty motor is being driven?
Suggest you tell us more about the circuit. Mmm- let me guess - a grunty motor is being driven?
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Re: "I have found a nice but hard to get TLL 74LS595", Microzed sell the HC595 for $5 for 5 = $1 each. http://www.microzed.com.au/chips.html
Historically there were TTL chips, then I think the F series, then LS and then HC and each generation used less power. HC also uses less volts as pointed out by manuka. I think most of the 4000 cmos series have now been absorbed into the HC series, so there are now chips like the 74HC4016 and this series combines all the benefits of the 4000 series and the HC series. So you can even choose a 74HC4094. It may come down to price in the end.
What are you driving?
Historically there were TTL chips, then I think the F series, then LS and then HC and each generation used less power. HC also uses less volts as pointed out by manuka. I think most of the 4000 cmos series have now been absorbed into the HC series, so there are now chips like the 74HC4016 and this series combines all the benefits of the 4000 series and the HC series. So you can even choose a 74HC4094. It may come down to price in the end.
What are you driving?
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@Hippy,
"If 3V3 is too low, that can probably be jacked-up to 3V9 by putting a diode between regulator Common and 0V, pointy-end to 0V"
Have been thinking of how to set up a SuperCap/GoldCap to keep the PICAXE running long enough to shut things down gracefully and save temp data to eeprom when the main supply fails.
The issue is that GoldCaps below 10F value will only supply uA where the 10F and above will supply mA (different construction). The 10F mA types are much more expensive and a single cell has a working voltage 2.3V. So even if 2 were stacked the voltage rating would still be exceeded at 5V supply. A stack of 3, apart from having way too much capacity and being more expensive, also requires balancing to ensure 1 does not see too much voltage. The 5V supply is required as some devices need 4.5V and more, and will be supplied with less than 5V when fed from the GoldCap.
Having read your post I was wondering about using some similar technique to 'lift' the GoldCap 2.7V above ground.
The idea is to charge/discharge a single GoldCap via parallel reversed Schottky diodes. The diode's current/resistance reduces as the GoldCap approaches supply V and the GoldCap charges to the full supply voltage over time. When the supply disappears, the GoldCap will supply the PICAXE through the other Schottky but as the load is sufficient, the output will be around 0.3V less than normal and continue to drop as the charge is bled from the GoldCap.
CalibADC10 would be used in the main program loop to detect this change and shut down outputs, and save variables to eeprom. The program would then stay in a loop checking CalibADC10 until either power is restored, or the supply from the GoldCap runs out.
The problem is how to raise the "0V" of the GoldCap to 2.7V as the low leakage current through most semiconductors will allow the 0V side of the Gold cap to slowly approach PICAXE 0V over time and the GoldCap would then be subjected to excess voltage..
The other option is to use a boost circuit to raise the 2.3V to near 5V but this is not trivial either.
Any thoughts/suggestions appreciated.
"If 3V3 is too low, that can probably be jacked-up to 3V9 by putting a diode between regulator Common and 0V, pointy-end to 0V"
Have been thinking of how to set up a SuperCap/GoldCap to keep the PICAXE running long enough to shut things down gracefully and save temp data to eeprom when the main supply fails.
The issue is that GoldCaps below 10F value will only supply uA where the 10F and above will supply mA (different construction). The 10F mA types are much more expensive and a single cell has a working voltage 2.3V. So even if 2 were stacked the voltage rating would still be exceeded at 5V supply. A stack of 3, apart from having way too much capacity and being more expensive, also requires balancing to ensure 1 does not see too much voltage. The 5V supply is required as some devices need 4.5V and more, and will be supplied with less than 5V when fed from the GoldCap.
Having read your post I was wondering about using some similar technique to 'lift' the GoldCap 2.7V above ground.
The idea is to charge/discharge a single GoldCap via parallel reversed Schottky diodes. The diode's current/resistance reduces as the GoldCap approaches supply V and the GoldCap charges to the full supply voltage over time. When the supply disappears, the GoldCap will supply the PICAXE through the other Schottky but as the load is sufficient, the output will be around 0.3V less than normal and continue to drop as the charge is bled from the GoldCap.
CalibADC10 would be used in the main program loop to detect this change and shut down outputs, and save variables to eeprom. The program would then stay in a loop checking CalibADC10 until either power is restored, or the supply from the GoldCap runs out.
The problem is how to raise the "0V" of the GoldCap to 2.7V as the low leakage current through most semiconductors will allow the 0V side of the Gold cap to slowly approach PICAXE 0V over time and the GoldCap would then be subjected to excess voltage..
The other option is to use a boost circuit to raise the 2.3V to near 5V but this is not trivial either.
Any thoughts/suggestions appreciated.
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74HC595 are available in singles for 50 cents from futurelec. I wouldn't bother with the 4000 series part. My writeup on using the 74HC595 for output expansion is at:
http://profmason.com/?p=151
If you look carefully, there is a 7805 under the big heatsink(Silly). I generally run the 7805 from a 9V. By the time the 9V has dropped to 7, the battery is pretty much dead. Standard 9V are about 600mAh which is hours and hours and hours at a typical logic draw.
Also 9V + battery snap + 7805 is CHEAP!
http://profmason.com/?p=151
If you look carefully, there is a 7805 under the big heatsink(Silly). I generally run the 7805 from a 9V. By the time the 9V has dropped to 7, the battery is pretty much dead. Standard 9V are about 600mAh which is hours and hours and hours at a typical logic draw.
Also 9V + battery snap + 7805 is CHEAP!
What am I driving.
Hi Folks
Long term project is animatronics.
The reason that I am running 7.5V is so I can drive a bank of servo's. the PicAXE also provides some housekeeping as well as controling servos.
I was fortunate to get hold of some 1.5V 11000mAh (Yes that IS 11amp) D cells that I have built into a battery pack. I have also taken a 5V run off of the batteries for the PicAXE and other electonics.
The animatronis will be a combination of central and independant control. Independant being a PicAXE controlling a couple of servos and using a serial highway (Many thanks Hippy for that)
The Central will be run from a SD-21 which will sit on a I2C bus together with a 18X picAxe.
I will also be using a HM2007 with a parr to serial convertor system (also using a picAxe) so to interpret commants and interrupt the 18X to tell it what sequence to run.
The whole thing is set up so to be fully independant of outside control so that the wearer does not need to have controllers/helpers to perform.
There are some aspects of the cooling like the fans etc that will also be under picAxe control and I am looking at a graceful functional shutdown of capabilities depending on voltage levels.
i.e. if I (the wearer ) am performing and I know that I will need another 3 hours out of the batteries before I have a change to get out of the suit I can let the PicAVE know that and it will cut down on some of the more power hungery seqences whichout affecting some of the overall performances.
in essence - if you have watched Underworld or Underworld Evolutions and seen what they did with the werewolve's heands to get the emotions etc - I am taking that to a more full bodied aspect but including tail wags (hence the independant control)
This is all a hobby for me but a few of my experiments and idea;s got out to the general costume making community for them to pick up and run with.
Take Care
Marc
Hi Folks
Long term project is animatronics.
The reason that I am running 7.5V is so I can drive a bank of servo's. the PicAXE also provides some housekeeping as well as controling servos.
I was fortunate to get hold of some 1.5V 11000mAh (Yes that IS 11amp) D cells that I have built into a battery pack. I have also taken a 5V run off of the batteries for the PicAXE and other electonics.
The animatronis will be a combination of central and independant control. Independant being a PicAXE controlling a couple of servos and using a serial highway (Many thanks Hippy for that)
The Central will be run from a SD-21 which will sit on a I2C bus together with a 18X picAxe.
I will also be using a HM2007 with a parr to serial convertor system (also using a picAxe) so to interpret commants and interrupt the 18X to tell it what sequence to run.
The whole thing is set up so to be fully independant of outside control so that the wearer does not need to have controllers/helpers to perform.
There are some aspects of the cooling like the fans etc that will also be under picAxe control and I am looking at a graceful functional shutdown of capabilities depending on voltage levels.
i.e. if I (the wearer ) am performing and I know that I will need another 3 hours out of the batteries before I have a change to get out of the suit I can let the PicAVE know that and it will cut down on some of the more power hungery seqences whichout affecting some of the overall performances.
in essence - if you have watched Underworld or Underworld Evolutions and seen what they did with the werewolve's heands to get the emotions etc - I am taking that to a more full bodied aspect but including tail wags (hence the independant control)
This is all a hobby for me but a few of my experiments and idea;s got out to the general costume making community for them to pick up and run with.
Take Care
Marc
hippy
Ex-Staff (retired)
Conceptually it will, but in reality it won't. You'd be effectively adding the voltage dropping resistor on the 'wrong side of the circuit', but it would be the same as dropping the voltage were it added to the top.
The diode trick only works with regulators because they are active components balancing current ( or something like that ).
@Hippy,
Resistors were not proposed for this.
I have tested the above scheme with regular capacitors and it works fine EXCEPT for the the fact that there is the constant very low current flow through the active device (diode(s) in tests).
The cap charges and the voltage level stabilises in the cap and across the diode(s). But over time - i.e. 10s of minutes, the voltage levels keep changing by 1/100ths of a volt as that very small leakage current keeps on flowing - e.g. a diode does not completely block all current flow.
I guess what is required is the 'ideal' device which fully blocks current flow when at the critical voltage.
Resistors were not proposed for this.
I have tested the above scheme with regular capacitors and it works fine EXCEPT for the the fact that there is the constant very low current flow through the active device (diode(s) in tests).
The cap charges and the voltage level stabilises in the cap and across the diode(s). But over time - i.e. 10s of minutes, the voltage levels keep changing by 1/100ths of a volt as that very small leakage current keeps on flowing - e.g. a diode does not completely block all current flow.
I guess what is required is the 'ideal' device which fully blocks current flow when at the critical voltage.
hippy
Ex-Staff (retired)
Sorry, I meant "diode" not "resistor", that was a misleading mis-typing :-(
While it seems to works 'under power', does it when the caps alone are powering the circuit ? Another way to look at it is; when cap powered only, the complete circuit is cap (battery), diode, then chip. It doesn't matter where the diode is, it's still a voltage drop in the circuit. Maybe I'm not visualising what you have correctly ?
While it seems to works 'under power', does it when the caps alone are powering the circuit ? Another way to look at it is; when cap powered only, the complete circuit is cap (battery), diode, then chip. It doesn't matter where the diode is, it's still a voltage drop in the circuit. Maybe I'm not visualising what you have correctly ?