Is there a chip with multiple N & P Fets for fast 5V to 12V conversion?

In summary, the conversation discusses the need for a chip with several buffers built in for converting hi-speed signals from 5V to 12V within .05us to 1us. The user has tried using N & P Fets and open collector with a pull up but found it too slow. They have also experimented with using the UDN2982 and ULN2803 with a pull-up, but are still experiencing issues with power consumption and ideal edge specifications. The conversation also addresses the issue of excess capacitance when using long wires and the suggestion is made to use open-collector buffers on the cmos board to avoid running the signals across long wires. The user also mentions the need for a good and tight 12V signal
  • #1
Jmayes
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Hi, new to these forums and glad I found them!

I am working on a project that requires conversion of fairly hi-speed signals (10us pulses) from 5V to 12V. The usual open collector with a pull up is much too slow getting from 0 to 12V. I really need the ramp to happen between .05us to 1us. I can achieve this with N & P Fets (push-pull) but really want a chip with several of these buffers built in as I have many signals to convert. At slower speeds using a UDN2982 worked great (with pull-ups) and if there were another chip with a similar pin-out that would be ideal for me.

http://www.allegromicro.com/en/Products/Part_Numbers/2981/2981.pdf

Any help will be greatly appreciated!
Jmayes
 
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  • #2
That IC you linked is a relatively high power driver and is much slower than what you require for simple buffer/level shift.

The 50 to 100ns edges you require and not the least bit difficult to achieve with a simple open collector buffer as long as you keep the pull up resistor value reasonably low (1k to 2.2k). Use the linked open collector buffer (7407) with a 1.2k to 1.5k pull up resistor and you'll easily achieve rise/fall times around 50ns into a single 12V cmos input.

http://www.google.com.au/url?sa=t&s...nbizAw&usg=AFQjCNEP0FbmogM-5bZax5dPlOpYmWh0HA
 
  • #3
In fact I tried that exact combination of 7407 & pull-up, I had to go down to 330 ohms and the edge was only just fair, I could use it but it was not ideal and power consumption was high (most of the time the signal is at Gnd level). I know the UDN part is a high current driver with no pull-down, I also worked with the ULN2803 (reverse part) and a pull up. Also with just NPN transistors but everything left me on the hairy edge. I really need something with a push-pull output and many drivers in one chip if one exists. Looking at perhaps trying a LM3900 or opamps but don't want the extra outboard parts.

Thankx for the help!
Jmayes
 
  • #4
Jmayes said:
In fact I tried that exact combination of 7407 & pull-up, I had to go down to 330 ohms and the edge was only just fair, I could use it but it was not ideal and power consumption was high (most of the time the signal is at Gnd level). I know the UDN part is a high current driver with no pull-down, I also worked with the ULN2803 (reverse part) and a pull up. Also with just NPN transistors but everything left me on the hairy edge. I really need something with a push-pull output and many drivers in one chip if one exists. Looking at perhaps trying a LM3900 or opamps but don't want the extra outboard parts.

Thankx for the help!
Jmayes

What type of input where you driving Jmayes. Are you trying to drive multiple inputs or do you have a very long wires or some other source of excess capacitance? You understand that this all comes down to RC right. The rise time is approx 2.2 RC. The capacitance of a single 74Cxx (or CD4XX) cmos input is only about 5 to 10 pf. Add another 5 to 10 pf of stray capacitance and you've got your RC time constant and rise times.

If the input capacitance is 15pf and the pullup 1.5k then the time constant is approx 22.5ns and the rise time approx 50ns (that's how I arrived at the suggested pullup).

If you needed 330 ohms just to get it down to 100ns then your input capacitance is out of control (>100pF). This is the thing you need to address. Are you using a breadboard?
 
  • #5
BTW. One other quick thought. You wouldn't be the first person to ruin their rise times by connecting a CRO probe! I've even seen noobs trying to measure this kind of thing with a piece of 50 ohm coax connected instead of even using proper probe (haha probably several hundred pF!)

If you're observing this on a scope make sure you have low capacitance probes (<15pF)! Even this will double your observed rise times (compared to what they really are when you're not looking at them). Otherwise only observe at the cmos output and just deduce the input rise time.
 
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  • #6
Yes, using a proper scope probe :) and Yes, I am driving a length of cable (about 3 feet), using standard ribbon with every other conductor grounded. So yes the capacitance is what is killing me so hence the need for a stronger active pull-up and pull-down. Sorry I did not think to add that info to my OP. (Yes , I have played with termination too)

Thankx again for your help,
Jmayes
 
  • #7
Jmayes said:
Yes, using a proper scope probe :) and Yes, I am driving a length of cable (about 3 feet), using standard ribbon with every other conductor grounded. So yes the capacitance is what is killing me so hence the need for a stronger active pull-up and pull-down. Sorry I did not think to add that info to my OP. (Yes , I have played with termination too)

Thankx again for your help,
Jmayes

Yeah that's about an extra 100pF right there, just what I was expecting. Can you run 0,+5V over the 3' wires and place your open-collector buffers (plus bulk and bypass capacitor) on the cmos board. There should be some way you can alter the layout to avoid having to run the open collector outputs across the 3' wires.
 
  • #8
Some of the circuits end up going around several boards after the entry point so I really just need a good tight 12V signal that can withstand probably 6 feet of wire. This is a retrofit to a 80's model device so most of the signals go to plain transistors and diodes. It is a MUX buss type of circuit that already puts 3 items on the buss and I am expanding it to go to four so the edge specs need to be better then original. I have seceded with external push-pull logic but really want to find a chip to make this easier over 60+ circuits. A drop into the 2983/2903 is the ideal device as I have already wired for those chips.

Thankx again,
Jmayes
 
  • #9
OK you could just buffer it (not open collector) after you do the level translation but before you go routing it all around the place. But I see what you're looking for now, you want the level shifting and the buffering (push pull) all in the one package to keep it simple. Maybe someone else will suggest a suitable part.
 
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  • #10
Which brings me to the OP, I need a solution that is in one chip, If I convert and buffer again it's two chips + glue.

Thankx again,
Jmayes
 
  • #11
How about just a simple level shifter?
 

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  • #14
Thank you for the suggestions! I remember finding the 4504 but did not have one in my parts kits to try and forgot about it, it looks very promising. I have over 60 circuits to buffer so the hi-speed MD1810 would be more then I could afford and requires outboard inverse drivers.

I really like the CD4504, I need to get some in and test them. If anyone knows of something equivalent that the inputs are on one side and the outputs on the other (dip part) that would even be better as I have already wired a prototype for that type of part.

Thankx again all!
Jmayes
 

1. What is the purpose of converting 5V TTL to 12V CMOS?

The purpose of this conversion is to allow devices that operate on different voltage levels to communicate with each other. TTL (Transistor-Transistor Logic) operates on 5V, while CMOS (Complementary Metal-Oxide-Semiconductor) operates on 12V. By converting the voltage levels, these devices can understand and process each other's signals.

2. How does the conversion process work?

The conversion process involves using a voltage level shifter or converter. This device takes in the 5V TTL signal and adjusts it to 12V CMOS levels. It typically uses ICs (integrated circuits) or transistors to perform this task.

3. What are the potential risks of not converting 5V TTL to 12V CMOS?

Without converting the voltage levels, devices may not be able to communicate properly. This can lead to errors, malfunctions, and even damage to the devices. It is important to ensure that the voltage levels are compatible to avoid any potential risks.

4. Are there any limitations to converting 5V TTL to 12V CMOS?

The conversion process may introduce delays in the signals, which can affect the overall performance of the devices. Additionally, the voltage shifter may have a limited capacity and may not be able to handle high-frequency signals. It is important to consider these limitations when selecting a voltage shifter for the conversion.

5. Can the conversion be reversed?

Yes, it is possible to convert 12V CMOS back to 5V TTL using a similar voltage shifter. This may be necessary if the devices need to communicate bidirectionally. However, it is important to ensure that the voltage levels are properly matched to avoid any potential risks.

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