Need a breakout board that can supply 25ma

[damae]

I'm using a CNC4PC breakout board now and it seems very nice, except that it cannot supply more than about 6ma for the TTL signal. The problem is that I'm interfacing with a servo amp (Emerson FX455) that requires 25ma on the pulse/direction signal lines!

Looking over the schematic for the Emerson controller, it uses the TTL input to power opto-isolators directly. So I am guessing it needs the 25ma to light the internal LED in the opto-isolator.

Anyway, are there boards readily available that can supply 25ma?

If not, how would I modify this board to get the 25ma? I only need it on 2 outputs, step/direction.

All help will be greatly appreciated..

[gar]

050711-1852 EST USA

damae:

Take a look at the 2N700 with input protection and capable of being driven from TTL. Vishay/Silconix 2N700KL. This is a preferred one but not stock at Mouser.com.

Use mouser.com to access the data sheet. Key in 2N7000KL, then select datasheet.

Another option is a bipolar transistor type 2N4400. This would require maybe a 2.2 k resistor to the base from the input.

.

[damae]

Originally Posted by gar 050711-1852 EST USA Take a look at the 2N700 with input protection and capable of being driven from TTL. Vishay/Silconix 2N700KL. ...

Thanks for the tip. I downloaded the spec sheet and will try to find something equivalent at digikey or somewhere that stocks it.

I'm really rusty with electronics, so I guess it's back to school for me. I don't know how to calcuate the resistor value I'd need. Does this resistor, the 2.2k that you suggest, limit the current to 25ma? Or is it for some other purpose?

Thanks!

[damae]

Someone just suggested to me to use a TTL buffer IC. Any thoughts as to if that will work?

[gar]

050711-2040 EST USA

damae:

For the 2N4400 I sort of pulled 2.2 k out of the hat. For a TTL gate output you should be somewhat over 2 v out for hi without additional pullup, and you are maybe .7 on the 2N4400 base at saturation of the collector. These are a guess without checking. So you can approach 1 ma into the base. If we assume a minimum beta of 100, then you can easilly drive a 25 ma load. You could go to 1 k for the base drive with no problem.

Note you will have a phase inversion from your current output to the 2N4400 unless you use a pullup resistor to the load and shunt the load with the 2N4400 output.

You could use a Fairchild 2N7000 and these are readily available, but I do not believe these have the input protection and reversed biased output diode that the Siliconix 2N7000KJ has. Same phase inversion problem. This might also be solved in your software.

.

[Al_The_Man]

Originally Posted by damae Someone just suggested to me to use a TTL buffer IC. Any thoughts as to if that will work?

If you have to buffer more than a couple of TTL outputs, a compact way is by the ULN2003 high voltage, high current darlington inverter, there is 7 in the one IC.
BTW the National DS2003 may be more readily available.
Al.

[smarbaga]

with driver chips like the uln2003, you can parallel the drivers to increase the current even more.
each buffer can supply half an amp , so all the drivers paralleled
0.5 amps x 7 , not bad !
your controller will only draw as much current as it needs and there maybe some current loose due to your wire as well,

[damae]

Originally Posted by gar 050711-2040 EST USA For a TTL gate output you should be somewhat over 2 v out for hi without additional pullup, and you are maybe .7 on the 2N4400 base at saturation of the collector. These are a guess without checking. .

I measured 5.1v out of the breakout board when the signal is hi. Are you saying I need to drop that to around 0.7v at the mosfet's base? Sorry if my questions seem uneducated (and I know only enough to know that they must sound dumb) but I really am a neophyte when it comes to electronics. It's been a long time since I learned this stuff in school. But I am determined to learn enough to be dangerous again. =)

Originally Posted by gar So you can approach 1 ma into the base. If we assume a minimum beta of 100, then you can easilly drive a 25 ma load. You could go to 1 k for the base drive with no problem. .

So the beta is the factor that determines how many times more current output the collector can handle compared to the input at the base?

Originally Posted by gar Note you will have a phase inversion from your current output to the 2N4400 unless you use a pullup resistor to the load and shunt the load with the 2N4400 output. .

Does this mean that when the TTL signal is hi (at the base) then no current will flow through the 2N4400?

Originally Posted by gar You could use a Fairchild 2N7000 and these are readily available, but I do not believe these have the input protection and reversed biased output diode that the Siliconix 2N7000KJ has. Same phase inversion problem. This might also be solved in your software. .

I am using TurboCNC right now, just to generate a pulse to turn the motors. (that's my first objective). There is an option to set active = hi or low for both step and direction outputs. I hope this would solve the inversion problem.

[damae]

Originally Posted by smarbaga with driver chips like the uln2003, you can parallel the drivers to increase the current even more. each buffer can supply half an amp , so all the drivers paralleled 0.5 amps x 7 , not bad ! your controller will only draw as much current as it needs and there maybe some current loose due to your wire as well,

Wow, 500ma! Just one chip should be enough to supply the current I need...

[damae]

Originally Posted by Al_The_Man If you have to buffer more than a couple of TTL outputs, a compact way is by the ULN2003 high voltage, high current darlington inverter, there is 7 in the one IC... .

The ULN2003 also looks good, more than able to handle the 25ma current I need. So does the 2N4400. However, it looks like the 2N700 is the only one that will have some kind of circuit protection built-in? How neccessary is this?

[gar]

050712-0643 EST USA

damae:

At this point you are going to get confused.

The 2N4400 is a bipolar transistor, in other words an original type transistor. This is a current controlled device in contrast to a voltage controlled device which the 2N7000 FET is.

With the 2N4400 the collector current is controlled by the current to the base, and the change in current to the base is small compared the change at the collector. This ratio is the current amplification and is called beta. A normal order of magnitude for beta is 100. With a beta of 100 a change of 1 ma at the base will produce a change at the collector of 100 ma. It is not this simple, but this is the concept.

If I have zero current to the base then I have zero collector current, and therefore no load current. This occurs if the lead to the base is opened, or if the base is connected to the emitter. If a resistor is connected between the base and emitter, then this is part way between an open and a short to the base. In all these cases the base current is zero except internally there is some leakage current to the base, and in your case we assume this to be zero.

If I apply 1 ma to the base then with a beta of 100 I will get anywhere from 0 to 100 ma of collector current depending upon the collector resistance and source voltage. Once I make the collector load resistance low enough that 100 ma flows, then the collector circuit looks like a constant current source of 100 ma. See bipolar transistor data sheets.

By putting a resistor between a source voltage and the transistor base I can control the base current by the applied source voltage. The base to emitter junction has characteristics very similar to a solid state diode. In the forward direction this has a voltage limit in the 1 volt range without destroying the device. In otherwords you can not apply 5 v directly between the base and emitter.

There is more that you need to know in this area, but enough for now.

In your application you would be overdriving the base with more current than is need to just switch the load to 25 ma. In this case you are saturating the transistor, and the load current is primarily controlled by the load resistance.

I made some actual measurements this morning at room temperature on a sample of one device each. In otherwords this experiment in no way indicates a worst case condition.

The source is a 74LS00. Its high output is 3.5 volts with no load. With a 2.49 k load into a 2N4400 base the 3.5 dropped a very small amount. Basically it is still 3.5 v. With the 2N4400 collector open the base voltage was 0.68 v.

With a constant current source of 23 ma to the collector the collector to emitter voltage was 0.1 v and the base voltage changed to 0.75 v.

Next with 90 ma collector current the collector voltage was 0.7 v and the base was 0.77 v. We are not getting good saturation at 90 ma.

If we change the base resistor to 1 k, then the 74LS00 output drops to 3.4 v. At 25 ma collector current its voltage is 0.05 v the base is 0.75 v, and at 87 ma the collector voltage is 0.15 v and the base is 0.8 v. Much better saturation for the 90 ma.

The 2N7000 is a different story.

The other suggestions are good and produce the desired result with fewer loose components. The above comments were to provide you with a little background.

.

[Al_The_Man]

The ULN2003 also has diode and output protection if you look at the internal schematic of the IC. By the way both of the options of ULN2003 and 2N7000 are classified as TTL buffers, except that the 2N7000 is a descrete device.
Both invert and are open collector output, which means a high on the input will cause the output to be low or conducting.
The ULN device actually has four different versions if you look at the spec sheet.
Al.