PRT TTL Output Values

[pat_wardell]

Hi, I opreate and do engineering on a PRT for someone and I was thinking of controlling the On/Off of the spindle motor (3/4 hp single phase 3450 rpm, they just do foam) with one of the outputs. I was thinking of using an SSR (solid state relay) but they typically require 10ma to turn on.

I read that the outputs were TTL. Are the outputs standard TTL? I was wondering what the actual sink and source specs were for the outputs? Are the open collector with pull up resistors?

Thanks,
Pat

[Gary Campbell]

Pat...
I am not sure on the PRT, but on the PRS the outputs are sourced. Depending on the board version, you may also have other options.

IF Mike R. sees this he will give the best answer.

[ken_rychlik]

The manual of the prt had a schematic for a simple relay system with all the parts available at radio shack.

It uses a 12 volt relay driven by a 5 volt source. It is very simple and easy to make.

I built several and they are working fine on my machine.

Dig through your manual or you can download the old manuals on the shopbot web site also.

[ken_rychlik]

Here is a copy of the shopbot page with the relay on schematic.

Kenneth

[dana_swift]

Patrick- TTL stands for "Transistor-to-Transistor-Logic". Actually TTL is effectively extinct. What the spec is is "LS-TTL" which stands for "Low-power Shottky TTL". I expect that distinction didn't help you immediately. However that is a specific definition of the output levels and current capabilities. What is specifies is: The output will "pull down" as much as 8 milliamps at 0.35v. (logical zero) The output will "source" as much as -0.4 milliamps at 2.7 v. (logical one) Source (Digikey/Fairchild Semiconductor DM74LS00N-ND) The general characteristics of a logic family may be exceeded by any given device. What is important to notice is that the capability to "pull to logical zero" is much greater than the ability to "push to logical one". This asymmetry causes some odd design characteristics. It is preferred that a TTL circuit be in the "one" state most of the time, because it takes more power from radio interference to convert a logical one to a logical zero than the other way around. There are many ways of hooking digital signals to each other, here is a page listing many of them, just "FYI": http://en.wikipedia.org/wiki/Logic_family Solid state relays are available that accept LS-TTL logic levels directly. If you have not found one keep looking. Be sure your solid state relay is designed for an inductive load, some aren't. Hope that helps- D

[richards]

The problem is the TTL specification. As far as I can tell, on my PRT-Alpha, each output can source or sink 24mA. True TTL can sink 8-10mA, but it can only source about 10% of that value.

All of the SSRs that I've used (solid state relays) have worked with as little as 4mA. The switching speed is slower as the voltage drops, but all of them have been reliable at 4mA.

All that really happens inside a SSR, on the control side, is that you turn an internal LED on and off. The output side of the SSR detects the LED and uses that detection to turn on the outputs.

To test the circuit using less than $2 in parts, just hook up a standard RED LED wired in series with a 330 ohm 1/4 watt resistor. Connect one end of the resistor to +5VDC. Connect the other end of the resistor to the LED's anode. Connect the Cathode end of the LED to the controller's output. If the LED turns on and is fairly bright, then the circuit will drive a SSR. That test uses the controller output to sink current.

The second test is to connect one end of the resistor to the controller's output , the other end of the resistor to the anode on the LED and the cathode of the LED to the ground. If the LED turns on brightly, then the controller is able to source sufficient current to operate an SSR.

Note than in both tests, the Anode end of the LED is pointed towards the higher voltage and the Cathode end of the LED is pointed towards the lower voltage. The flat side of the LED is the Cathode. If the LED has no flat side, the shorter leg is the Cathode.

If the LED only turns on dimly, then you will need to amplify the current with a transistor. It would be easier to Google simple circuits than to explain the connections, but if you email me, I can draw up a simple schematic that would show how to use a transistor to amplify the current.

[pat_wardell]

Here is a copy of the shopbot page with the relay on schematic.

Kenneth


Kenneth.... Thanks.

Looking at the schematic I see they are using Mosfet which is a very low power input to power the relay. This tells me the outputs on the PRT are low current outputs. I'm glad to see they are using a diode accross the relay coils to control the inductive spikes.

Pat

[pat_wardell]

Patrick- TTL stands for "Transistor-to-Transistor-Logic". Actually TTL is effectively extinct. What the spec is is "LS-TTL" which stands for "Low-power Shottky TTL". I expect that distinction didn't help you immediately. However that is a specific definition of the output levels and current capabilities. What is specifies is: The output will "pull down" as much as 8 milliamps at 0.35v. (logical zero) The output will "source" as much as -0.4 milliamps at 2.7 v. (logical one) Source (Digikey/Fairchild Semiconductor DM74LS00N-ND) The general characteristics of a logic family may be exceeded by any given device. What is important to notice is that the capability to "pull to logical zero" is much greater than the ability to "push to logical one". This asymmetry causes some odd design characteristics. It is preferred that a TTL circuit be in the "one" state most of the time, because it takes more power from radio interference to convert a logical one to a logical zero than the other way around. There are many ways of hooking digital signals to each other, here is a page listing many of them, just "FYI": http://en.wikipedia.org/wiki/Logic_family Solid state relays are available that accept LS-TTL logic levels directly. If you have not found one keep looking. Be sure your solid state relay is designed for an inductive load, some aren't. Hope that helps- D


Dana thanks for al the work...

Yes TTL is out of date but I think the manual I read online for the PRT stated TTL. I was hoping it was open collector which would mean they had a resistor to the collector of the TTL ouput transistor allowing the designer to provide a higher output (source) current. Yes TTL can sink (pull to ground) more than they can source (pull high).

I think the "asymmetry" you were speaking of is considered hysteresis which is beneficial in a nosiy environment unlike CMOS which changes state typically at the same voltage (midpoint of 0V and the supply voltage).

I guess I should go to where the machine is and see if I can determine which chips they are using for the output drivers.

Thanks for the input on the SSR. My concern with the SSR is that they are opto-isolated meaning they use an LED to turn on the SSR. Typical mfg. specs are ~10ma to turn on the SSR and TTL is typically .4ma output. This is what prompted my question.

I have done some communication with major SSR manufacturers for another project. Here are their recommendations for a capacitor start single phase 3/4-1 HP 120V motor if anyone is interested. For an inductice load such as an electric motor they suggest a "Random" type SSR and not a "Zero Crossing" type. A 1 HP 120V motor runs at full load around 12-15amps but they suggested a 50 amp SSR due to the inrush current that may be 50-100amps when the motor starts.

Thanks again, Pat

[pat_wardell]

The problem is the TTL specification. As far as I can tell, on my PRT-Alpha, each output can source or sink 24mA. True TTL can sink 8-10mA, but it can only source about 10% of that value.

All of the SSRs that I've used (solid state relays) have worked with as little as 4mA. The switching speed is slower as the voltage drops, but all of them have been reliable at 4mA.

All that really happens inside a SSR, on the control side, is that you turn an internal LED on and off. The output side of the SSR detects the LED and uses that detection to turn on the outputs.

To test the circuit using less than $2 in parts, just hook up a standard RED LED wired in series with a 330 ohm 1/4 watt resistor. Connect one end of the resistor to +5VDC. Connect the other end of the resistor to the LED's anode. Connect the Cathode end of the LED to the controller's output. If the LED turns on and is fairly bright, then the circuit will drive a SSR. That test uses the controller output to sink current.

The second test is to connect one end of the resistor to the controller's output , the other end of the resistor to the anode on the LED and the cathode of the LED to the ground. If the LED turns on brightly, then the controller is able to source sufficient current to operate an SSR.

Note than in both tests, the Anode end of the LED is pointed towards the higher voltage and the Cathode end of the LED is pointed towards the lower voltage. The flat side of the LED is the Cathode. If the LED has no flat side, the shorter leg is the Cathode.

If the LED only turns on dimly, then you will need to amplify the current with a transistor. It would be easier to Google simple circuits than to explain the connections, but if you email me, I can draw up a simple schematic that would show how to use a transistor to amplify the current.


Mike thanks! I was at home asking this question and was hoping someone had the answer on the PRT specs off the top of their head. I have designed quite a few circuits in my time but was hoping an actual spec would allow me to shortcut digging into the machine or designing a circuit to fit my needs. I am working off the KISS methodology (Keep It Simple Stupid). I guess I will take my DMM to the site and do some current measurments. I don't want to over tax the TTL so I can maintain reliability.

My research indicates ~10ma to fully turn on the opto-isolation of an SSR which the TTL may not provide.

I will post my test results on what the outputs can source when I'm done.

Thanks..... Pat

[richards]

Pat,

You'll probably find that you don't need 10mA to operate an SSR. A few years ago when I visited Opto-22, a company that makes Solid State Relays and other opto-coupled devices, a designer told me that many people drive SSRs directly from the outputs of micro-controllers that could only sink 2mA. He cautioned me that turn-on time might take a few milliseconds, but that the SSR would work.

I tried some tests using 5VDC and a 2.2K resistor in series with the control side of an SSR. Everything worked fine.

[pat_wardell]

Thanks Mike, you just got my brain in gear. :confused:

I'm going to run a 500 ohm resistor from +5 to output 1 for example. This should give me 10ma + what the TTL device can source. The TTL should be able to sink 16ma so it should pull to a 0 state no problem. Then I'll run the my control wire from output 1 to the SSR. Should work.

I'll use a 50A Random switching SSR with a 3-32VDC input SSR.

Note: I am going to test what the PRT outputs can source before putting the resistor in place to make sure I don't overload the TTL device when it tries to pull to a 0 state. I will adjust the resistor size depending on what the device can source.

I'll post my results when I get it done. This would also be good for vacuum systems on the old PRT.

Pat

[richards]

Keep in mind that you don't need to add a resistor in series with an SSR. They have current limiting built in. Also, don't forget the dreaded 'diode drop'. An LED 'consumes' about 1 volt, sometimes more. When I build process control computers, I normally use 330 ohm or 390 ohm resistors when I need 10mA through an LED that is sourced with 5V. Because of the 'diode drop', the circuit really only has 3.5V to 4V, which means that a 330 to 390 ohm resistor is needed to allow 10mA to pass through the circuit.

[pat_wardell]

Keep in mind that you don't need to add a resistor in series with an SSR. They have current limiting built in. Also, don't forget the dreaded 'diode drop'. An LED 'consumes' about 1 volt, sometimes more. When I build process control computers, I normally use 330 ohm or 390 ohm resistors when I need 10mA through an LED that is sourced with 5V. Because of the 'diode drop', the circuit really only has 3.5V to 4V, which means that a 330 to 390 ohm resistor is needed to allow 10mA to pass through the circuit.

Mike, I'm not planning on a resistor in series with the SSR, they should have built in contant current control on the input for the wide 3-32vdc input.

Yes a 330 or 390 ohm resistor should work fine but no more that 330 because the TTL may not be able to pull it to a low state. 330 ohms off of the +5v supply would put you about 15ma through the resistor + what the TTL device will source. Keep in mind I am planning to use the resistor as a pull up to the PRT output.

I learned my digital stuff back when TTL and CMOS was the thing to use. Mid '70's. Have to dust off the cobwebs to remember ohms law:).

Thanks,
Pat

[ken_rychlik]

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[richards]

I guess I'm having trouble visualizing your circuit.

When I design an Active Low (Sinking) Input that uses TTL, (assuming that the device sending the input signal to the controller is external to the controller board), I would normally tie the Input signal on the controller board to +5 volts through a 4.7k resistor. That assures that the signal from the external device going to the input on the controller board will always be seen as inactive until the external device pulls the signal Low.

When I design an Active Low Output circuit for an external SSR, I connect the (+) Control Terminal on the SSR directly to +5VDC on the controller board and the (-) Control Terminal on the SSR to the controller board's TTL output signal. When the controller board's output signal goes Active (0vdc), the external SSR will turn on.

When I visualize your circuit, I think I'm seeing a 500 ohm resistor in series with the SSR control logic. That would cut down the current available to the SSR.

Ohm's law tells us that the maximum current available is computed by dividing the voltage by the resistance, so, even though the device could be capable of sinking 16mA, the resistors internal to the SSR and external to the SSR will determine how much current can actually flow in the circuit.

To keep things simple. Let's assume that the SSR is being driven from a 5VDC circuit and that it has an internal 500 ohm resistor. (We'll ignore diode drop in this example.) The internal 500 ohm resistor will limit the usable current to 10mA, i.e. 5 Volts / 500 ohms = 0.010 Amps. Adding the external 500 ohm resistor would restrict the usable current to 5mA, i.e., 5V / 1000 ohms = 0.005 Amps.

[pat_wardell]

I guess I'm having trouble visualizing your circuit.

When I design an Active Low (Sinking) Input that uses TTL, (assuming that the device sending the input signal to the controller is external to the controller board), I would normally tie the Input signal on the controller board to +5 volts through a 4.7k resistor. That assures that the signal from the external device going to the input on the controller board will always be seen as inactive until the external device pulls the signal Low.

When I design an Active Low Output circuit for an external SSR, I connect the (+) Control Terminal on the SSR directly to +5VDC on the controller board and the (-) Control Terminal on the SSR to the controller board's TTL output signal. When the controller board's output signal goes Active (0vdc), the external SSR will turn on.

When I visualize your circuit, I think I'm seeing a 500 ohm resistor in series with the SSR control logic. That would cut down the current available to the SSR.

Ohm's law tells us that the maximum current available is computed by dividing the voltage by the resistance, so, even though the device could be capable of sinking 16mA, the resistors internal to the SSR and external to the SSR will determine how much current can actually flow in the circuit.

To keep things simple. Let's assume that the SSR is being driven from a 5VDC circuit and that it has an internal 500 ohm resistor. (We'll ignore diode drop in this example.) The internal 500 ohm resistor will limit the usable current to 10mA, i.e. 5 Volts / 500 ohms = 0.010 Amps. Adding the external 500 ohm resistor would restrict the usable current to 5mA, i.e., 5V / 1000 ohms = 0.005 Amps.

Mike,

I have attached .jpg of what I think your circuit is and mine. Both will work but yours would be my prefered method. Yours uses less parts/power and will only sink what is required by the SSR when the TTL pulls to a low state.

The reason I don't want to do it this way is that the Shopbot PRT controller comes up with the outputs in a low state. With the 'pull low to turn on' method the motor would turn on when the controller is turned on. Mine may require some resistor changes but nothing below 330 ohms. The resistor needs to be large enough to turn on the SSR but small enough that the TTL can pull to a low state. The resistor is to supplement source current of the TTL device. I think it should work. SSR's should be in soon, then I can test it out.

The only change I would make to your circuit is add a small fuse to the +5v before is goes to the wire to the SSR in case there is a short to ground. Don't want to blow the supply in the controller.


Thanks,
Pat

[pat_wardell]

Finally got to do some testing on the outputs of the Shopbot PRT.

I turned a couple of outputs ON and checked the current on the outputs, they were at ~24ma.

With that said the output current at a high condition (logical 1) will easily supply enough current to turn on an SSR. This is good news. The resistor I had in the schematic in the previous post will not be needed. I will just run the + side of the SSR directly to the PRT output. The output of the PRT will not need a resistor because the SSR has an input voltage range of 3-32 volts and constant current control that will keep the current to no more than what is needed (~10ma).