Electronics > Beginners
74LS14 testing circuit pls
<< < (13/15) > >>
rstofer:

--- Quote from: Chriss on April 18, 2018, 02:00:44 pm ---What is wrong with this circuit?

I use the P832 Photo interrupter and the SN74LS14 to transform the signal from the P832 to a square signal.
When I interrupt the photo signal with a card board the interrupter actually goes low aroun 0.25V but the 7414 won't
invert the input.

This SN74LS14 is a good one now.  :D

Here is my circuit what I use:


I also tried the R3 to change from 1K up to 10K but no big difference...

What does I miss?

Thank you...

--- End quote ---

That circuit just isn't going to work.

Assume the transistor is in saturation, call it  0V drop between collector and emitter (it'll never really be that low) then CALCULATE the output voltage going to the LS14.  You'll only get about 5V / 11 or around .46V.  That 30k/3k voltage divider is killing you.

On the sensor, put a 10k resistor between the Vcc pin and 5V - connect this collector junction to the LS14

Remove the resistor in the emitter and ground it.

The datasheet uses 0.5 mA when they specify VCE and, for 5V, this requires 10k Ohms.

The 10k resistor will drop, at most, 10k * 20 uA = 0.2V when driving the LS14 input high so the gate will see about 4.8V and it only needs to see 1.9V to get above the threshold.  In other words, 10k is fine.

Use the circuit in Reply 59.  I would still go for the 10k resistor...
Chriss:
Thank you guys!
I really appreciate your help.

Actually, the P832 I salvaged from an old printer pcb.
The schematic above in my post #52 what is not working is taken from the printer pcb.
But the on the junction where I connected my 7414 is connected to a pin froma uC on the original pcb.

So, however, that is in my case not working with the 7414.

Please check this if I understand in the right way what is standing in the datasheet for the 7414:


IOH does it mean when the input is feed with low signal, the Imax on the output is the max current of -0.4mA what the 7414 can delivery.
IOL does it means when the input is feed with a high signal, the Imax on the output is the max current of 8 mA what the 7414 can delivery.

Does I understand it correct?
I ask just to clear the meaning because of my English language understanding...

I trying to understand the datasheet and wish to find the mayor parameters to set up the 7414...
rfengg:
IOH does it mean when the input is feed with low signal, the Imax on the output is the max current of -0.4mA what the 7414 can delivery.
IOL does it means when the input is feed with a high signal, the Imax on the output is the max current of 8 mA what the 7414 can delivery.

You are correct.
When the input is low , as its an inverter, the outout goes high.
At this time , the maximum current that the TTL gate can deliver  is 400uA to reliably hold the output at the high state.
The 400uA also determines the fanout of the logic family i.e., the number of other devices you can connect to the output.

When the input is high  , as its an inverter, the outout goes low.
At this time , the maximum current that the TTL gate can sink is 8mA to reliably hold the output at the low state.
Please also remember the IOH and IOL are dependent on the exact logic family (74XX vs 74SXX vs 74LSXX etc etc.....) and also for the military versions starting with 54, so you need to be diligent to look up these on the data sheet. For eg, on the TI SN7414, the recommended IOL is 16mA while for the SN74LS14 its 8mA and 4mA for the SN54LS14.
rstofer:
I => current
O => output
H => high

This is the maximum output current when the pin is high.  It is shown as (-) because the current is coming out of the pin. Source current, the pin is sourcing the current.

I => current
O => output
L => low

This is the maximum output current when the pin is low.  It is shown as (+) because the current is going into the pin. Sink current, the pin is sinking the current.

You can now see why we don't try to pull LEDs up!  400 uA just won't work.  Even 8 mA of sink current won't provide the 20 mA that a normal LED would like to see.  The LED will light up but not to full brightness.
rstofer:
OOPS!

The transistor in the opto device has to sink the collector resistor current PLUS the IIL of the 74LS14 and this total current should be about 0.5 mA.  The LS14 has a max IIL of 0.4 mA so that only leaves 0.1 mA for the resistor current.

5V / 0.1 mA = 50k Ohms, not 10k  -  52k should work.

The 52k resistor times 20 uA (IIH) will drop 1.04V so the LS14 will see 3.96V on the pin.  Far above the threshold voltage of max of 1.9V so everything should work fine.

Navigation
Message Index
Next page
Previous page
There was an error while thanking
Thanking...

Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod