Hi,

I came across a problem yesterday, and I wonder how one does to find the good answer.

Using a voltage divider configuration, I took a 9V battery to provide a logical 5V (Vcc) to a Schmidt trigger IC.

But to calculate the good value of R1 to pick for the voltage divider, I needed to know the equivalent resistance of the IC acting as R2 in the VD. (real part of the impedance, because I use a DC source).

Nevertheless, this IC is not a passive piece of hardware, so as long as it is not "polarised", not giving the IC its Vcc, a physical ohmic measurement will show a huge resistance value.

So here it is:

Question #1: How does one measure the equivalent resistance of a chip if you don't have the proper source voltage to "polarize" it, and therefore see its operating equivalent resistance?

Question #2: Would this information be provided on the datasheet...? If so, what is the usual name given to this spec..? I didn't investigate yet. Maybe I should.

Question #3: In simulation, should I polarise the IC properly, then measure its resistance and then proceed with my calculations to find R1 or again... use the datasheet?

Thanks,

zeta_no

p.s: I ended up making R2 with the IC in parrallel with a small resistance (around 100 Ohm). Therefore, I minimized the effect of the IC load, what ever its value, on the voltage divider. Nevertheless, this way I didn't get a sharp 5V.

p.s.2: Sorry for the length...

 

Have you looked at a data sheet for the device in question? For instance, something like a 74LS14, which is a common Schmitt trigger hex inverter.

Notice that the input currents are stated for high and low inputs at the maximum level of Vcc. Ohm's law lets you get the input resistance.

For linear devices, the input impedance is explicitly stated in the data sheet.

Try two 4.7K resistors. That way, your weenie 9 volt battery doesn't get loaded down.

 

Thanks beenthere for the quick answer,

Exactly, I use the 74ls14. I checked on a NS datasheet. I found "Input Current @ Max Input Voltage". Here they say 0.1 mA and max Vcc is 5.25V.

Should be it... right?

Using V=RI, we find R = 5.25/0.1x10^-3 = 52.5 kOhm.

With my 100 Ohm, I mostly remove the effect of the IC in the VD but I gave the current a highway to ground... k.

zeta_no

p.s: If I'm wrong somewhere, please let me know.

 

That should be correct. Just remember that a 9 volt battery is happier with supplying a current in the single milliamp range.

 

Thanks beenthere for the quick answer,

Exactly, I use the 74ls14. I checked on a NS datasheet. I found "Input Current @ Max Input Voltage". Here they say 0.1 mA and max Vcc is 5.25V.

Should be it... right?

Using V=RI, we find R = 5.25/0.1x10^-3 = 52.5 kOhm.

With my 100 Ohm, I mostly remove the effect of the IC in the VD but I gave the current a highway to ground... k.

zeta_no

p.s: If I'm wrong somewhere, please let me know.

You should use a regulator to provide 5V for the chip (the chip current varies too much to reliably generate 5V simply with a dropping resistor).

Once you have 5V, simply tie inputs to the 5V rail with a resistor, e.g. quoting from http://www.nxp.com/acrobat_download/various/HCT_USER_GUIDE.pdf

7.4 Termination of unused inputs
To prevent any possibility of linear operation of the input circuitry of an LSTTL device, it is good practice to terminate all unused LSTTL inputs to VCC via a 1.2 kW resistor. Inputs should not be connected directly to GND or VCC, and they should not be left floating.​

 

it is good practice to terminate all unused LSTTL inputs to VCC via a 1.2 kW resistor

- possibly you meant 1.2 Kohm?

I find that 4.7 K works fine (it is well-known that any circuit with at least one 4.7K resistor will work very well).

 

- possibly you meant 1.2 Kohm?

I find that 4.7 K works fine (it is well-known that any circuit with at least one 4.7K resistor will work very well).

I was just concerned that he was getting the idea that a resistor to +9v was OK.

Do we have some sort of font incompatability?