enter image description hereIs this wiring and resistor the best choice as a pull-down resistor?
Is it necessary to put the pull-down resistor(s) on all inputs and outputs?

I put a 10K Ohm resistor connected to the chip ground. The IC is a HD74LS32P, and it is a positive OR gate. I connected two inputs to the other end of the resistor.

With power to the chip, the multimeter reads:

  • Vcc to ground is 4.92V
  • A2, V=1.02
  • B2, V=1.03
  • Y2, V=0.74

I have a PWM driving device, and when that is powered and set to the lowest setting:

  • A2, V=0
  • B2, V=0
  • Y2, V=0.17

When the PWM is set to max value:

  • A2, V=4.3
  • B2, V=4.3
  • Y2, V=0.42

2 Answers 2


The pulldown resistors that you are using is on the high side for one LSTTL pin and probably beyond acceptable when used to pull down two pins at once (see explanation at end). A lower value of resistor should be used. However - this does not properly explain the problems that you are seeing. THe following procedures will help you establish whether the IC is working OK.
Once you know that the IC works OK you can go on to see what happens with PWM drive.

A troubleshooting template:

As well as addressing your questions and showing you how to find what is wrong, the following serves as a troubleshooting template for finding what is wrong with simple circuits. It takes a step by step approach that "must" work if followed*.

  • All inputs or outputs that are not relevant to the immediate tests are disconnected.

  • The power supply voltage is checked at the power supply (NOT on its leads).

  • The power supply to the IC is checked ON the IC pins.

  • A meter polarity and lead-connection sanity check is carried out.
    This can sometimes avoid major disasters - such as plugging a meter into mains AC with one test lead plugged into the 10A DC socket. Or not. (Ask me how I know :-) ).
    (After enough decades you are allowed a certain %age of stupid actions - and with time learn to usually not take up your allotment).

  • Basic test conditions are set up, and then measurements are made ON the IC pins to ensure that the conditions appear on the pins. This avoids things like broken wires, misconnected wires, bad breadboard and other connections and fault conditions which load signals so they are not what is expected.

  • Results of the applied inputs are checked AT THE IC pins.

  • Input sets and outputs are worked through logically.

*The above USUALLY works. It is not in fact 100% foolproof. Murphy says that nothing is ever 100% certain, and knowing this is an essential part of understanding things. In the sequence above, no mention was made of eg grounding or terminating unused inputs in the 3 other OR gates in the IC. In a TTL IC this is usually not going to cause problems as TTL has implicit input pullup action. With some ICs, failure to terminate all unused inputs in a defined and proper manner may lead to failure of other IC sections to function properly, or at all. Occasionally it can lead to magic smoke and a dead IC and occasionally to dead other things as well. In this case. hard connecting all unused inputs to Vcc OR to ground would do no harm and is good practice. In this case it is not liable to affect the results.

What you are doing is essentially correct but the result is confusing either because

  • PWM duty cycle max is very low.
    This needs to be checked. or

  • Vcc is not applied correctly to pin 14 or

  • IC is dead or

  • One of the wires running to off-stage where we cannot see does not connect to what you and we think

Here is a quick sanity check

Not as long as it may seem at first glance.
This is (if I have not had a brain fade) a classic bullet proof troubleshooting sequence that assumes nothing.
It "must" work [tm].

When I mention measuring AT a pin, put meter probe ON THE PIN on the IC.
When I say to connect eg Vcc to a pin it is OK to do this via the breadboard sockets as the next step checks that it has happened.

  • Disconnect wire to pin 4 (PWM input) - not needed or wanted for these test.

  • Set voltmeter to 20V range. (10V to 40V OK)
    If using a typical DMM with plug in leads, ensure black lead is in COM (common)and red lead is in correct socket for measuring volts.

  • Measure power supply voltage AT power supply terminals - NOT on connected wires.
    About 5V from what you say.

  • Measure voltage between IC pin 14 (Vcc) and IC pin 7 (Ground) with probes ON ic pins (not on wires.)
    Red lead to pin 14, black lead to pin 7.
    This avoids being fooled by wrong feed, bad wires, bad breadboard contact, Murphy.
    Should be about +5V with pin 14 more positive than pin 7.
    If not, remedy.
    V = 0 = bad connection.
    V < say 4.8V = find out why.

  • Reverse meter leads - red to pin 7, black to pin 14.
    Measure volts.
    Should read -5V (negative 5). ENSURE meter shows a -ve sign in this mode. If it shows -% in the prior test and +5 here the power supply is reversed.
    It happens.

    • Connect wire from pin 4 to ground. Confirm that pin 4 AT IC to pin 7 = 0 V.
      Confirm that pin 5 AT IC to pin 7 = 0 V.
      (You say you have joined pins 4 & 5, and the photo suggests that you have. This ensures that the voltage is the same on both pins.)

    • Measure pin 6 to pin 7 - should be <1V.
      Report value.

    • Disconnect ground wire to pin 4.
      Connect wire from pin 4 to Vcc (eg to level on pin 14) Confirm that pin 4 AT IC to pin 7 =~ 5 V.
      Confirm that pin 5 AT IC to pin 7 =~ 5 V.
      If not, find out why.

    • DISCONNECT the wire currently going to pin 6 (wherever it goes) Measure pin 6 to pin 7
      IF you have done all the above correctly, then
      V pin 6 wrt pin 7 = high = >= 2.4V.
      Probably about 3.5V.
      See 74LS32 data sheet for levels.

IF pin 6 is not "high in the above step then

  • IC is dead or

  • You did not follow instructions above correctly or

  • IC is dead

IF pin 6 IS high, reconnect wire to pin 6 that was connected above.
If pin 6 falls to under 1 V (or ever under say 2.3V) then there is an "illegal " load on the other end of the wire that you did not tell us about.

Report back.

Is it necessary to put the pull-down resistor(s) on all inputs and outputs?

For ICs that do not have internal pullup or pulldown 'resistors', Pull downs (or pull ups as appropriate) are needed on all input pins that you wish to be low (high for pullups) when they are not being driven. If a driver is ALWAYS connected a pulldown is not needed.

Normal "push pull" or totem-pole outputs that actively drive high and low do not need pullups or pulldowns.
Open drain or open collector outputs (no active high drive) need something to drive them high when they should be high. This is often a pullup BUT depends on the circuit design.

In the case of LSTTL ("Low Power Schottky") ICs the inputs have internal pullUP resistors.

These are nominally 20k - so two inputs joined effectively have a 10k pullup.
If you pull 2 pins down with 10K you can expect > 2.5 V on the input due to these resistors. Vinhi min is 2V and Vinlomax is -.8V (see data sheet). To get 0.8V max with 5V Vcc and one pin you need about 4k pulldown - so 2K for 2 pins together - so a 1K is probably safe. If you are driving these pins with PWM you may not need a pulldown at all. Deciding that is the next stage after this.

enter image description here

  • \$\begingroup\$ The TTL is not always connected, so the resistor is needed. The 20K for a TTL line made the IC work properly. \$\endgroup\$
    – Allen
    Jun 27, 2015 at 7:36
  • \$\begingroup\$ @Allen When you said "properly" you really meant "sometimes under some conditions". You still need to find that out. | What I said (or meant) was that if the PWM provides a low and high impedance drive when working then it may drive the gate OK BUT if the PWM is not connected then a single 20k as shown is TOO LARGE does not meet data sheet spec and will probably cause problems in some cases - for the reasons that I gave above. \$\endgroup\$
    – Russell McMahon
    Jun 27, 2015 at 8:58

If a pull-down is required (I don't see a schematic) then 10K is too high a value.

Maximum input voltage for a '0' is 400mV (for 400mV noise margin, as usual). Maximum input current IL is 0.1mA (see the datasheet). Therefore the pull-down resistor for a single 74LS inout should not exceed 4K. If 'n' inputs are connected to the same pull-down the maximum pull down should be 4K/n.

Your signal should be between 0V and 400mV for logic '0' and between 2.4V and Vcc (5V) for a logic '1'. The gate is guaranteed to interpret less than 800mV as logic '0' and more than 2V as logic '1'.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.