Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
Hi there,

I have been working on an AM radio test generator. It is from the book ”building your own transistor radios”.

I would provide a schematic of the circuit although I believe the copyright would prevent me from doing so.

The two ICs used in the circuit are the 74HC14 hex Schmidt trigger and the 74HC05 hex inverter. My circuit construction consists of a copper clad board and the components soldered on the board suspended above the board except for ground connections. The ICs are soldered dead bug style with the leads also suspended except for ground connections.

The circuit does not work and the investigations with the DMM show that the schmidt trigger although being supplied with a regulated 5 volts just produces noise instead of oscillating H and L outputs in a feedback loop configuration.

I assume I damaged the IC through my soldering. Essentially I would heat up the pins and apply solder, then heat the wire to be connected to the pin. Would this damage the IC?

A second question is that the data sheet shows a thermal impedance of 127 degrees C. Would this mean that a temperature above this threshold would do damage?

Look forward to your reply.
 

jpanhalt

Joined Jan 18, 2008
10,068
A picture of your board might help.

People copy and paste schematics here all the time. It's virtually impossible to discuss and resolve such problems without the schematic. If posting the whole thing bothers you, then post a significant portion of it that shows where you think the problem is.

Thermal impedance is usually written as a temperature increase per watt of power dissipated by the device. It has nothing to do with maximum allowable operating temperature, except you can calculate from the power dissipated whether that will lead to a temperature in excess of the maximum allowable.
 

DickCappels

Joined Aug 21, 2008
6,532
(Some text removed for clarity)
I assume I damaged the IC through my soldering. Essentially I would heat up the pins and apply solder, then heat the wire to be connected to the pin. Would this damage the IC?

A second question is that the data sheet shows a thermal impedance of 127 degrees C. Would this mean that a temperature above this threshold would do damage?
It is unlikely that the IC was damaged by the soldering. You can test the Schmitt trigger by manually changing the input voltage and watching the output. I have soldered lots of IC's (mostly in DIP packages) dead bug style over manyyears and have not had a failure of a semiconductor device from the soldering yet.

1596810693269.png


The thermal resistance is the temperature rise over ambient in °C per watt of power dissipation. That means that a part with 127 °/watt thermal resistance in a 30° environment when dissipating 0.1 watt would rise to 30° + 12.7° = 42.7°. Often the specification also notes what points are considered in the measurement, for examples, junction to ambient temperature rise, junction to case, and junction to seating plane.





Just a guess: Does the schematic include the circuit below?

1596811126534.png
 

crutschow

Joined Mar 14, 2008
25,247
Heating the pin and applying the solder to tin it should not take more than a second or two.
If longer, you can overheat the IC.
You should also tin the wire before soldering to the pin.

Apply a small amount of rosin flux to the pin before you tin it, which will speed up the process.
 

Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
Thank you all for your prompt replies.

DickCappels: Correct!, that is the first part of the circuit, the output of the trigger is feedback to itself, and the output is also connected to the hex inverter. This creates two test signals.

I will proceed to test the trigger and go from there.
 

jpanhalt

Joined Jan 18, 2008
10,068
@Anonymous User 30
I think a picture would be very helpful in solving the problem you describe, if it hasn't been solved already. The resistor+capacitor+Schmidt inverter is a very robust oscillator. If you look at @DickCappels example, notice some parts are upside down. Legend has it that is why it is called "dead bug" instead of just Manhattan.

Anyway, I think the likelihood of overheating an IC pin is low, probably very low. The probability that something is misconnected is much higher. That is not a criticism. Everyone makes that type of mistake, particularly when parts are upside down.

BTW: If you just want a simple AM receiver, there is a simple single chip solution.

Edit: We cross posted. Sorry.
 

Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
Thank you all for your prompt replies.

DickCappels: Correct!, that is the first part of the circuit, the output of the trigger is feedback to itself, and the output is also connected to the hex inverter. This creates two test signals.

Crutschow: I believe my technique is leaving a bit to be desired at the moment. I will do more research on this.

I will proceed to test the trigger and go from there.
 

Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
AM Radio Test Generator.jpg

I've tested both ICs with a function generator and they both work. Made progress thanks to the discovery of a dodgy connection.

So far i've been able to get the 535KHz CW working. I noticed the connection from U1A to U2A results in the inverter not inverting the signal and I get nothing at AM 535KHz output. I believe it is because I am not reaching the H and L V thresholds for the inverter. The output of pin 2 of the 74HC14 trigger is an oscillation that has a H of 5v and a L of 3.5ish V. I believe this means that it is not triggering the inverter when it goes L.

Any help would be appreciated.
 

jpanhalt

Joined Jan 18, 2008
10,068
NB: Pin # is in parenthesis after device label.

Output of U1A(2) is input to U1B (3) and U2A (1). How can U1B be working to give 535 kHz CW, but U2A does nothing?

They should both be getting exactly the same signal. Is U1C(6) giving an output?

Suggestion:

The 74HC05 is open drain (Y in diagram):

1596873802641.png
The diode to VCC clamps maximum voltage to VCC; it does not supply current.

What happens if you disconnect R6 at U2A(2) and put a pull-up resistor (1 k to 4.7 k) to VCC? This is from the TI datasheet for the device:
1596874159233.png
The 74HC14 is not open drain/collector. It can source current.
 

Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
Hi jpanhalt,

Thanks for your help. Im going to try and digest you've written and report back. Please note, i'm just involved in electronics as a hobbyist and do not have any formal education in the field.

What you said regarding the open drain inverter might be the issue as I realise now the ICs are infact the 74HC04 which are not the open drain types.
 

Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
Hi jpanhalt,

Please see below printout from the oscilloscope. Please note that IC is indeed 74HC04. I managed to disconnect R6 from U2A and add a pull up resistor to VCC of 1k.

CH1 is pin 1 and CH2 is pin 2

DSO00001.jpg

Below is the oscilloscope readout for pin 6 of U1C as requested.

DSO00002.jpg

Thank you, AU30.
 

jpanhalt

Joined Jan 18, 2008
10,068
The 74HC04 should not work as a replacement for the 74HC14. If you use it instead of the 74HC05, then you do not need the pull-up resistor. Try that experiment without R6 and you should see the 535 kHz at U2A(2).

As for the rest of the circuit, that may not work as shown without the open collector/drain.

A capacitor between U1A(2) and U2A(1) will get rid of that voltage offset, which may allow U2A to operate as expected..
 

Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
The 74HC04 should not work as a replacement for the 74HC14. If you use it instead of the 74HC05, then you do not need the pull-up resistor. Try that experiment without R6 and you should see the 535 kHz at U2A(2).

As for the rest of the circuit, that may not work as shown without the open collector/drain.

A capacitor between U1A(2) and U2A(1) will get rid of that voltage offset, which may allow U2A to operate as expected..
Sorry to confuse, im using a 7HC04 instead of the 7HC05 (im definately using the 75HC14). Would this still pose an issue? Thanks for your help. I'll try a capacitor between the two ICs and report back during the week. Much appreciated for your time.
 
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jpanhalt

Joined Jan 18, 2008
10,068
The 74HC04 output provides both a push and a pull. It can source current based on VCC. The 74HC05 depends on the load.

In thinking about the oscilloscope tracings, I think you need to got back to the 535 kHz oscillator (U1A) and find out why its output has so much DC offset. Can you set up on a breadboard that oscillator just by itself and see what you get? Be sure sure all unused inputs on the chip are grounded (or pulled high). How is your scope attached?

A CMOS oscillator should give something like this:
1596975195073.png
Source: AN-118 (Fairchild)

EDIT:

Maybe this will help. Uding Nexperia datasheets.

74HC05 (open drain):
1596976884204.png


74HC04:
1596976967438.png

Notice that the HC05 state with a low input is described as "off" or high impedance. The on state (high input) is low impedance, i.e., the output transistor is on. Whereas, for the HC04, the respective output voltages are high and low, respectively. For the HC05, one might question whether the output states should be described high Z and low Z to ground, respectively. Calling low Z to ground "L" is justified as both devices do that. See this circuit for the output of the 74HC05:

1596977424646.png

In contrast the output of the 74HC04 is like this (Source: https://www.ti.com/lit/an/scla007a/scla007a.pdf?ts=1596977941039&ref_url=https%3A%2F%2Fwww.google.com%2F ):

1596978022827.png
 
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Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
Hi jpanhalt,

Managed to get the circuit on a breadboard and that solved the DC offset issue from the CW test signals.

The first shot is the CW signals with CH1 being the 455kHz CW output and CH2 being the 535kHz CW. Note that I have used 2 1k resistors in lieu of the 1k pots. This explains why CH2 is less than 535kHz. I can't explain CH1 (supposed to be 455kHz CW) has very fast oscillations, would this be some sort of noise? I'll check connections on breadboard and report back in a couple of days.

Same situation with the AM outputs in image 2. CH1 being the 455kHz AM and CH2 being the 535kHz AM. The AM test signal for IF amplifiers did not work.

I'll check the connections and report back in the next few days. Trying to balance work with all of this.

Thank you.

DSO00006.jpg


DSO00007.jpg
 

jpanhalt

Joined Jan 18, 2008
10,068
I looks like you have some high frequency noise. Moreover, the fine oscillations in CH2 roughly match the oscillations in CH1. Have you tried to match them up?

Usually a soldered board works better than a solderless breadboard. It is probable something on the soldered board in not correctly connected.

Work on ONE oscillator at a time. Most likely, you are picking up noise from the power supply. As one section of a chip oscillates, it produces alternating high and low currents drains from the supply. That affects every other chip attached to the supply.

To reduce that effect, one adds "decoupling" capacitors between the supply pins of every chip. In your schematic, C10 is for decoupling U1 and C9 is for U2. Sometimes it take more decoupling, like 10 uF, 1 uF and 0.1 uF capacitors (3 total) in parallel as close as practical to the power pin and ground.
 

Thread Starter

Anonymous User 30

Joined Sep 4, 2017
35
Hi,

Loose connections and inferior connections on the breadboard have been fixed, i've been working on one oscillator at a time and managed to get the 2 AM waves and 2 CW signals working. The 455kHz test sig does not work and I'm assuming im using wrong components:

Both AM signals:

DSO00008.jpg

Both CW signals:

DSO00009.jpg

See below traces for UC1(6) (more like after after R5) and AM test sig. Test sig is definately something I have to investigate and UC1(6) I believe to be correct. The book states this signal to be a 1kHZ triangle wave for amplitude modulation.

I see that the AM signals are not receiving any amplitude modulation from the 1kHz triangle wave (U1C[6]). It states to vary the modulation R6 and R7 can be varied. I'll try this and report back.
 

DickCappels

Joined Aug 21, 2008
6,532
The tendency for those plastic breadboards to have poor connections is why I stopped using them nearly 40 years ago. It does not take much to throw a microprocessor into the HCF. "Halt And Catch Fire" mode.
 

dl324

Joined Mar 30, 2015
11,231
I've been using solderless breadboards for about 45 years and have never had problems with poor connections. I've even used some of the low quality boards from Ali Express for less than $2 each (shipped). I haven't had the "hard to insert" or loose connection problem with any of the half dozen I bought (3 each from 2 different sellers).
 
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