Help me troubleshoot this simple design

[whiz115]

Remove R3 and rest of circuit functions. go for functional tests of ESR measurement .

why it is not needed?!

also do you know what these "buffer amps" are?
do they amplify the output from the relaxation oscillator and seperate the
oscillator from the rest of the circuit?

There is one other thing that might cause error in your measurement. That is that since the analog multimeter is probably a very high resistance, the time required for voltage to bleed off of C6 may be too large. .

this time i was using my digital multimeter...

The value of 2.15 V at the transistor collector also seems to be good. However, this assumes that you have a capacitor across the test terminals TP1 and TP2. If you don't then this voltage is not correct. I will assume that you have a good capacitor there. .

no i don't have any capacitor connected...


P.S did you noticed that i'm not getting anything on R7-R8 this is normal?!

 

[mvs sarma]

It doesn't matter even if one inverter IC1F and the concerned resistor are not there and rest of buffers can do enough buffering.

 

[RadioRon]

why it is not needed?!

also do you know what these "buffer amps" are?
do they amplify the output from the relaxation oscillator and seperate the
oscillator from the rest of the circuit?

Yes, these buffer amps are limiting amplifiers. Their job is to drive a 5 volt square wave into a low resistance load with enough current. They also isolate the oscillator from this load too. The original designer decided to use 5 of these amplifiers simply because that is how many he had available. It would be better if you had all 5 operating correctly but you can get away with losing one. The value of R2 to R6, 680 ohms is designed assuming 5 amplifiers so technically we should adjust the value. I suggest that you get IC1F working instead.

this time i was using my digital multimeter...



no i don't have any capacitor connected...


P.S did you noticed that i'm not getting anything on R7-R8 this is normal?!

Your envelope detector circuit requires some voltage to begin showing an output because of the forward voltage drop of D1-D4, which is nominally about 1.2 volts. The expected AC pp voltage at R7 is only about 400mVpp so this would not be enough to give a reading. So I am not surprised that there was 0 volts, but I am surprised that you measured a voltage there. When you are recording values such as you did a couple of posts ago, when your meter reads zero at R7 (for example) you should write down "R7 0 " instead of "R7 ". When you write nothing instead of zero, I understand that there was no measurement made, not that you measured 0 volts. So I believed that you had not measured there. Next time, write down a zero.

So, if there was no capacitor at TP1 and TP2 I am wondering why you got a strong AC voltage measurement at Q1 collector. You should only get a reading like this if there is a capacitor connected. So, I am wondering if you could please check the actual resistance of R18 and report it back to us.

 

[whiz115]

without using the probe we made.. i can see ~220mV from the esr meter output... R18 is 980K

i'm doing something wrong... i don't understand what's going on with the circuit and behaves so weird!

 

[RadioRon]

without using the probe we made.. i can see ~220mV from the esr meter output... R18 is 980K

i'm doing something wrong... i don't understand what's going on with the circuit and behaves so weird!

Let's be sure that we aren't having problems because your meter at the output has very high resistance. Place a 100K ohm resistor (anything from 10K to 100K is ok) across C6 and then tell us what the output is once more.

If you are still seeing 220mV at the ESR meter output, then we must trace backwards by alternately disconnecting things until the reading goes to zero. Start by disconnecting C3 from Q1 collector. Now the meter should show zero.

Now, reconnect C3 to Q1 collector. Then disconnect C2 and leave it out. Your ESR meter should now read about zero volts.

Please report your readings.

 

[whiz115]

[double post]

 

[whiz115]

Let's be sure that we aren't having problems because your meter at the output has very high resistance. Place a 100K ohm resistor (anything from 10K to 100K is ok) across C6 and then tell us what the output is once more.

across means parallel?! if i put the 100k (98,5k) resistor parallel with the C6 i get 18.5mV at the output.

If i remove R3 doesn't help on anything.

with disconnected C3 the output is still 220mV
with disconnected C2 and C3 the output remains 220mV
If i connect C3 and let C2 disconnected the output is still high.


Btw.. i have a question! why one of these amplifiers doesn't work? i don't get it, is it burned?! if yes how come only one of them while they all are connected togather?!

 

[whiz115]

Hi

i can't understand what's going on, i think that if i disconnect C3 then i shouldn't be able to see any output! if this is true then it's ridiculous what is happening here

i'm checking all the time the PCB for bridges but i don't see any!

what should i do?

 

[RadioRon]

across means parallel?! if i put the 100k (98,5k) resistor parallel with the C6 i get 18.5mV at the output.

If i remove R3 doesn't help on anything.

with disconnected C3 the output is still 220mV
with disconnected C2 and C3 the output remains 220mV
If i connect C3 and let C2 disconnected the output is still high.


Btw.. i have a question! why one of these amplifiers doesn't work? i don't get it, is it burned?! if yes how come only one of them while they all are connected togather?!

Yes, across means "in parallel with". If you put the 100K resistor in parallel with C6 then what you are doing is loading the detector circuit with a known resistance and insuring that the voltage charge on C6 can discharge in a reasonable amount of time.

You can leave R3 out for now if you like.

If one of the amplifiers does not work, then the most likely cause is that you don't have electrical connection to the IC pin for some reason or that there is a short circuit to ground on the output pin. It is unlikely that the circuit inside the IC is broken just in this one amplifier, but it is possible. Since you have pin 5 also connected to pins 2, 3, 9, 11, and 13 and the other amps work ok, then we can assume that pin 5 does not have a short. However, we don't know if pin 5 is actually getting the signal. One way to check this is to use your digital voltmeter to measure resistance in ohms and, with the power removed from your board, measure resistance from the IC lead pin 5 (right where the leg goes into the black plastic of the ic, not underneath where it is soldered) back to the ic lead pin 2. It should be very low resistance.

Next, measure from the IC leg pin 6 (where the leg goes into the plastic, not underneath) to the junction of R7 and R18. You should see 680 ohms. Also measure from IC leg pin 6 to ground. You should see about 690 ohms.

 

[RadioRon]

Hi

i can't understand what's going on, i think that if i disconnect C3 then i shouldn't be able to see any output! if this is true then it's ridiculous what is happening here

i'm checking all the time the PCB for bridges but i don't see any!

what should i do?

As I said before, don't despair, we will find the problem.

Ok, so if when you completely disconnect C3 but you still see a significant voltage on your ESR meter, this could be because:
- the voltmeter is faulty or poorly calibrated
- the wiring to your board is acting as an antenna for local AM broadcast stations and putting significant AC voltage into your bridge rectifier
- your voltmeter or ESR board is picking up 60 Hz magnetic field from nearby mains wiring and converting that into a DC voltage

The first thing to do is to simply disconnect the meter from the ESR circuit and observe what voltage it shows with its input wires unconnected to anything. Is it still 220mV? What is the value shown?


When you left R17 out of the circuit, you have invited problems. By using a digital or analog multimeter instead of the specified 50uA panel meter, you may also be inviting problems. But we can cope with this as long as we understand what is going on. The original circuit was designed using a 50 uA meter along with a 25Kohm zero pot built into the meter (as far as I can tell). Such analog meters will have an internal resistance of about 1600 ohms (typical). Because it is an analog panel meter and with such low resistance, the meter itself is not sensitive to nearby electric or magnetic fields and so won't pick anything up by itself. Digital multimeters are more complex with internal circuits plus they are much higher input resistance, so it is relatively easy to pickup a voltage due to stray fields. Analog multimeters are simpler and so less prone to problems, but they still have very high input resistance when measuring volts and so still are not ideal.

Note that the original design uses internal batteries. This is an advantage compared to your design because it keeps the wiring short and isolates the circuit from the mains supply, which means there is much lower chances of picking up outside noise or signals.

In order to use a current meter, the current must be limited by an external resistance and that is why the schematic uses R17. R17 is critical to setting the total load resistance placed on the bridge rectifier. We can estimate that the load across C6 due to R17 plus the meter is about 11.6K ohms. If you are using a digital multimeter in place of the panel meter and measuring volts instead of uA, then you should have a resistor across C6 to take the place of the missing 11.6 K ohms load. Your digital multimeter does not do this because it's input resistance is probably more like 1 M ohm or higher (when measuring volts. If you use the multimeter to measure microamps, then you would need R17 only).

So, if you will be using a volt meter for this ESR device, then add a 10 K ohm resistor in parallel with C6. You mentioned that when you added a 100K ohm resistor at that point, the reading on the voltmeter went down to 18.5 mV. This is good. With no connection of a capacitor or short circuit across TP1 to TP2, we expect to see zero or nearly zero on the meter. 18.5 mV is nearly zero.

Then, when you short TP1 to TP2 with a capacitor or simply with wire, the reading on your meter should go up to some higher voltage. Please let us know what that voltage is. I expect it to be 0.7 volts.

 

[whiz115]

However, we don't know if pin 5 is actually getting the signal. One way to check this is to use your digital voltmeter to measure resistance in ohms and, with the power removed from your board, measure resistance from the IC lead pin 5 (right where the leg goes into the black plastic of the ic, not underneath where it is soldered) back to the ic lead pin 2. It should be very low resistance.

Next, measure from the IC leg pin 6 (where the leg goes into the plastic, not underneath) to the junction of R7 and R18. You should see 680 ohms. Also measure from IC leg pin 6 to ground. You should see about 690 ohms.

pin 2-5 resistance is infinite!!
pin #6 to R7 is 690ohm
pin #6 to R18 is 680ohm
pin #6 to ground is 690ohm

So, if you will be using a volt meter for this ESR device, then add a 10 K ohm resistor in parallel with C6. You mentioned that when you added a 100K ohm resistor at that point, the reading on the voltmeter went down to 18.5 mV. This is good. With no connection of a capacitor or short circuit across TP1 to TP2, we expect to see zero or nearly zero on the meter. 18.5 mV is nearly zero.

Then, when you short TP1 to TP2 with a capacitor or simply with wire, the reading on your meter should go up to some higher voltage. Please let us know what that voltage is. I expect it to be 0.7 volts.

yes... i'm going to use the esr meter with a analog voltmeter... but for the above you said i'll try with the digital so i can see if it is going to be 0.7V

 

[RadioRon]

pin 2-5 resistance is infinite!!
pin #6 to R7 is 690ohm
pin #6 to R18 is 680ohm
pin #6 to ground is 690ohm



yes... i'm going to use the esr meter with a analog voltmeter... but for the above you said i'll try with the digital so i can see if it is going to be 0.7V

If pin 2 - 5 is infinite, then this explains why that channel isn't working. It tells us that you have no connection so the amplifier is getting no input signal. This is either because the solder joint below pin 5 is no good, or it could be because the ic pin is not making contact in the socket. This is unusual if the pin is not bent, but if the pin got bent while going in, this could be the reason.

The other readings tell us that the connections on the output side of the buffer amp are OK.

 

[whiz115]

Vcc ~5.80V
resistor 9.87k

output without resistor ~244mV
nothing connected 2.1mV
with capacitor 2.2mV
bridged 2.4mV

 

[RadioRon]

Vcc ~5.80V
resistor 9.87k

output without resistor ~244mV
nothing connected 2.1mV
with capacitor 2.2mV
bridged 2.4mV

We need to be sure that the circuitry from C3 to the meter itself is all working correctly now and I think the best way to do that is to repeat what we did earlier, that is disconnect the left side of C3 and use this point, with a wire, to probe IC1D pin 2. If everything is correct, we should see about 1.6 volts DC or perhaps a bit more. Last time you measured, what was it, over 2 volts, so this time should be similar or a bit lower.

I made a small error in my previous calculations. I think that the resistance of the meter in his original design was not 1600 ohms exactly, but rather I now think that he was using a 25 K ohm potentiometer in series with the meter to calibrate his full scale reading on his meter. So, guessing that it might have taken a value of about 15K on that pot, that means the overall resistance that his meter presented to the detector circuit was 26.6K ohms not 11.6 K ohms. I don't think it makes much difference, but to be precise, we probably should change the resistor that you added across C6 from a 10K ohm to something higher like 22K or 27K ohms. This value isn't really all that critical so don't bother making this correction until we've debugged the system.

 

[whiz115]

i did once again the test but i used the 100k resistor and the results was


nothing connected 22.5mV
with capacitor 22.3mV
bridged 22.6mV

i must get a stabilized power supply...

We need to be sure that the circuitry from C3 to the meter itself is all working correctly now and I think the best way to do that is to repeat what we did earlier, that is disconnect the left side of C3 and use this point, with a wire, to probe IC1D pin 2. If everything is correct, we should see about 1.6 volts DC or perhaps a bit more. Last time you measured, what was it, over 2 volts, so this time should be similar or a bit lower.

probably the IC said "bye bye" to all of us.....
i measure 4.70V to pin #2 maybe i did an awkward move and it got burned... should i change IC?

 

[RadioRon]

i did once again the test but i used the 100k resistor and the results was


nothing connected 22.5mV
with capacitor 22.3mV
bridged 22.6mV

i must get a stabilized power supply...



probably the IC said "bye bye" to all of us.....
i measure 4.70V to pin #2 maybe i did an awkward move and it got burned... should i change IC?

Hmmm, I wonder what is going on here? When you measure 4.70V at pin 2, do you mean using our circuit of C3 and the rest, or do you mean when you simply measure with your DC multimeter on the volts range?

 

[whiz115]

Hmmm, I wonder what is going on here? When you measure 4.70V at pin 2, do you mean using our circuit of C3 and the rest, or do you mean when you simply measure with your DC multimeter on the volts range?

i have disconnected the left leg of the C3 and i have connected to it's place the probe...

don't get surprized...probably the ic got burned...because i measure nothing between the resistors and most of amp buffers.

Should i use another IC? i got a good one from Texas instruments lol..

 

[RadioRon]

I am still somewhat surprised that you measure 4.7 volts at pin 2. I expected something closer to around 1.5 to 2 volts if things are working normally. And if you measure 0 VDC when probing any pin 4, 6, 8, 10 or 13 then something is very wrong with the IC. But didn't you just try doing some repairs around pin 5? Perhaps you caused a new short circuit by bridging some solder across pads or something? This type of IC is difficult to burn out, so changing is our last resort. I suspect that something is shorted or faulty around pins 2, 3, 5, 9, 11, 13 or even around pin 1. Please check those carefully. One way to check them besides simply looking is to do an ohms measurement from each pin to pin 14 (power supply) and also to pin 7 (ground). This must be done with the power OFF. If any of these measurements are low resistance, then you have a short circuit that needs to be found.

Perhaps the most likely fault is that pin 1 has been somehow shorted to ground since this might cause you to get 4.7 volts at pin2 and also 0 volts at pins 4, 6,8, 10, 13.

 

[whiz115]

Hi RadioRon! i have good news...

the problem was at pin 2,3,5,7,8 some joints wasn't so good and i discovered it with the multimeter's continuety test. the IC is fine (it wasn't burned as i was thinking...) so i didn't replaced it with new one.

You are great and very patient guy! and i want to thank you for all the help
you provided so i can locate the problem!

The device works very good and i'm very satisfied from the results, altough
i would like to improve the output a bit because the multimeter niddle does
not go full range... can i double the output current by tweaking the collector-emitter resistors at the transistor?

 

[RadioRon]

Thank you for your kind words. I am happy that it is working as that was a lot of work to debug, wasn't it? The easiest way to get full scale reading on an existing multimeter is to add that series resistor R17 using a variable resistor (poteniometer) and then change your multimeter back to measure current. If you get a reading of, for example, 1.3 Volts on the output multimeter for a good capacitor (or short bridge) across TP1 TP2, then the way to do it is to put a R17 into the circuit using a variable resistor and then change your analog multimeter to the 0.05 mA scale and attach it to the output side of R17. As you may recall, Ohm's law says that if you have 1.3 volts DC on the left side of R17 and if we assume that the multimeter when in 0.05 mA scale has a resistance of about 1600 ohms (guess), then we can tune the position of the R17 potentiometer to get exactly full scale on the meter by adjusting its resitance to 1.3/(R17+1600) = 0.05 mA and then solve for R17. This is exactly what the original designer has done. I estimate that R17 would be about 24K ohms. The original designer found it convenient to break this up into one fixed value of 10K and one variable resistor of 25K. You could do the same.

If you do it this way, try to use the most sensitive current range on the meter as this keeps the R17 higher which is better because it doesn't load down the detector as much.

If you want to continue just using the voltmeter, the easiest way is to put a potentiometer at the output, with one end to ground, the other end to C6 (top) and the wiper to your voltmeter. Then just choose a 1V scale and adjust your potentiometer for a maximum output of 1V. In this case, any potentiometer of 20K ohms up to 500 K ohms would work ok.