Freezer Temperature Monitor

[ljh]

Being a dummy ain't fun

Ok - I think I may NOT be wiring according to the schematic. I did not know that "VDD" meant negative supply voltage. I thought "VDD" was positive supply voltage and "GND" was negative supply voltage.

It has been years since I have played with any of this and the terminolgy has obviously changed. Used to be so simple with "+" and "-"

Is "+ 12 rail" referring to a DIN rail ( I assume a method of mounting components) and is that the same as "GND"? Or does "GND" just mean a common connection?

It appears that "GND" has to be the "VCC" or "positive supply voltage" - otherwise where would it come from? If that is the case, I have had everything reversed. Good thing I have duplicate components.

Once I have a better (read "some") idea what I am doing, can probably get it to work.

Please clarify the above and I should be good to go.

Thanks again - sorry to be wasting your time on dumb things.

 

[dknguyen]

Nononono. FOr MOSFETs, there is drain (Vdd) and source (Vss). For NMOS MOSFETs, Vdd should be at a higher voltage than Vss. So Vdd should be connected closer to +12V, and Vss should be connected closer to GND.

For NPN-type BJTs, the collector (Vcc) should at a higher voltage than the emitter (Vee).

Vxx does not mean + or - power supply so much as it means which terminal has to be at a higher voltage and which has to be at a lower voltage.

So depending on "which end" you want to place your transistor, one terminal will be directly connected to GND or +12V, while the other terminal will be connected "closer" to the other power terminal. But since you want to use it as a switch, you want current to flow in series through the load and the transistor, so you cannot connect the both ends directly to the power supply (or else it would be in paralle with the load and not be able to switch the load current). Rather (for the case where the MOSFET is connected on the low-side of the circuit- as close to ground as possible. High side is as close to +V as possible) you connect one terminal directly to ground and since you need the load current to flow through transistor (transistor in series with the load), then you can't connect the other end of the transistor straight to +12V. Instead you must connect it to the negative end of the load (the positive end of the load is the one that goes straight to +12V).

Of course, the terminal which must be at a higher/lower voltage (Vcc, Vee, Vdd, and Vss) becomes reversed when we talk about PMOS and PNP transistors which are "opposite".

Look at Philba's schematic. Both transistors are on the low side (connected closest to ground). So one terminal is connected to ground, while the other terminal is not connected straight to the power supply. Instead it is connected to the end of the load to make a series connection with the load so load current flow can be controlled by the transistor.

 

[ljh]

Reference Philba's TIP120 Schematic

If I interpret the prior message correctly, GND = +12vdc applied to this particular schematic and the positioning of the Darlington Transistor - correct??

+12 vdc to pin 3, Emitter of the TIP120, + 12vdc to GND pin of the TC622 and - 12vdc to VDD and R-2 on the TC622 circuit?

 

[Hero999]

Isn't the term Vdd meaningless with CMOS, then?

The drains of the N-channels go to +V and gates for the P-channels go to 0V.

 

[dknguyen]

If I interpret the prior message correctly, GND = +12vdc applied to this particular schematic and the positioning of the Darlington Transistor - correct??

+12 vdc to pin 3, Emitter of the TIP120, + 12vdc to GND pin of the TC622 and - 12vdc to VDD and R-2 on the TC622 circuit?

No. Look at Philba's schematic It has all the connections you need. (For the unlabelled MOSFET in the circuit, the drain is the top terminal and the source is the bottom terminal.

Where are you getting the idea to apply +12V to the GND pin and GND to the Vdd pin of the TC622? You destroyed the IC if you did. Where are you getting the idea that to connect all the voltage in reverse to what we have said? Just look at Philba's schematic and you'll see that for some reason you are reversing the polarity across everything.

Vdd on the TC622 goes to +12V. GND on the TC622 is Vss for the TC622 (since it had a Vdd for the other power terminal)- so GND on the TC622 goes to 0V. Why would you connect Vdd to GND and Vss/GND to -12V? THe polarity is corrrect (Vdd is higher than Vss) but you don't even have a -12V in your circuit. You should be saying +12V and 0V instead of 0V and -12V.

Isn't the term Vdd meaningless with CMOS, then?

The drains of the N-channels go to +V and gates for the P-channels go to 0V.

Correct. Vdd and Vss actually refer to the individual transistor, but if they are all the same type (either NMOS or PMOS), then I guess it doesn't really matter and you can use it "circuit-wise". But people do it anyways. When in doubt about a CMOS blackbox (like a PIC), assume Vdd is positive and Vss is negative - due to the fact that NMOS are more popular and it is more likely NMOS convetion was used.

 

[ljh]

I am talking about the Darlington Circuit utilizing the TIP120. The lower schematic.
Is GND common?
Is GND +12vdc ?
Is GND - 12vdc?
Is VDD on both the relay coil and the TC622 -12vdc?

From prior messages, I think I understand that the connections can vary depending on the component and the objective. I am primarily concerned with how to power this particular circuit.

It appears that as far as the "power supply" connections, both circuits would be the same. I just need the basics as indicated by the questions above.

 

[dknguyen]

First things first. Where is -12V coming from? There should only be +12V and 0V.

GND is where all the voltages in your circuit are referenced to, so it is neither +12V or -12V. GND is 0V (for data signal circuits, things like sinusoids that jump above and below 0V between +12V and -12V still have 0V for a reference.) Yes GND is common for this same reason. GND is used where the voltage is unipolar and common is used when the voltages are bipolar.

The Darlington uses BJT transistors, there is no Vdd or Vss. It is Vc and Ve. The Vdd and Vss (actually labelled as GND on the TC622) are because there are MOSFETs inside the IC. The power supply could probably have been better labelled as +V instead of Vdd (since both BJTs and MOSFETs are present in the circuit)

Vdd_<power supply> goes to both Vdd<TC622> and + relay coil.
+V__<power supply> goes to both Vdd<TC622> and + relay coil.

Same thing, just different names for the power supply. It goes to both (it is powering both the relay and TC622 after all).

 

[philba]

stop. pull the plug. let's start over.

Vdd is +12V in this case - I used that because the temp sensor datasheet uses it for +V. Vdd and Vcc are often used interchangably. different manufacturers use them to mean the same thing on datasheets. yes, it can be confusing, as demonstrated here. microchip uses Vss for gnd in some cases, to further complicate things. of course, gnd usually really means 0V. heh heh, clear as mud?

"rail" in this case simply means supply. as in "rail to rail" op amps. replace Vdd with +12V to remove confusion.

 

[ljh]

confusion

I looked up VDD and it was defined as negative voltage connection - I guess that is where the confusion originated. It appears that was a simplistic answer not including the variations of NPN and PNP or the positioning of the component in the circuit.

If someone can tell me where to connect positive and negative in this particular schematic, I should be ok.

The 12 volts is coming from the generator battery -

 

[dknguyen]

In your specific circuit:

*Vdd pin on the TC622 goes to +12V
*GND pin on the TC622 goes to 0V
*Emitter (pin 3?) on the TIP120 goes to 0V

Don't forget to place a small capacitor (0.1uF to 1uF) from +12V to 0V (aka across the rails) very near to the TC622.

There should be no -12V anywhere. Perhaps you were thinking "opposite of +12V" and actually meant GND. The difference between +12V and 0V is 12V which is what you want. Between +12V and -12V is 24V which is not what you want.

 

[ljh]

trying to see through the particles of mud, it might be becoming a little clearer.
It appears transistors have changed nomenclature substantially. Seems like a lead acid battery should have a positive and a zero terminal rather than a negative. If I were using a regulated bench top power supply, I guess I would have the option of +12 and 0V AND -12 and OV.
I guess I should be referring to the connection to the negative side of the battery as "the other connection" and that would eliminate the plus and minus voltage scenario.

BUT a question - if I understand this circuit at all, there is a +12v coming from the high out of the sensor (approximate - a percentage of the input voltage) - this goes through the transistor to the relay coil. Graphically, there should be + 12v between the transistor and the relay coil and then slightly less than +12v at the top of the relay coil (accounting for the coil resistance) reading between that point and GND or OV.

However, the current after R3 (20K) would only be 0.6 ma (12 divided by 20,000 = .0006 amps) - how do we get the 75ma current to operate the relay?

I know the Darlington Transistor is supposed to do this.

I am gong to go through the debugging procedure Philba recommended. I think I needed debugging more than the circuitry. Will now have to ask people "was that a negative comment, was that a zero comment or was that a positive comment" - am I catching on??

 

[philba]

OK, here is the procedure in more graphical form. take the sensor out of the circuit and test the transistor. see the first diagram. Vdd = +12V. use a dmm and measure across the the transistor. with R3 going to gnd, the DMM should read somewhat less than 12V. if it's zero, you've got a wiring problem - make sure there is power to the circuit and the relay is really connected. Then connect R3 to Vdd (+12V). the relay should pull in ("click!") and the dmm should read a bit less than 2V.

You can do the same with the mosfet circuit though I would add a gate resistor just to be safe. 20K should be fine. but, again, debug one at a time.

If that works, then add the tc622 back into the circuit. see the second diagram. measure at point 2 in the diagram and you should see around 10V when the sensor is below the setpoint temperature. note, you should see the opposite on /OUT. so if OUT is 0, /OUT should be about 10V.

Phil

 

[dknguyen]

Yeah, in my mind a +12V and -12V battery is actually a two cell battery with 3 terminals: +12V, 0V, and -12V. Depending on how it was connected you would have either a have 12V or 24V.

Anyways, what happens is this (look at Philba's schematic while reading this).

1. Tiny TC622 current turns on the right transistor in the darlington.
2. The right transistor in the darlington passes a larger current to the left darlington transistor (much the same way a smaller current can switch on and off a larger current in a relay).
3. The left darlington transistor passes and even larger current which flows through the relay coil terminals.
4. THe relay coils magnetize and switch the relay contacts thus passing an even larger current.

It's like using a bunch of amplifiers in a row. You didn't have enough initial signal to drive the final load, so you passed it through an switch that would pass a larger current, but this current was still not enough. So you passed it through yet another switch to pass an even larger current, etc. until you had enough to drive the final load. You can think about it has 3 relays instead of 1 relay and a darlington. A darlington is actually two bipolar transistors. A smaller one and a larger one. The smaller one drives less current than the larger one, but takes less to turn on. But the current that the smaller one can pass is enough tot urn the larger one on. So you connect them so that the smaller one drives the larger one. What you get is something that can drive the same current as the larger BJT, but requires the same current as the smaller one to switch on and off.

The resistor is just to bias the current through the transistors properly to turn them on. FOr NPN transistors, when they are on there is a somewhat constant voltage drop across the base-emitter terminals when the trasistor is fully conducting. Very similar to a diode. So your calculations is incomplete. Your calculation should look like R3 and a diode in series with 12V (like how a resistor is used to limit current in an LED to stop it from burning out). Either way, the 0.6mA doesn't matter too much as long as it is enough to turn on the right darlington transistor. If this happens the transistor will behave as a switch for a larger current for the left darlington transistor. This will pass an even larger current which should be ~75mA when in series with the coils. But now I am repeating myself.

 

[philba]

you posted while I was doing the same. to answer your question, the transistor is an current amplifier. the darlington has a DC gain (Hfe) of 1000. so you .6 mA (actually more like .5) coming out of the tc622 will get multiplied by 1000 or .6A. Of course, the coil resistance limits this to the 75 mA.

the mostfet acts a bit differently - think of it as a variable resistor. super high resistance when 0 is at the gate, super low (.4 ohm) when +10V is at the gate. in both cases, you get enough current to cause the relay to pull in.

 

[ljh]

Thanks - will go try that suggestion now. I was typing when you sent your reply so mine is above yours. Even I should be able to follow the most recent instructions.

 

[philba]

to respond to an earlier point that the 510 is overkill. well, yes but the tc622 produces so little current that a small signal bipolar transistor is marginal at best (you'd need an Hfe of around 150 which may not always be there). the cost of the 510 in quantity 1 is 4X the cost of a 2N7000 but we are still talking well less than a buck here so it's kind of a moot point. the cheapest solution, by the way, is to use 2 small signal transistors in a darlington configuration. I think our designer here needs simple more than cheap.

 

[dknguyen]

Yeah, doesn't matter right now, but when you build thousands or millions, those expenses add up. My prof said he worked on a team designing some microprocessor circuitry or something and their goal was to reduce the cost by 1 cent. They produced about 100 million a year so that's $1 million savings. They worked and worked and worked and managed to get rid of one op-amp...which they thought was pretty damned good.

I just like MOSFETs because I understand them better. I tend to avoid BJTs like the plague in my personal projects where the cost of little things like that is not really a concern.

 

[philba]

the whole area of cost engineering is fascinating. more often than not the real issues are board space and package count rather than the price of a package. smaller board and fewer chips make for cheaper costs. a low cost mosfet maybe a be cheaper solution when you factor in the lower part count.

On one project I did, cost reduction on the HW caused me to have to completely change the software driving it. savings - $.85 or so per unit. cost of the software changes and subsequent QA - about a manyear or $100K. the cost of keep some stupid bean-counter happy - priceless.

 

[philba]

sigh. I got a bit currious about this sensor and found an app note that talks a bit about it. https://ww1.microchip.com/downloads/en/AppNotes/00762b.pdf It's worth a read to understand what's available. one thing that does come out is keeping the output current low so that the chip does not self-heat and cause premature generator shut down. you could further reduce the base resistor.

The appnote has a diagram that I doubt is correct. fig 8 on page 4 shows using a 2N4401 directly connected to a tc622. I really don't think this will work and keep the TC622 within spec.

One other thing to point out that the setpoint resistor formula uses degrees kelvin. so a) make sure you add than in and b) use celcius, not fahrenheit.

 

[ljh]

I played with the trigger set chart and came up with 91K to give me a trigger point of -5 Celsius which is approximately 23 F. Probably playing it safe and not going up to 32 F but because the normal is 0 to -5 F, the 28 degree F range should not take very long with the freezer running to get back down to temperature.
I did some rewiring and am pretty sure I finally got it working with the Darlington circuit. Just one other thing on the schematic that threw me a curve - the direction of the arrow above the relay pointing to VCC - I thought that was an output of VCC or +12. Once I applied +12 to it, voila, it seemed to work.
Thanks for everybody's assistance - I learned a lot and think I have my project close to completion - finally. I guess a week is not too long a learning curve. Now I want to play with the MOSFET circuit as well and see if I can get that to work.
This forum was great - don't know where I would have turned to get the info you guys shared.
It appears I also provoked some meaningful dialogue among the more knowledgeable of you which is positive or is it zero? Don't think it is negative. .
After all of this, does anybody want to see the complete circuit or have you had enough?