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Old 05-18-2017, 07:38 PM   #41
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Default Re: First circuit design needs to be checked please?

That should work too... test, observe...
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Old 05-18-2017, 10:07 PM   #42
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Default Re: First circuit design needs to be checked please?

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Originally Posted by eccerr0r View Post
Bipolar transistors don't care about floating base, all the charge on the base will bleed off very quickly. People must be thinking about FETs which should have them, but the input impedance of BJTs are fairly low in common emitter configurations.
Nope, not about the FET, but from experience and good practice not leaving input floating without termination, it also has to do with Base-Collector leakage.
Also if the transistor has high beta such as Darlington, it can be problem with fault triggering.
https://www.quora.com/What-is-the-ne...-of-transistor
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Old 05-18-2017, 11:09 PM   #43
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Default Re: First circuit design needs to be checked please?

I've never really seen an issue with it on bipolars, and a lot of TTL gates. Anything with FET inputs, however, input termination is necessary due to impedance.

I've never seen any B-C leakage enough to turn a (silicon) BJT on except bad transistors... That would sure make biasing very hard if it leaked that way.

BTW: I do not trust arbitrary web pages including quora. Reading some of those "explanations" they are even answering the wrong problem (a lot of biasing solutions), and even the top response one wasn't even sure by how it's written.
Biasing is the only plausible answer that definitely would require the pulldown but we don't need biasing here in this thread's problem. If that were a Darlington and if noise was an issue, it would help somewhat. But here we don't have a Darlington either.

Perhaps for high voltage transistors it becomes an issue, but 12V is not high voltage... and leaving a nonisolated high voltage input...

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Old 05-19-2017, 12:20 AM   #44
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Default Re: First circuit design needs to be checked please?

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That should work too... test, observe...
OK, I made one minor change. I decided to use a MJE182 transistor mainly because its smaller, and the specs are very similar. It has half the hfe of the 6487, but the on voltage is only different by a tenth of a volt and since you seem to be OK with me essentially saturating the base emitter junction with a lot of voltage (presumably because the 220 ohm resistor is what defines the physics in that junction), I didn't see a big deal going with the smaller transistor. Its rated at 80 volts max Vce, and its got a higher Vbe at 7 volts instead of 5, but the voltage divider that you said was ok, is providing 10 volts across the 2.2k resistor anyways and you didn't seem to think that was an issue so Im assuming the fact that the new transistor can handle more voltage is only a better situation overall...

Anyways ... it works just fine, but I have a question. Of course when the circuit is off, the voltage across CE is 12 volts (it's weird to me, to realize that the coil is has a measurable flow of electrons in it such that I can read a voltage that is being passed through it ... but yet its not technically consuming any energy because the transistor is fully off and not passing current ... that seems counter intuitive to me ... to a point .. I do realize that the coil is nothing more than a whole lot of wire ... and it must be passing a very small current when I have the volt meter across CE because Im reading SOMETHING inside the meter ... but I digress ...

When the circuit is on, Vce drops initially to about 1.08 volts and the longer I leave the circuit on, the lower that voltage gets until it evens out at roughly .7v

So I know that the voltage drops because the solenoid now has the voltage across it being applied to the iron core and now doing work in the form of electromagnetic force ... but why does the voltage drop slowly over time on the CE junction if I leave the circuit on? And we're talking like 20 seconds ... maybe a lil less ... does the solenoid "absorb" more voltage, or does the transistor get warm maybe and repel more current? Or something else??? I'm having a hard time understanding what might be going on there...

I do know that after 20 seconds of leaving it on, the transistor is definitely warm ... but not too hot to touch, which was one of our goals ... so thats a good thing. Under every day use, the circuit should never be on for more than couple of seconds at a time anyways, so we're well within tolerance there.



Also, the current passing through the magnetic reed switch when the circuit is on, is negligible at 3.8 milliamps which is well within spec by quite a bit as I believe its rated at a full 1.5 amps (if I remember correctly) ... so it looks like all of the goals have been accomplished ... minimal heat dissipation by the transistor, minimal current going through the reed switch (the main reason I started this project in the first place), and now zero drain on the battery when the circuit is idle, which was a welcomed surprise.



Mike
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Old 05-19-2017, 12:24 AM   #45
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Default Re: First circuit design needs to be checked please?

There's two reasons to have a resistor R2 from E-B on the transistor.
First is for leakage current ICBO, which you don't want wasting battery power by turning the transistor on any amount at rest.

Second, the wiring to the reed switch makes a lovely antenna, not sure how long the cable is but it will pick up RF, AM radio, AC hum etc - not enough to turn on the solenoid but the transistor will amplify. Do a finger test and only touch the B and you might hear the solenoid making some sound. Adding resistor from E-B or 10nF cap stops all that. It's just good practice that costs less than $0.05
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Old 05-19-2017, 03:33 AM   #46
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Default Re: First circuit design needs to be checked please?

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Second, the wiring to the reed switch makes a lovely antenna, not sure how long the cable is
The reed switch is mounted on the circuit board directly connected to the power source pin and the base resistor.
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Adding resistor from E-B or 10nF cap stops all that. It's just good practice that costs less than $0.05
Well, there is already a resistor on the EB junction ... or are you suggesting that I do something like this?
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Old 05-19-2017, 03:33 AM   #47
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Default Re: First circuit design needs to be checked please?

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Second, the wiring to the reed switch makes a lovely antenna, not sure how long the cable is
The reed switch is mounted on the circuit board directly connected to the power source pin and the base resistor.

Quote:
Originally Posted by redwire View Post
Adding resistor from E-B or 10nF cap stops all that. It's just good practice that costs less than $0.05
Well, there is already a resistor on the EB junction ... or are you suggesting that I do something like this?
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Old 05-19-2017, 06:41 AM   #48
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Default Re: First circuit design needs to be checked please?

Any idea why my last response got echoed? im seeing it twice...
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Old 05-19-2017, 09:14 AM   #49
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Default Re: First circuit design needs to be checked please?

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Originally Posted by EasyGoing1 View Post
OK, I made one minor change. I decided to use a MJE182 transistor mainly because its smaller, and the specs are very similar. It has half the hfe of the 6487, but the on voltage is only different by a tenth of a volt and since you seem to be OK with me essentially saturating the base emitter junction with a lot of voltage (presumably because the 220 ohm resistor is what defines the physics in that junction), I didn't see a big deal going with the smaller transistor. Its rated at 80 volts max Vce, and its got a higher Vbe at 7 volts instead of 5, but the voltage divider that you said was ok, is providing 10 volts across the 2.2k resistor anyways and you didn't seem to think that was an issue so Im assuming the fact that the new transistor can handle more voltage is only a better situation overall...

Anyways ... it works just fine, but I have a question. Of course when the circuit is off, the voltage across CE is 12 volts (it's weird to me, to realize that the coil is has a measurable flow of electrons in it such that I can read a voltage that is being passed through it ... but yet its not technically consuming any energy because the transistor is fully off and not passing current ... that seems counter intuitive to me ... to a point .. I do realize that the coil is nothing more than a whole lot of wire ... and it must be passing a very small current when I have the volt meter across CE because Im reading SOMETHING inside the meter ... but I digress ...

When the circuit is on, Vce drops initially to about 1.08 volts and the longer I leave the circuit on, the lower that voltage gets until it evens out at roughly .7v

So I know that the voltage drops because the solenoid now has the voltage across it being applied to the iron core and now doing work in the form of electromagnetic force ... but why does the voltage drop slowly over time on the CE junction if I leave the circuit on? And we're talking like 20 seconds ... maybe a lil less ... does the solenoid "absorb" more voltage, or does the transistor get warm maybe and repel more current? Or something else??? I'm having a hard time understanding what might be going on there...

I do know that after 20 seconds of leaving it on, the transistor is definitely warm ... but not too hot to touch, which was one of our goals ... so thats a good thing. Under every day use, the circuit should never be on for more than couple of seconds at a time anyways, so we're well within tolerance there.



Also, the current passing through the magnetic reed switch when the circuit is on, is negligible at 3.8 milliamps which is well within spec by quite a bit as I believe its rated at a full 1.5 amps (if I remember correctly) ... so it looks like all of the goals have been accomplished ... minimal heat dissipation by the transistor, minimal current going through the reed switch (the main reason I started this project in the first place), and now zero drain on the battery when the circuit is idle, which was a welcomed surprise.



Mike
1) As the temperature goes up the junction Voltage goes down.
2) "is providing 10 volts across the 2.2k resistor" You have E-B junction in parallel with the 2.2K, so you are not going to have 10V on the 2.2K resistor, it will be = to forward Voltage of the E-B junction, we are dealing with parallel circuit here.
3) "Also, the current passing through the magnetic reed switch when the circuit is on, is negligible at 3.8 milliamps" How did you come up with that 3.8mA? That is barely enough to turn on the Transistor. Put the meter in DCA mode and then hook up the two probes of the meter across the reed switch, that will show you how much current is flowing through the reed switch and what the approximate Bias current will be. Or you can short out the reed switch and take the Vdrops of R1 then you can calculate the current draw.
http://www.onsemi.com/pub/Collateral/MJE171-D.PDF

Last edited by budm; 05-19-2017 at 09:18 AM..
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Old 05-19-2017, 03:32 PM   #50
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Default Re: First circuit design needs to be checked please?

Why are you attached to using the 2.2k base-drive resistor, we're all saying it's 10x too big lol.

If you learn Kirchoff's voltage law KVL and analyze the circuit as two loops, it may be much easier.
Loop 1: battery, R1, R2, E-B of transistor
Loop 2: battery, solenoid, C-E of transistor

Now you are using a smaller transistor so current/heat/temperature have a greater side-effect than the previous big guy 2N6487.
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Old 05-19-2017, 06:30 PM   #51
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Default Re: First circuit design needs to be checked please?

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Why are you attached to using the 2.2k base-drive resistor, we're all saying it's 10x too big lol.
The design I ended up with is a design that BUDM and I worked out together. I completely replaced the two resistors I originally speced in exchange for the two I have now which is the 2.2k base to ground resistor then the 220 ohm base to Vcc resistor ...

Whats wrong with the design as it is now? Here is it again for your convenience.

It works well... the transistor no longer heats up and it no longer drains the battery when its not in use.
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Old 05-19-2017, 07:43 PM   #52
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Default Re: First circuit design needs to be checked please?

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Originally Posted by redwire View Post
Why are you attached to using the 2.2k base-drive resistor, we're all saying it's 10x too big lol.

If you learn Kirchoff's voltage law KVL and analyze the circuit as two loops, it may be much easier.
Loop 1: battery, R1, R2, E-B of transistor
Loop 2: battery, solenoid, C-E of transistor

Now you are using a smaller transistor so current/heat/temperature have a greater side-effect than the previous big guy 2N6487.
If 2k2 is 10x too big then why do you have 10K in the drawing? Or how about not having no R2 at all then how many time too big will that be?
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Old 05-19-2017, 07:50 PM   #53
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Default Re: First circuit design needs to be checked please?

lets complicate things and use a darlington part
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Old 05-19-2017, 07:53 PM   #54
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Default Re: First circuit design needs to be checked please?

Let says right now Vbe is 1V (nice even number), then the Voltage on R1 = 12V - 1V = 11V for VR1, so the current flowing R1 is 11V/220 Ohms = 50mA.
Since VR2 is 1V due to Vbe of 1V) that means the current through R2 is 1V/2200 Ohms = 0.45mA which is really small!
The Base current is then 50mA - 0.45mA = 49.55mA, you can drive it even more to ensure saturation.

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Old 05-19-2017, 07:55 PM   #55
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Default Re: First circuit design needs to be checked please?

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lets complicate things and use a darlington part
Already suggested in post 31.
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Old 05-19-2017, 08:34 PM   #56
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Default Re: First circuit design needs to be checked please?

Base drive also contributes to total power dissipation too, so there's a sweet spot. Granted since more current flows through the collector, any overdrive in the base will reduce the chance of not saturating, and not saturating will hurt more.

Fortunately we don't have to worry about speed so deep saturation is fine as long as you don't fry the base doping by exceeding max base voltage/current.

And no, omitting the 2K2/R2 resistor will not turn on the transistor with broadcast noise because you're not going to get enough current through the base to make significant difference in collector current due to its low gain. And I sure hope modern silicon transistors will not have C-B leakage at such low voltages, else our quick go/no-go multimeter check from C-E will become invalidated.
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Old 05-19-2017, 09:14 PM   #57
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Default Re: First circuit design needs to be checked please?

http://www.onsemi.com/pub/Collateral/MJE171-D.PDF
Per page 2 of the spec you can have Ic of 3A with Base current of 600mA with Vb-e sat of 2V, so we are not even close to the limit of the Transistor.
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Old 05-19-2017, 10:53 PM   #58
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Default Re: First circuit design needs to be checked please?

Right, we just don't want to dissipate 1.2W on the base for no reason.
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Old 05-20-2017, 01:01 AM   #59
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Default Re: First circuit design needs to be checked please?

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Originally Posted by EasyGoing1 View Post
The design I ended up with is a design that BUDM and I worked out together. I completely replaced the two resistors I originally speced in exchange for the two I have now which is the 2.2k base to ground resistor then the 220 ohm base to Vcc resistor ...

Whats wrong with the design as it is now? Here is it again for your convenience.

It works well... the transistor no longer heats up and it no longer drains the battery when its not in use.
That looks good part of the fun in electronics is learning by theory and changing parts is easy.
budm, the E-B resistor being 2.2k I was mixing up with the original 2.2k base-drive resistor. Lost 0.6% of base-drive to a good cause.

hFE is higher at lower collector currents, so a few uA of radio I can see amplified with a scope. You can prove it by putting a loudspeaker in place of the solenoid and touch the base or add wire to it. I have picked up the weirdest interference with these kinds of switch circuits so I add parts to stop that. But most people never notice and OP is ok since the reed switch is close, instead of a long cable.

Transistors partially on due to their own leakage currents, I encounter in builds, usually not much but a hot transistor can have high enough leakage current to keep itself on and make heat and stay on.... yuck.

Changing to a Darlington such as TIP122 (or... make one using BOTH MJE172 and 2N6487...), sure it has much more gain but I think higher losses. VCE(sat) is high at 2V which means 10V for the solenoid, and (2*1.7A) 3.4W of losses so it will heat up fast. I don't see a payoff to using a Darlington.
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Old 05-20-2017, 09:36 AM   #60
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Default Re: First circuit design needs to be checked please?

You guys are funny! lol
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