First circuit design needs to be checked please?

[redwire]

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.

[budm]

Re: First circuit design needs to be checked please?

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.
https://cdn.badcaps-static.com/pdfs/...0959f24a0b.pdf

[EasyGoing1]

Re: First circuit design needs to be checked please?

Any idea why my last response got echoed? im seeing it twice...

[EasyGoing1]

Re: First circuit design needs to be checked please?

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.

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?

[EasyGoing1]

Re: First circuit design needs to be checked please?

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.

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?

[redwire]

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

[EasyGoing1]

Re: First circuit design needs to be checked please?

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

[eccerr0r]

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...

[budm]

Re: First circuit design needs to be checked please?

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

[eccerr0r]

Re: First circuit design needs to be checked please?

That should work too... test, observe...

[EasyGoing1]

Re: First circuit design needs to be checked please?

No, connect one end of the 2.2K to negative, another end to the Base of the Transistor.
What you show is Emitter and GND (negative).

OH, you mean like this ...

[eccerr0r]

Re: First circuit design needs to be checked please?

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.

[budm]

Re: First circuit design needs to be checked please?

So then you recommend doing it like this ... won't that 2.2k reduce our Vbe current?

No, connect one end of the 2.2K to negative, another end to the Base of the Transistor.
What you show is Emitter and GND (negative).

[stj]

Re: First circuit design needs to be checked please?

that wont work, the 2k2 will kill the current to the solenoid

[EasyGoing1]

Re: First circuit design needs to be checked please?

That will give about 50mA of Bias current, try it and see what Voltage you will get between E and C of the transistor to see if it is in saturation mode or not. I would put a 2.2K resistor between GND and Base so the Base circuit will not be floating when the switch is not closed

So then you recommend doing it like this ... won't that 2.2k reduce our Vbe current?

[budm]

Re: First circuit design needs to be checked please?

That will give about 50mA of Bias current, try it and see what Voltage you will get between E and C of the transistor to see if it is in saturation mode or not. I would put a 2.2K resistor between GND and Base so the Base circuit will not be floating when the switch is not closed

[EasyGoing1]

Re: First circuit design needs to be checked please?

You're running as a saturated switch, more base current is needed to saturate! Reduce R1 to around 150R to 220R and no need for R2 (remove it) or Rb (short it out). Don't worry about the base voltage limit, the base voltage won't go past 5V just like an LED won't go past its characteristic voltage as long as the current is limited. You should connect the resistor directly to the base though, just for safety.

Ideally you want to completely saturate the switch and thus the voltage across Vce will be as low as possible, decreasing the dissipation. For your transistor it's a max of 1.3V at 5A - but I suspect it will be typically lower, and even lower for 1-2 amps - so at an amp of collector current it'll still dissipate up to 1W or so, which can get warm but it's fine for an unheatsinked TO220.

Increasing Ib is can be a problem too beyond the transistor's base current limit: what's the maximum current of the reed switch? I suspect it will take 100mA just fine so perhaps this isn't a problem after all.

So is this what you're saying I should be using as the design?

[eccerr0r]

Re: First circuit design needs to be checked please?

You're running as a saturated switch, more base current is needed to saturate! Reduce R1 to around 150R to 220R and no need for R2 (remove it) or Rb (short it out). Don't worry about the base voltage limit, the base voltage won't go past 5V just like an LED won't go past its characteristic voltage as long as the current is limited. You should connect the resistor directly to the base though, just for safety.

Ideally you want to completely saturate the switch and thus the voltage across Vce will be as low as possible, decreasing the dissipation. For your transistor it's a max of 1.3V at 5A - but I suspect it will be typically lower, and even lower for 1-2 amps - so at an amp of collector current it'll still dissipate up to 1W or so, which can get warm but it's fine for an unheatsinked TO220.

Increasing Ib is can be a problem too beyond the transistor's base current limit: what's the maximum current of the reed switch? I suspect it will take 100mA just fine so perhaps this isn't a problem after all.

[budm]

Re: First circuit design needs to be checked please?

By the way, the solenoid is rated at 12 volts and 1 amp (thats what the sticker says anyways)

You can easily verify that by using DCA to measure the current draw of the solenoid when 12VDC is applied to the solenoid through the DC AMP meter.

[budm]

Re: First circuit design needs to be checked please?

YES, its working fine, but the transistor does heat up pretty quickly if I leave the solenoid engaged for more than a few seconds.

That is happening because the transistor is not fully turned on as explained. R1 needs to be lower value to provide bias to put that transistor in saturation mode, or you can use Darlington Transistor which has a lot more hfe.