TIP120 Question

[EasyGoing1]

I was messing around with a TIP-120 NPN transistor, and I wired it up just like this



But at NodeA, with respect to ground, I'm only seeing around 4.3 to 4.5 volts, and I'm wondering why? Shouldn't I be seeing a lot more voltage than that?


And when I have it connected like this, I'm seeing 12 Volts at NodeB with respect to ground where I would expect to see none, but I do see zero volts across the resistor until I turn it on, where I then only see 10 volts. And the weird thing about this second config, is that when I measure the current going through the resistor, I'm only seeing around 4ma, which is about what I would expect, but in spite of that, the transistor heats up right now to where within a few seconds its far too hot to touch - that is when I apply 5 volts to the base.



So when I have it like this, there is only 4ma going through the resistor, but the transistor gets extremely hot within seconds.



OK, so in that configuration, I'm seeing 2.7 amps going through the base node ... I think it needs a resistor.

[EasyGoing1]

Re: TIP120 Question

OK, I think I figured it out ... when I use this configuration, I get almost 11.6 volts across the 2k resistor and about 5ma through it, where the base has no measurable current going through it... I think I was oversaturating the base-emitter junction?

[sam_sam_sam]

Re: TIP120 Question

OK, I think I figured it out ... when I use this configuration, I get almost 11.6 volts across the 2k resistor and about 5ma through it, where the base has no measurable current going through it... I think I was oversaturating the base-emitter junction?

In this circuit diagram is the typical way that you would do this type of setup

[eccerr0r]

Re: TIP120 Question

Case 1 with netlist
.model Q1 TIP120
.model R1 R(2K)
Q1 E=NodeA B=+5 C=+12
R1 N1=NodeA N2=GND

I'm surprised NodeA even gets to more than 4V, I'd expect around 3.8 volts or so. You are sort of turning it on so I guess that's why you're getting more than 4V but it's not really turned on, and definitely not saturating like a lot of emitter follower/source follower circuits.

For case 2 with netlist
.model Q1 TIP120
.model R1 R(2K)
Q1 E=GND B=GND C=NodeB
R1 N1=NodeB N2=+12

This should keep Q1 in cutoff. As Q1 is completely shut off, NodeB should rise to the +12 rail as no current flows through Q1 and thus no drop across R1, and therefore 12V-0V=12V

In case 3 with netlist
.model Q1 TIP120
.model R1 R(2K)
Q1 E=GND B=+5 C=NodeB
R1 N1=NodeB N2=+12

You are forcing 5V into the E-B junction of the transistor which may damage it. As you have two forward biased junctions here, its characteristic voltage is 1.4V and you're forcing 5V into 1.4V with no current limiting, and thus it will draw a lot of current, P=V*I where V=1.4 and I=huge (say 10A) and you'll easily get 14W of power dissipated here up until you destroy the transistor.

And as stated above post2 (case 4) is the proper connectivity for bipolar transistors in common emitter configuration. Base current would be (5-1.4)/1k6(R2) = a bit over 2mA.

Due to the gain of the TIP120 you can reduce the resistance of R1 if you need more current. Typically TIP120s have gains of 1000 or so, you can easily control a 2A circuit with 2mA of base current. However you will end up with around 1V of drop across C-E of Q1 at higher currents.

[EasyGoing1]

Re: TIP120 Question

In this circuit diagram is the typical way that you would do this type of setup

You are forcing 5V into the E-B junction of the transistor which may damage it. As you have two forward biased junctions here, its characteristic voltage is 1.4V and you're forcing 5V into 1.4V with no current limiting, and thus it will draw a lot of current, P=V*I where V=1.4 and I=huge (say 10A) and you'll easily get 14W of power dissipated here up until you destroy the transistor.

This makes sense. Something I realized after working through this little experiment, was that I've been in digital switching mode - mentally - for quite a long time now (years) and so I completely dozed over the fact that was actually setting up an analog circuit where I am driving a load with a transistor... In my mind, initially, I just figured since I was connecting the base to ground, I was essentially applying a ZERO to the base and thus turning it off... kinda silly now when I think about it ... but that's where my head was at anyway when I initially tested this.

I'm actually glad I worked through it via experimentation because it "clicked" quite well.

disclaimer: No transistors were harmed in the process of this exercise :-)

And as stated above post2 (case 4) is the proper connectivity for bipolar transistors in common emitter configuration. Base current would be (5-1.4)/1k6(R2) = a bit over 2mA.

I wasn't able to measure the current in the base because the damn mA / µA fuse in my fluke was blown... I ordered a set on Amazon and got them today so I'm going to re-check it. Those damn fuses are EXPENSIVE!

Edit: I measured 2.04 mA in the base, just as you predicted.

Due to the gain of the TIP120 you can reduce the resistance of R1 if you need more current. Typically TIP120s have gains of 1000 or so, you can easily control a 2A circuit with 2mA of base current. However you will end up with around 1V of drop across C-E of Q1 at higher currents.

I'm actually going to be using this circuit to drive one of these, which, according to the web page, needs 430 mA to actuate. And, I am aware of the need for the flyback diode across the solenoid to protect the transistor.

Edit: After thinking about using a current limiting resistor in series with the solenoid, I calculate that I would need a 30Ω resistor to get .4mA ... but since VCC is 12 Volts, I would need that sucker to be a 5 Watt resistor and I don't happen to have any 30Ω 5 watt resistors laying around ... so perhaps just skip the resistor entirely and just do something like this? (The solenoid will only be on for short durations of time ... like three seconds or less.




And I have a question: When I hook that circuit up on a breadboard, everything works as you'd expect ... and I get around 11.3 volts across the solenoid when the transistor is on ... but what I don't understand, is that when the transistor is off, why do I read 12 Volts from the collector with respect to ground (I have it connected WITHOUT the flyback diode at the moment if that matters)?

[eccerr0r]

Re: TIP120 Question

The solenoid supposedly rated to be run at 12V and it will draw 430mA at 12V (apparently 1 minute max duration) so you don't need a resistor as you're only running at 11.3V. Theoretically it will be less than 430mA since the voltage is lower.

This is about right that you measure 11.3V across the solenoid when the transistor is on... and 12V across the transistor when the transistor is off.

[EasyGoing1]

Re: TIP120 Question

The solenoid supposedly rated to be run at 12V and it will draw 430mA at 12V (apparently 1 minute max duration) so you don't need a resistor as you're only running at 11.3V. Theoretically it will be less than 430mA since the voltage is lower.

This is about right that you measure 11.3V across the solenoid when the transistor is on... and 12V across the transistor when the transistor is off.

Actually, I get 10.6 across the solenoid for some reason ... not sure where the 2+ volts are going ... I assume in the transistor somewhere...

OK, so you say that it's right that I get 12 volts AT THE COLLECTOR with respect to ground, but my question is, WHY? This would imply to me that there is a current flowing through the solenoid ... but I know there isn't, so how is there 12 volts at the collector when the transistor is off?

Also, I corrected the schematic... I had the diode backward, as the solenoid would not actuate with the cathode facing the collector... but that begs the question ... isn't the flyback diode supposed to protect the transistor by preventing any backflow current from reaching the collector? In this setup, that doesn't seem like the situation ... seems to me that I should have the cathode facing the collector.



The only way I can make it work with the diode in the correct orientation is to do it like this ...



Though in this config. it would seem to me that one of the functions of the flyback is not being met where the coil would be discharging through the dioed.

[eccerr0r]

Re: TIP120 Question

yes you should get a voltage drop across the transistor when it's on, this is why people don't use darlingtons as switches anymore - they're slow and they burn more power than mosfets.

It's 12V at the collector because a solenoid is a piece of wire at DC... Since no current is flowing, no voltage drop across the solenoid, so 12V.

The snubber diode should be in parallel with the solenoid, inverse biased during normal operation (cathode to V+). If you had cathode to the transistor it sure won't work, would bypass all energy that is needed to power it. Having the diode in series is pretty much a waste diode, might well omit it.

[stj]

Re: TIP120 Question

why is D1 in series with the coil instead of in parallel with it?
your causing volt-drop

[lti]

Re: TIP120 Question

isn't the flyback diode supposed to protect the transistor by preventing any backflow current from reaching the collector? In this setup, that doesn't seem like the situation ... seems to me that I should have the cathode facing the collector.

That current flows in the opposite direction.

[EasyGoing1]

Re: TIP120 Question

why is D1 in series with the coil instead of in parallel with it?
your causing volt-drop

Because the solenoid will not actuate when it's in parallel with it. I'm not going to use the flyback diode anyways ... I'm building this on a pegboard with very limited real estate in the first place and I could GAF if the transistor somehow fries in that one in a million chance when a silly 12-volt solenoid that gets turned on for a lousy 2 seconds a few times a day manages to kill it ... I say if it does.... then it deserved to die ... Jimmy cracked corn ... and I don't care! lol

[eccerr0r]

Re: TIP120 Question

The diode should not affect operation unless it's hooked up wrong or is defective. As usual, it's the disengage that kills. 1 on a million is too optimistic, frying could happen much more frequent than that. Also frying could end up causing the transistor to stick on, leaving the door unlocked...

[EasyGoing1]

Re: TIP120 Question

I'll tell you what really upsets me throughout this project is the fact that I spent $30 on 4 stinking fuses for my Fluke ... WTF is going on that they can charge so much for some lousy fuses?

[EasyGoing1]

Re: TIP120 Question

The diode should not affect operation unless it's hooked up wrong or is defective. As usual, it's the disengage that kills. 1 on a million is too optimistic, frying could happen much more frequent than that. Also frying could end up causing the transistor to stick on, leaving the door unlocked...

I literally tested at LEAST 4 different brand new diodes in both directions and I could not get the solenoid to actuate when the diode was in parallel with it.

And it's ok if the latch gets stuck "on" ... it's not even for a door ... it's for a small little lock-box. I'll upload photos once its done printing.

Also, the TIP-120 is rated for 8 AMPS! peak current ... if a 12 volt .4 amp solenoid could generate more than 8 amps long enough to kill this transistor ... I'll send each of you a $5 bill.

[stj]

Re: TIP120 Question

you sucker, FF-rated fuses cost a couple of $
you dont need the fat 1kv fuckers when the 600v ones will do just fine - your not even going to put more than 400v in the meter anyway.

[stj]

Re: TIP120 Question

[QUOTE=EasyGoing1;1127934if a 12 volt .4 amp solenoid could generate more than 8 amps long enough to kill this transistor ... I'll send each of you a $5 bill.[/QUOTE]

nice - thanks.
the coil is inductive, the inrush could be 10 times the rated current until the field saturates.

[eccerr0r]

Re: TIP120 Question

test the diodes manually, something is wrong. Maybe you destroyed them when forward biasing them. And eh, not the inrush that kills - again, it's the release...

[redwire]

Re: TIP120 Question

TIP120 does lose about 0.75V so at 0.4A it will dissipate 0.3W and warm up a bit, solenoid will not get 12V but maybe 11V no big deal.

Putting a reverse-diode across an inductor/solenoid/relay coil does clamp back-EMF but it also greatly slows down the magnetic field collapsing.
It might be why the solenoid has poor return performance, depending on the spring too.
For a snappy (fast) return, you use put a HV zener+series diode, up to about 45V instead, across the solenoid coil. Enough to protect the TIP120 (60V rated) minus the 12V power.

If there is a chance the solenoid can get stuck powered on, a PTC fuse is a good idea, so the solenoid at 5W does not burn up.

[EasyGoing1]

Re: TIP120 Question

test the diodes manually, something is wrong. Maybe you destroyed them when forward biasing them. And eh, not the inrush that kills - again, it's the release...

OK, I was remembering incorrectly ...

THIS - DOES NOT WORK




But THIS DOES!

[eccerr0r]

Re: TIP120 Question

Yep, cathode to V+ to reverse bias it to be correct. Connecting cathode to the transistor, the diode will just heat up and the inductor won't work. Indeed you said in post #7 that you drew it wrong. Bad me for not scolding you on post #5

BTW need to be CIVIL folks, voltage leads current for inductors... Capacitors do inrush, and stalled motors do inrush (this is a mechanical "capacitance" problem).