Hey guys, I'm working on a bachelor project and we need to use transistors (NPN switch) in our prototype, unfortunately none of us have had any past experience and we've been struggeling to get enough current out of it.

From what I read again and again, I should multiply current flowing to base by the gain factor (Beta) of the transistor. It's gain is 1000 Beta or Hfe as it says in the datasheet. I'm looking to get 250 mA. So I put on 25K resistance to base... 5v/25 = 0.25 mA, 0.25 mA * 1000 = 250 mA. From what I understand I don't need resistance on the collector (unless for safety?), it should output 250 mA anyway? And if I've got 12V or 5V through collector - emitter it should still be 250 mA (doesn't make sence, but that's my understanding from everything I've read). Can anyone help me figure out what I'm missing?

If you could point me in the right direction of figuring out if a transistor is compatible from the datasheet that would be great (tips? reading material?), I've read for countless of hours but I can't seem to grasp how to make it work.

The transistor I'm using is a TIP120

 

hi G3,
Welcome to AAC
Do you have a diagram to post.?
E

 

See


You need to take into account the base-emitter voltage (~ 1.4V). With an amplification of 1000 you will not have enough current base to get 250 mA.

 

Battery is 12V not 9V

I measured the transistors hfe and it was 850. So I should get about 280mA (850 * (5v/15)) from the collector right? How do you calculate or take into account the base-emitter voltage? How does it relate to IC = Ibase * Beta?

 

Your Base supply is from the Arduino which is 5V, the Base/ Emitter voltage for a Darlington is 1.4V, so your resistor is 5v- 1.4v= 3.6V @25K = 144uA.
With an hfe of 1000 gives 144mA Collector current .
If you want 250mA collector current you need a 14K Base resistor .

 

hi G3,
Consider that the TIP is a Darlington, is the Base current really equal to, 5v/15k.?
May I ask the purpose of the experiment.?
E

 

hi G3,
Consider that the TIP is a Darlington, is the Base current really equal to, 5v/15k.?
May I ask the purpose of the experiment.?
E

So the base current is really (5v - 1.4v) * Beta for the TIP120? I didn't know the transistor lowered the voltage. Is it fixed to 1.4V for the darlington or is the value parallel to IC or some other value? What symbol do I look for in the datasheet to find it (or calculate)?

We use a timer which activates the microcontroller every 2 hours and then checks if it has gone 6 hours and if it's ready to send sensory information over SMS to our backend. If it's ready to take measurments it activates the rest of the components, where some of them needs 12V or a higher amperage than the arduino can supply.

 

The datasheet is here: https://www.onsemi.com/pub/Collateral/TIP120-D.PDF

If you look at figure 10, for a 2 A (picked) you need between .8 and 8 mA for the base. Naturally, you don;t pick the extremes. You don't have to design exactly for your load. I your load is 250 mA, you can still design for a max of say 1 amp. You want your transistor is saturation or a very low Vce. This makes the transistor dissipate the lowest power. Vce is subtracted to get the load voltage. There is a temperature dependence here too.

Base current (Ibe) in a darlington has to go through 2 diode drops. Silicon (P-N junction) has a diode drop of about 0.6 to 0.7 V and varies with temperature. A Shockley diode (metal-semiconductor) has a voltage drop of around 0.2 V. You need to take these into account when computing Ib.

Also remember that Hfe varies all over the place. Some transistors give you a different value when the transistor is saturated.

 

One thing to be aware of, and I think people starting out don't realize this, is that BJT transistors are very (ie. *VERY*) sensitive on the base. The reason the gain is so significant is because the base will respond to exceedingly small levels of current. Far smaller than most starters think, so they tend to saturate the base immediately.

 

Do you need exactly 250mA? You usually don't pick the value to be exact if you just want do drive something. It says that the min gain is 1000, and there are to Vbe drops in a darlington, so that would give you (5v-1.4v)/250mA*1000=14.1k. This would give a current of 250uA, which is a current the arduino has no trouble of supplying. So we take the nearest common resistor value, which would be 12k(we don't want to go higher than 14.1k). That will always do the job for a 250mA load, but you can set a lower resistance, say 10k(because it's easy to find and it doesn't load the arduino output a lot) or even 3.3k which would give you a base current of 1mA and a drive capability of 1A, just in case the gain dips or your load needs more.
But there is a problem with darlingtons, the reason why we have a higher Ib*Bf than the Ic is so that the transistor stays in saturation mode, where the voltage between collector and emitter is lowest(~0.1v) and so the dissipation is lowest. Because of the construction of a darlington this voltage is significantly higher, at least 1v which isn't that big of a problem for dissipation for 250mA(250mW which is no problem for to220) but may cause a problem for your load, you will get 11v on your load instead of 12v(Could be even less, the datasheet specifies Vces up to 3v, but this is for higher currents). You can make a darlington which does not have this problem with a discrete darlington without the collectors connected together.

 

An N-channel logic level MOSFET might be a better choice, but this might be classified under homework.
i.e. You need to use a TIP120.