NPN Transistor as a switch: what's the use of the emitter resistor? - Page 1
Forum section: electronics / beginners

author: Moriambar
author_rank: Unknown
timestamp: March 19, 2019, 08:27:46 am
content: Hi,as I stated elsewhere I'm a noob with BJT transistor, since I almost always used FETs (which I find way easier).Regardless, I've seen some design of npn transistors used as switches and they almost always feature a 100R resistor from the emitter to ground.What's the purpose of it?Thanks

author: bd139
author_rank: Super Contributor
timestamp: March 19, 2019, 09:05:46 am
content: That's not a plain switch then. It's an emitter follower.

author: Moriambar
author_rank: Unknown
timestamp: March 19, 2019, 09:20:14 am
content: Quote from: bd139 on March 19, 2019, 09:05:46 amI thought that in the emitter-follower configuration the load should be connected to the emitter. Mine is connected to the collector.The emitter has just 100R to ground.I provided an example to clarify what I'm saying

author: bd139
author_rank: Super Contributor
timestamp: March 19, 2019, 09:24:47 am
content: The load should be connected between emitter and ground with the emitter follower. This is used for linear devices and current sources rather than switching though. It may switch but it'll be incredibly inefficient.To switch with a transistor, the emitter should be grounded, the load connected across the positive rail and the collector. The base current should be appropriate to switch the transistor on fully ((Ic / beta) * 2 approx to jam the transistor on hard)

author: Moriambar
author_rank: Unknown
timestamp: March 19, 2019, 09:26:51 am
content: Quote from: bd139 on March 19, 2019, 09:24:47 amOk so either I found wrong examples (I've not them at hand) or I did not understand the purpose of the circuit. 100% they were not emitter followers, though.

author: bd139
author_rank: Super Contributor
timestamp: March 19, 2019, 09:30:49 am
content: Yep. Could be either.Sometimes it's impossible to work out if the design is rubbish, the designer was drunk or the circuit does what is intended but the reason isn't obvious.

author: Moriambar
author_rank: Unknown
timestamp: March 19, 2019, 09:35:51 am
content: Quote from: bd139 on March 19, 2019, 09:30:49 amThanks. Well at least I have reasons to doubt the design, then.

author: leeatljs
author_rank: Contributor
timestamp: March 19, 2019, 09:40:53 am
content: Sometimes you put a resistor in the emitter to provide a constant current or limit.

author: Peabody
author_rank: Super Contributor
timestamp: March 19, 2019, 01:46:22 pm
content: Yes, I think it's a current limiting circuit. Current will flow through the transistor and the voltage across the emitter resistor will rise. But at some point that voltage will be too high for the base current to continue to flow, so the collector current will drop. When it settles out, you should get a fixed current through the load determined by the voltage driving the base. Something like that. Anyway, I think the emitter resistor is there for a reason.

author: Benta
author_rank: Super Contributor
timestamp: March 19, 2019, 05:15:14 pm
content: Yep, constant current (if the input voltage is known), this works perfectly for LEDs. Otherwise just a simple overcurrent protection.

author: soldar
author_rank: Super Contributor
timestamp: March 19, 2019, 05:25:43 pm
content: Not an emitter follower. The resistor limits the current in the device and prevents the bipolar transistor from going into saturation. Switching out of saturation takes longer time so fast counters and other switches will often prevent going into saturation by using such a resistor. Today we just use FETs, which don't have this problem.

author: bionic
author_rank: Newbie
timestamp: March 19, 2019, 07:51:25 pm
content: The emitter resistor will decide what Beta or amplification for that specific transistor will be!Remember that:IC/IE = AlfaWhere Alpha gives:Beta=(alpha/alpha-1)=IC/IB

author: Benta
author_rank: Super Contributor
timestamp: March 19, 2019, 07:56:47 pm
content: bionic, I don't think you've quite understood this circuit.But welcome to the Forum.

author: gf
author_rank: Super Contributor
timestamp: March 19, 2019, 08:36:50 pm
content: Per se - as drawn - it just looks like an amp with emitter feedback...But where/how is the load supposed to be connected?Or if really anything is supposed to be switched with the transistor, where/how is it connected to the given circuit?[ If nothing else is connected to the given circuit, then I'd consider it rather useless. ]EDIT: Or is RL itself supposed to the load? If yes, what kind of load is it?

author: Benta
author_rank: Super Contributor
timestamp: March 19, 2019, 09:00:03 pm
content: Quote from: gf on March 19, 2019, 08:36:50 pmRL is the load, yes. And it could be anything. Another input. A pullup resistor. An LED.The collector of the transistor is a current source.Let's say you have a 3.3 V logic output driving the transistor base. 3.3 - 0.7 = 2.4 V across the emitter resistor, equals 24 mA. Wonderful for driving an LED as load.Or another input through a cable that might be shorted to plus through a connector short cut: no damage to the transistor. And, and, and...Forget the emitter follower picture you have in your mind. It's a switch.

author: gf
author_rank: Super Contributor
timestamp: March 19, 2019, 11:13:12 pm
content: I did not call it emitter follower. I said "amp with emitter feedback" (which is eventually a voltage controlled current source, driving RL). With proper biasing at the input, the circuit could well act as amplifier with a closed-loop voltage gain of approx. RL*beta/(R1+R2*beta). But agreed, if the input voltage is low enough, then the current through RL can be turned off completely - then it becomes a switch. Contrary to the "traditional notion of a switch" (-> low impedance in the "on" position -> low power dissipation at the switch), the transistor and R2 do well dissipate power, when the "switch" (i.e the current source) is turned on.

author: Jwillis
author_rank: Super Contributor
timestamp: March 20, 2019, 08:37:26 am
content: Emitter resistor is a way of biasing the transistor to prevent thermal run away.Some times there will be a base resistors with paralleled transistors to prevent load hogging because not all transistors of like kind will turn on at exactly the same voltage.Look up transistor biasing for further explanation because the way a transistor is biased depends on application.

author: Audioguru
author_rank: Super Contributor
timestamp: March 20, 2019, 04:28:03 pm
content: The transistor circuit shown is called a "phase splitter". The phase at the emitter output is the same phase as the base input but the phase at the collector output is -180 degrees.

author: Benta
author_rank: Super Contributor
timestamp: March 20, 2019, 05:00:57 pm
content: Why do you guys keep going on about this? The headline clearly states is it to be used as a SWITCH.As an amplifier stage, it's far too simple and has no biasing provisions.

author: atmfjstc
author_rank: Regular Contributor
timestamp: March 20, 2019, 05:28:31 pm
content: I think @soldar has it right. It is indeed a switch configuration, with Re there to ensure maximum switching speed. I remember reading about this trick in school. There is some detail here:http://www.engineering-bachelors-degree.com/electronic-components/uncategorized/bipolar-junction-transistor-circuitshigh-speed-bjt-switching/

author: magic
author_rank: Super Contributor
timestamp: March 20, 2019, 06:20:52 pm
content: This isn't guaranteed to prevent saturation, Baker clamp should be used for that.

author: Benta
author_rank: Super Contributor
timestamp: March 20, 2019, 06:29:00 pm
content: Quote from: magic on March 20, 2019, 06:20:52 pmCorrect. But it depends on the load, if it has too high resistance, saturation may indeed occur.

author: David Hess
author_rank: Super Contributor
timestamp: March 20, 2019, 07:05:13 pm
content: As shown in Moriambar's example, the emitter resistance limits the current to (Vin-Vbe)/R2 where Vbe is about 0.6 volts.Base series resistor R1 has only a minor effect because the transistors current gain (hfe) should be much greater than the ratio between base series resistor R1 and emitter series resistor R2. It however should be included because if the load is missing, then the drive signal will see an effective short to ground through the emitter series resistor.Nothing about emitter series resistor R2 prevents the transistor from going into saturation if the collector voltage falls and again, this is why R1 is included to protect the driver.

author: Moriambar
author_rank: Unknown
timestamp: March 20, 2019, 07:18:29 pm
content: Quote from: Audioguru on March 20, 2019, 04:28:03 pmThe output (shown as RL) is at the collector, not at the emitter

author: bson
author_rank: Unknown
timestamp: March 21, 2019, 12:08:42 am
content: The emitter resistor makes it a current switch rather than a voltage switch. Some loads, particularly semiconductors but also solenoids, are current operated and need a particular current rather than voltage. With an emitter resistor you don't want a base resistor; base current is low as the resistance looking into the base is RE*hfe = high for high hfe, and a base resistor just makes life more complicated. You don't use a collector resistor with a current switch.

author: KE5FX
author_rank: Super Contributor
timestamp: March 21, 2019, 01:34:37 am
content: Quote from: bson on March 21, 2019, 12:08:42 amWin Hill actually just posted anifty example of this practiceas an appetizer for the 'X-chapters' volume for AoE 3. Some grad students needed to dump 200 amps into an LED in a hurry, and they came to him for advice when their first attempt failed to achieve the leading-edge speed they needed. It made for an interesting case study.Arguably, he's not trying to avoid saturation on the transistor's part, so much as he's trying to ensure that the LED sees the same current across the entire pulse as it heats up over time. Similar basic motivation, though.

author: Zero999
author_rank: Super Contributor
timestamp: March 21, 2019, 11:10:56 am
content: As far as using this circuit to switch a load, rather than as a linear amplifier, the only thing I can think of is an NPN transistor controlling the current to a PNP transistor's base, or the voltage across a MOSFET's gate, which in turn switches the load.R1 holds Tr1 off, preventing any leakage through Tr2 from turning it on. It also helps to speed up the off time of TR1 by discharging its base, when Tr2 is turned off. It be a fairly low value, say 1k.VR2= VINPUT- VBEIR2= VR2/R2Assuming Hfe for Tr2 is big and hardly any current flows through R1Tr1_IB= IR2Example:Suppose we need to switch a 200mA load and Tr1 has a low saturation voltage when IB=1/20IC. The minimum input voltage is 4.5VIC= 0.2AIB= 0.2/20 = 0.01ACalculate R2IR2= IB= 0.01AVBE= 0.7VVR2= VINPUT- VBE= 4.5-0.7 = 3.8VR2 = 3.8/0.01 = 380R. Use 330R.

author: Peabody
author_rank: Super Contributor
timestamp: March 21, 2019, 02:04:53 pm
content: Another application using the emitter resistor would be to have the base driven by the PWM ouput of an MCU GPIO port pin, with appropriate filtering. The duty cycle of the PWM (the average voltage applied to the base) would control the current through the load - i.e., the brightness of an LED.

author: bsfeechannel
author_rank: Super Contributor
timestamp: March 21, 2019, 03:10:55 pm
content: Quote from: Jwillis on March 20, 2019, 08:37:26 amI'll bet on this horse.

author: jlmoon
author_rank: Unknown
timestamp: March 21, 2019, 09:16:28 pm
content: The original circuit looks to be a common emitter type amplifier minus the biasing scheme on the base connection. Not much of a switch unless you take out the emitter resistor, tying the emitter to ground and put the load in series (ie: small relay or such) with proper current limiting in the collector circuit.

author: soldar
author_rank: Super Contributor
timestamp: March 21, 2019, 09:50:48 pm
content: Related:https://www.eevblog.com/forum/beginners/understanding-transistors/msg1211771/#msg1211771

author: bson
author_rank: Unknown
timestamp: March 23, 2019, 03:53:26 am
content: Quote from: KE5FX on March 21, 2019, 01:34:37 am200A pulses into a LED... That's mad science territory!Can't wait to get a copy!

author: Zero999
author_rank: Super Contributor
timestamp: March 24, 2019, 06:32:02 pm
content: Quote from: bson on March 23, 2019, 03:53:26 amI've been there and done that. I wouldn't recommend a BJT. Use a MOSFET instead. If the duty cycle is low, then the MOSFET can be driven, via a small transformer. I posted some of the schematics for someone who asked about making a high powered LED flash.https://www.eevblog.com/forum/projects/ultra-short-ultra-fast-led-flash/