# emitter follower conofiguration

I need help with the following configuration , using LIRC for as the source signal for infrared led transmitter circuit with raspberry pi .

when putting the infrared led emitter led above the collector , I couldn't get enough range for led ( wasn't bright enough using a digital camera ) even with higher voltage and current ( 5 volt )

so I've chosen the following configuration to get current gain without voltage gain , the led was bright and got range about 6 meters and even can handle 2 IR leds in series .

my question :
1- I know the emitter follower configuration has a current gain and no voltage gain ( the IR led can handle up to 1.5 A for a short time of pulses ) , so how can i calculate current flowing from the emitter to the IR led .

2- if the emitter follower has no voltage gain , it should have voltage drop about .7 , so vout should 3.3-.7 = 2.6 volt . I have used a digital mulitmeter between the emitter and ground and shout that voltage is about .2 volt . so , is it a wrong reading ? how could the led be bright if the IR led typical forward voltage is 1.3 volt ?

how can I calculate current of Ie ?

3- how can I simulate IR pulses using ltspice (generting squrewave from independent signal source tool)

attachments : example for IR pulses http://bit.ly/1N4GHRv

• The current gain to a load in the collector or emitter is largely the same so maybe you have done something wrong? Nov 15 '15 at 11:17
• but how can I calculate current flowing from the emitter (Ie) anyway in this circuit ? Nov 15 '15 at 11:21
• "a digital mulitmeter between the emitter and ground and shout that voltage is about .2 volt" definitely doing something different than described. Nov 15 '15 at 11:22
• My assumption on your problem is that you have not included a base resistor, or a diode-resistor in your drawing, meaning many different bad things can be happening. Probably the IR diode in the emitter path limits the IO-pin drop caused by the "unlimited" base-current allowing the processor to keep the output in a nearly-defined state, whereas with the emitter to ground it will drop to 0.7-ish volts, causing who-knows-what state. Use resistors where they belong. Diodes and transistors have knees in their curve and need resistors to not die (or cause other things to die) Nov 15 '15 at 11:28
• The fact the LED's resistor is still missing, made me assume you did also not use one in the set-up before this one. Where it is very desperately needed. As still is the diode limiting resistor. Which will then let you estimate the current through the diode. You can do that now, but you need to look up the exponential curves of all parts included, rather than read off a point on a graph. And yet again, it will become a pissing match between the transistor and diode, making the MCU very much see a current drain that may go beyond its capabilities, because the diode has no resistor Nov 15 '15 at 13:17

I think you are seriously abusing the TSAL6100; it's not meant to take those 1.5A surges on a repeated basis, but only 200mA pulses. If you keep putting 1.5A through it with your remote application, you will probably damage it soon enough. And then no wonder it has no range.

I think you need to read Visay's guide their diode (=LED) datasheets.

Peak surge forward current, IFSM: The maximum permissible surge current in a forward direction having a specified waveform with a short specified time interval (i.e., 10 ms) unless otherwise specified. It is not an operating value. During frequent repetitions, there is a possibility of change in the device’s characteristic.

As badly written by [non-native speakers of English] interns as that doc might be, the message is clear I think.

I don't know exactly how Vishay determines IFSM for their stuff, but a different manufacturer does this (for rectifier diodes):

IFSM, Non-Repetitive Forward Surge Current

IFSM is the maximum allowable nonrepetitive half-sine wave surge current under the following conditions: TJ = 45°C and the base-width of the half-sine wave surge pulse is 8.3 ms. A sample of diodes is selected and one-by-one the diodes are tested to destruction. This is done by hitting the DUT with a single surge pulse and checking to see whether the diode was destroyed. If so, the peak value of the surge is recorded as that diode’s pulse-height capability, and the next diode is tested. If not, the junction temperature is allowed to return to 45°C, the peak value of the surge is increased, and the DUT is hit again. This process is repeated until all of the diodes in the sample have been destroyed. Then the pulse-height capabilities are averaged and IFSM is set equal to half of the average.

• thanks for these valuable info. , and sorry for my terrible English . Nov 16 '15 at 9:20

The simplest solution is to put the IR transmitter in the colelctor and have a small valued resistor in the emitter. Thus, if you drive the base with 3.3 volts, the emitter voltage will be at about 2.6 volts and this will appear across the emitter resistor. If the emitter resistor is 26 ohms then the current that flows is 100mA - this current (maybe 99%) flows through the collector and IR transmitter. There are more accurate and higher power versions of this but get the basic circuit working first.

• andy aka , I did that , it worked , but with short range (less than 50 cm away from the infrared reciever even with using lenses , mirrors , and filters ) . i decided to use the emitter follower configuration to get current gain (the IR led can handle up to 1.5A with short pulses ) it worked with more range 10 meter , but how can i calculate the current flowing from the emitter to calculate gain too ? Nov 15 '15 at 11:38
• The forward voltage to get 1A flowing can be as high as 3V for that device so I would consider increasing the 3V3 feeding the device to something a bit higher. Als, your diagram seems to imply that your 3V3 rail is limited to 50mA - is this true? Also can you re-specify what your transistor is - you have PN22220 and this doesn't tally. Nov 15 '15 at 11:48
• You MUST not use an emitter follower or you risk damaging your IR LED - and you cannot rely on hFE of the transistor to predict current gain - your only option is putting it in the collector and using a smaller emitter resistor but, going to an op-amp and transistor may be a more accurate approach. Nov 15 '15 at 11:53
• when putting the ir led above the collector , I replaced the 3.3 volt with 5 volt (200 milliamperes max. ) and i couldn't get the a single ir led bright enough to increase the transmitting range . nothing worked fine with enough range and even handled 2 infrared leds except the emitter follower configuration . and yes the 3v3 rail is limited to 50ma max in raspberry pi 2 model b according to its specifications . it actually worked , but i have no idea how much current is flowing from the emitter ? how can i calculate it without using a resistor in base ? Nov 15 '15 at 11:55
• Another name for an emitter follower is a voltage follower i.e. whatever voltage is on the base appears on the emitter minus about 0.7 volts hence there is no equation for emitter current that suits your type of load because the load is highly non-linear. Nov 15 '15 at 13:14