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I'm relatively new to electronics projects, as a hobby. So please bear with me. I want to build an Infrared Repeater as a small project. I've found a component which I think will work nicely for the job. TSMP4138. Its an IR sensor, which will output a modulated signal. In the datasheet (which is linked), there is an example application circuit. For the purpose of my own learning, I want to understand all the components in the schematic, and why they are there. There are 3 resistors, and 2 capacitors in the diagram which im struggling to understand their use. If anyone could help me with some explainations, it would be greatly appreciated.

I do understand that the two IR LEDs (TSAL6400) need to have a resisitor, but I cannot work out the calculation to understand why its a 3ohm resistor.

TSMP4138 Application Circuit

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    \$\begingroup\$ look at the LED forward voltage and the LED forward current \$\endgroup\$ – jsotola Jul 27 '20 at 22:02
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    \$\begingroup\$ There are 3 resistors, and 2 capacitors in the diagram which im struggling to understand their use In order to answer you, please tell me: do you understand that the main components in the repeater circuit are the transistors, and that those 3 resistors and 2 capacitors have just a supporting role in this circuit? \$\endgroup\$ – mguima Jul 28 '20 at 22:06
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It looks like a simple design if you can learn to choose a driver transistor for these parts.

The design for 300mA might appear simple with these values using the model I made for this IR LED nominal values.

enter image description here

But getting there takes some trial and error with good NPN switches and effort in simulating datasheets.

Necessary details

You won't be able to work out the LED current until you have lots of engineering experience with diodes and transistor properties.

Once that is accomplished the solution is simply to match the transistor to the design specs and tweak the resistor values as required.

I can estimate the diode's quadratic VI properties into a linear curve over the desired linear range of current.

Although the LED can handle 1A max it is also limited to an ABSOLUTE MAX of 160mW and 85'C so choosing a safe margin is wise. The duty cycle is expected to 50% with continuous 20~30kHz carrier and then reduced by the amplitude modulated data stream. Thus 320 mW peak might be a safe peak value and possibly with 50% 0's & 1's 640mW peak.

Looking the VI chart for this LED 1.6V @ 330mA = 528mW may be a safe peak limit pending review on the data pattern.

There is always going to be some error involved with LED variances, ambient temperature etc so having good design specs is critical to this process of choosing R values, currents and transistors. There can be a lot of errors made in assumptions.

Now digitizing the Log I vs Vf chart and converting to linear I get the following spreadsheet using a 2nd order Polynomial curve fit:

enter image description here

Note I computed the average of 3 Voltages and divided by 3 curre ents to estimate the tangent slope or series resistance of the diode for each current level. Then I tweaked some graph reads to smooth the curve.

The result is a rough estimate formula for diode resistance and voltage based on a curve fit with a very small residual error.

To use the power limits above I chose from 240 to 360mA $$V_f= 1.2+I_f*5.5 Ω$$

Now the same can be done for the power transistor and I expect an Rs or Rce value around 1.5Ω which I know from experience is about a package rated for approx 1 W to achieve that low Rs value. A PN2222A being quite low is around 4Ω might require a lower Re =0 and get too hot.

Two LEDs going to 5V simply does not make sense as the LED voltage is not 2V unless you go to 800mA peak and a lower duty cycle for power with a bigger power transistor.

hFE is also a huge variable because in order to get say 300mA I started from here.

Ve= 0.90V
Vbe=0.75V
Vb=Ve+Vbe=1.65V which means Ib= (3-1.65)/680Ω= 2mA
thus hFE=300mA/2mA=150

When Vce drops below 1 at 10% of rated current the hFE drops towards 10% of its peak current gain. So hFE = 150 is unrealistic for a power switch and 20 to 50 is more reasonable than 10.

Here is my proof of design

Vf= 1.46V @ 230mA Vce = 845 mV @ hFE=100 Vre = ~ 700 mV @ 230 mA across 3Ω

I won't complete the design with a transistor choice, but I just want to show you that it is not simply Ohm's Law unless you figure out the quasi-linear values for each part and choose binned high hFE value power transistors or low RdsON Nch FET which would be more expensive in high volume but much simpler for design.

enter image description here This design can be easily made to work yet hard to predict Ic and standard deviation since saturating transistors have extremely wide tolerances on Rce=Vce/Ice and hFE.

Added Notes

Let me be clear, the base and collector current is not well defined in this unregulated 3V design. The transistor is well into saturation yet will never operate at the same conditions defined in the datasheet for Vce(sat) @ Ie with Ic/Ib = 10 = hFE. It will end up with hFE being much higher than 10, yet less than the linear region with Vce=10 as defined in tables, but a nominal value may be found in figures, but not easily. So my strategy is to create a linear saturation algorithm for Vce(sat)= Vth+Ic*Rce exactly as I did for the LED in the desired quasi-linear bulk ohmic range.

After doing this for hundreds of designs, there is a relationship between the saturation resistance and Pmax for all diodes and transistors that is reliable when used over a limited range instead of the usual quadratic behaviour.

e.g. \$V_{cc}=V_{led}+V_{ce}+I_c*R_e\$
using my data for a MMBT2222A-7-F (NPN) Vf=0.05+Ic*2Ω

\$3V= {1.2V+If*5.5Ω} + {0.05+Ic*2Ω} +Ie*3Ω \$
3V= 1.25V+ I*(5.5+2+3=10.5) (simplifying, let I=If=Ic=Ie) and solving I=167mA means Re may have to be reduced from 3Ω to 2Ω for this device.

This is all from datasheets for T=25'C which won't be true if the device heats up and current will increase as Vbe and Vf reduce with rising temp.

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  • \$\begingroup\$ Can you clarify two points? 1) Does the above design keep the transistor below the LED in active region ? VCE is mentioned to be 845mV. I would assume that the transistor is designed to go to saturation and end up with VCEsat across it. 2) Such a design would be less sensitive to hFE (since the base current can be designed to be slightly higher than necessary to send the transistor to saturation). \$\endgroup\$ – AJN Jul 28 '20 at 8:30
  • \$\begingroup\$ 1) active linear? no ! but cannot use Vce(sat) since Ic/Ib is not =10 ... 2) nothing permits Ic/Ib=10 and when all the resistances are so low and similar, controlling Ic is hard and harder for high currents as expected. \$\endgroup\$ – Tony Stewart EE75 Jul 28 '20 at 12:53
  • \$\begingroup\$ Simulation Ic=230mA, Vce=834mV so roughly Rce=~3.4Ω using 50mV offset. \$\endgroup\$ – Tony Stewart EE75 Jul 28 '20 at 13:00
  • \$\begingroup\$ Some re-calc. are necessary to check all assumptions . For 300ma Rce needs to be 1.8Ω \$\endgroup\$ – Tony Stewart EE75 Jul 28 '20 at 13:03
  • \$\begingroup\$ Testing my LED model seems to be close enuf to the datasheet at 2V @ 1A tinyurl.com/yytvxb9v vs 2.1V@ 25'C \$\endgroup\$ – Tony Stewart EE75 Jul 28 '20 at 13:22
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100 ohm and 1 uF capacitor

The chip is probably having switching components inside it which will draw sudden spikes of current during switching. This sudden current draw from the main 3V or 5V supply may introduce corresponding dips in the supply itself. The 100 ohm ans 1uF capacitor will likely provide some de-coupling. The capacitor's stored charge will provide the additional current during switching spikes.

Alternately, the external transistors are also switching at high frequencies and they themselves will draw spikes of current from the supply. The dips in the supply due to this will be filtered by the resistor and capacitor from affecting the sensitive circuits inside the TSMP chip.

3 ohm

From the data sheet, the TSAL6400 is supposed to be operated between 100mA and 1A. It will drop between 1V and 2V depending on the current drawn. Even if the transistor is fully ON, the 3 ohm resistor will limit the current through the LEDS to (3V-1V)/3ohm < 1A. when the supply is 5V, you are supposed to put two LEDS which will drop between 1+1V to 2+2V and the current is still limited to (5V-2V)/3 ohm = 1A. The 3 ohm could have been kept on the collector side also. Keeping it on the emitter side will create slightly higher input impedance for this stage making it easier for the previous stage to drive it. This may be why it is not kept on the collector side.

680 ohm and transistor

The TSMP chip may not be able to directly drive the transistor connected to the LED. the internal 33kohm resistance may allow enough current through it to directly drive the transistor connected to the LED. Hence the 680ohm and its transistor act as an amplifier. The 680 ohm will be able to pass through enough current to turn the LED's transistor ON.

47uF

To provide sufficient current since this circuit will try to draw spikes of current while switching.

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