circuitsarefun

modified 4 years ago

Help needed on Transistor circuit

144

02:07:42

I understand  forward bias of LED is 2V
I understand beta is 100 so
190µA*100 = 19mA current of LED
I know that the transistor is triggered by a  voltage greater than 700mV
I need help in calculating 814mV between the resistor and the transistor
I need help in calculating 190µA
I thought I understood  Ohms Law but my  calculations do not seem to work
Thanks

published 4 years ago

Robert_Kidd

4 years ago

A fully turned on transistor will have approximately 0.2V between its collector and emitter. You have a 5V supply and, as you rightly say, the LED requires 2V across it. EC has a default LED current of 20mA. LED today are generally much brighter so in a real circuit you may want less current but let’s stick to 20mA for the moment. So, the resistor in series with the LED needs to drop 5V-2V-0.2V=2.8V. With a current of 20mA this gives a resistor value of 2.8/0.02=140 Ohms.

Robert_Kidd

4 years ago

As you say, default forward gain is 100 so base current would be 0.02/100=0.0002A. Assuming 0.7V across base emitter junction, base resistor would be (5-0.7)/0.0002=21,500 Ohms.

Robert_Kidd

4 years ago

EC transistors seem to be set for >0.8V for base emitter voltage so you may want to tweak slightly. It’s good to use preferred component values too. Hope this helps.

PrathikP

4 years ago

It's generally a good idea to supply more base current than required, just to ensure that the transistor is "properly on". In this case, you can supply twice the base current.

circuitsarefun

4 years ago

Thanks for the responses The purpose was to construct a simple circuit to show how me how a transistor works. The challenge I faced was to figure out what resistor to use. I start with what was known Vs 5 volts Vf of the diode 2 volts Voltage at the collector is 3V (5V-2V) Maximum current of the LED 0.02A The transistor beta value 100 Beta= Ic/Ib by definition Ib= Ic/Beta Ib=0.02/100 or 0.0002 Amps R=3/0.0002 R= 15KΩ Had 22KΩ so that is what I used EveryCircuit is such a joy and I am so thankful for the helpful responses. I am still struggling with how EveryCircuit came up with 814mV

Robert_Kidd

4 years ago

I thought you might have modified the circuit to reflect what you have been shown?

Robert_Kidd

4 years ago

Looking at your notes above note that the collector voltage should not be 3V. You are using the transistor as a switch in its ON condition. In this condition the collector voltage is not 3V, it is around 0.2V - for calculation purposes it is often taken as zero. You must use a resistor in series with the LED to drop the extra voltage above the 2V the LED requires.

Robert_Kidd

4 years ago

Regarding the 814mV, this is simply the base emitter drop you can expect in EC for a diode forward volt drop - you’ll appreciate that a transistor is structurally two diode junctions joined together (put simplistically - there’s a more to it than that since if you connected two diodes like that you wouldn’t get the gain that a transistor gives you).

Robert_Kidd

4 years ago

Why 800mV? Well, we all generally think and read about a forward biased diode junction (or base emitter junction) as being 0.7V. Some publications suggest 0.6V to 0.8V.

Robert_Kidd

4 years ago

For reasons I don’t fully understand EC forward biased diode junctions are defined as 800mV rather that 700mV. Note it’s approx. 800mV so will vary a little depending how hard you drive it.

Robert_Kidd

4 years ago

Now, there’s also a ‘magic’ value of current that EC applies when using zener diodes. If you select a zener diode to use in a circuit, let’s choose a 5.6V zener, then it’s knee voltage of 5.6V will be achieved at a current of 13.6mA. Why this value? We don’t know! Developers presumably had very good reasons. Now, if we connect up an ordinary diode so it’s forward biased (or a zener the ‘wrong ‘ way around) and pass 13.6mA through it, look what the voltage across it is - 800mV!!

Robert_Kidd

4 years ago

OK, lots of words but fundamentally all you need to do is accept that for a transistor that is turned on fully it’s base voltage will be around 800mV. Use that you help you calculate your base resistor. Does that help you?

Robert_Kidd

4 years ago

Just noticed! 800mV against 814mV. The transistor has gain of 100. If you increase the base resistor value until you have exactly 800mV across base emitter junction you’ll notice that base current is exactly one hundredth of the collector current. So the fact that there is a little more base current that EC’s defined value pushes the base emitter voltage up from 800mV to 814mV.

Robert_Kidd

4 years ago

........ and what is that base current? Well, it’s 13.6/100mA, one hundredth of that magic 13.6mA.

circuitsarefun

4 years ago

I think transistors are "magical". Robert, I am glad I included my notes and appreciate you taking the time to look at them. I accept the 800-814mV. I will continue to study your responses as it enhances my understanding of transistors. I am almost to the point of putting this circuit to rest so I can sleep tonight:) In EveryCircuit the voltage shown between the LED the the collector is 3.01V. My new struggle is to understand why the collector voltage is 0.2V. I may be a hopeless cause.

Robert_Kidd

4 years ago

In this circuit the transistor is being used in switching mode. You pull the base up to rail and it is turned on. You pull base down to ground and it’s turned off.

Robert_Kidd

4 years ago

When turned off, no current flows so collector voltage will be at rail (5V in your circuit). If there is no current flowing there is no volt drop across the LED so it’s all across the transistor collector-emitter. Think of the transistor as an open switch.

Robert_Kidd

4 years ago

When it’s turned on consider it again like a switch, but closed, so the collector emitter voltage is close to zero. Obviously an actual closed switch would be close to zero ohms. A transistor is not as perfect so you just have to accept that when closed there is a small volt drop across it of 0.1V to 0.2V.

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