Voltage drop across transistor (Vce) vs LED load forward voltage

Question

So I'm trying to simulate a BJT switch for an LED.

Here's the detail and how I setup the simatiomln.

• Vsupply = 4.2 V
• LED UV Vf = 3.6 V
• Hfe = 100

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Here's the simulation:

Falstad link, does it work?

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My calculations start from what I want to achieve: 20 mA for the LED. That makes Ic = 20 mA.

Calculate Ib needed for Ic = 20 mA:

Ic / Hfe = Ib = 20 / 100 = 0.2 mA.

Then calculate Rb for Ib = 0.2 mA:

Rb = Vrb / Ib = (Vsupply - Vbe) / Ib = (4.2 - 0.7) / 0.2 mA = 17k5

That's all.

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I'm assuming, because I set the forward voltage of the LED to 3.6 V, that the transistor (Vce) will take up the rest (4.2 - 3.6) = 0.6 V, and 0.6 V * 20 mA would burn into heat, but apparently it doesn't work like that.

What am I missing here? Am I setting it wrong, is there some variable that should be added to the simulation?

The goal is obviously 3.6 V @ 20 mA at LED.

One thing I notice though if change the Rb to 470R, I get 3.6 V drop in the LED, but Ic = 730 mA doesn't make sense.

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FIXED

I messed up my LED-UV model. turn out I need to set "Current at Above Voltage" to 20 mA. I don't even know what it means, I think it's related to LED working slope, it should be called "Current at Forward Voltage Above" so as not to confuse.

In my previous model, I just set the "Max Brightness Current" to 20 mA, and leaving "Current at Above Voltage" value to default 1 A.

Below is the final circuit (I would put the current limiting resistor later).

Answer 1

Hfe is an approximation for a transistor in the linear region, it doesn't apply to a BJT in saturation. You can use it to estimate a base resistance, but the characteristic curves will show you more detail.

You should not rely on the saturation voltage of a transistor to limit the current in an LED. It can vary quite a bit over temperature and from unit to unit (as will the steep LED I-V characteristic). You should add a current limiting resistor so that the LED current is less dependent on VCEsat, or design a constant current driver.

You want 20mA in the LED, so estimate or look up the saturation voltage of your BJT. Let's ballpark it at 0.3 V. You can start with the value of Rb that you calculated and adjust as necessary.

Next get the forward voltage of the LED at 20mA: 3.6 V.

So you have 4.2 V - 3.9 V or 300mV of headroom (as Spehro pointed out if this is a Li battery you don't have much room for discharging it.)

Pick a resistor of 0.3 V/.02 A or 15 ohms, add it in series with your LED. Now it should be easier for the BJT to saturate.

However, for a workable design you would want more voltage headroom.