Dear Experts...

I have an schematic here with Q1 configured as an emitter follower. The emitter lead on the transistor follows the base junction voltage less ~0.65V.

Q: How can I counteract this 0.65V drop and bring the output up to 2.2 volts as per the Vref from the LED?

 

Add a diode or transistor in series with the LED.

 

Add a diode or transistor in series with the LED.

Perfect. Easy greasy.

 

Perfect. Easy greasy.

However, the voltage at Vout will not track the LED voltage very well.

 

However, the voltage at Vout will not track the LED voltage very well.

What if I were to bias the emitter with a voltage?

 

However, the voltage at Vout will not track the LED voltage very well.

It depends on what, "very well" is needed. MCU88 is asking about a circuit with 3 parts. I could make it track deadly well, but it would need several more parts and be a lot more involved in the understanding of the circuit.

 

It depends on what, "very well" is needed. MCU88 is asking about a circuit with 3 parts. I could make it track deadly well, but it would need several more parts and be a lot more involved in the understanding of the circuit.

Keen to see it. Please post an schema and schematic for us...

 

Consider this:

 

That turned out better than I expected! Choose the right 5 volt op-amp and you might do this in 4 parts.

 

Consider this:

Ok so the OPAMP has the transistor in the feedback path and therefore regulates the output to the Vref of the LED - ~0.65V.

Why the FET?

 

The op-amp drives the transistor so that the output of the transistor is the same as the voltage at the LED, give or take errors in the op-amp. This can be as low as the microvolt range.

The fet is a constant current device. In this configuration it allows 5 ma to 15 ma to flow, depending on which transistor your fingers touch when you reach into the parts bin. This constant current ability replaces the resistor and does an excellent job of refusing to respond to the voltage variations of the power supply.

 

The op-amp drives the transistor so that the output of the transistor is the same as the voltage at the LED

Arh yes of course. The emitter is in the feedback path so there it compensates for the 0.65 volt drop.

I was sorting of hoping for an transistor only solution though...

 

Here is how the proposed circuit is effected by load current and temperature. My LED has a slightly different Vf. Note the temperature effect is largely a function of LED Vf being effected by temperature... which the opamp circuit does not solve.

 

Here, let me flip that around for you. It's just a current enhanced voltage follower. It will track the voltage of the LED religiously. If you want the variations of the LED voltage that are caused by temperature changes, you got it! If you want a steady reference voltage, the LED is a miserable choice in the first place.

 

I was sort of hoping for an transistor only solution though...

You can built a sort of op-amp with a differential pair of transistors.

 

Here, let me flip that around for you. It's just a current enhanced voltage follower. It will track the voltage of the LED religiously. If you want the variations of the LED voltage that are caused by temperature changes, you got it! If you want a steady reference voltage, the LED is a miserable choice in the first place.

In general I am looking for ways to get around the 0.65 volt drop of an BJT. That's it. Just interested in methods of getting around this limitation that is sometimes an useful aid.

 

You can built a sort of op-amp with a differential pair of transistors.

Keen to see that. Can you post an schematic with explanation(s) and practical uses in dot point form?

 

Boosting the output is often done with just a diode. You can use this trick with an LM7805, too.

Design you a differential amp with transistors at 5 am? I'm getting tired right now.
Maybe tomorrow.

 

Boosting the output is often done with just a diode. You can use this trick with an LM7805, too.

Design you a differential amp with transistors at 5 am? I'm getting tired right now.
Maybe tomorrow.

Sleep is for the weak. Knock off time is 8am. Go grab yourself an cup of coffee.

 

Boosting the output is often done with just a diode. You can use this trick with an LM7805, too.
Maybe tomorrow.

There is an beer on the table for you via PayPal buddy...