transistor voltage drop

[painstream]

Hello All,

I have a transistor setup in a Emitter-Follower type Circuit. I get a voltage drop. 1.34V DROP!.

12 V DC Power Supply > transistor > 10K POT > 12V DC 130 mA FAN. . I'm new to electronics. How do I overcome the VDrop?

Thanks for your help.

 

[MrAl]

Hi,


The voltage follower (common collector) is known for it's large voltage drop. The problem stems from the drive voltage however, not the transistor itself. With a DC supply voltage of 12 volts and a drive voltage of 12 volts (directly on the base) the emitter must be around 0.7v down from 12 volts. That's because the base emitter drop is around 0.7 volts. So the best you can do is around 0.7 volts that way. Since yours is 1.34 volts, you should be able to do better than that.

However, with a common emitter topology (PNP emitter goes to 12 volts or NPN emitter goes to ground and collector is output) you can get as low as 0.2 volts or even better with some modern transistors made for low Vce. Check out that configuration and see if it works better for you.
If you connect the NPN emitter to ground and drive the base with say 1k, the collector voltage will drop as low as 0.2 volts or so, so connecting the fan to the collector and to the positive supply rail (12 volts) you'll get a much larger voltage output.

The advantage of the voltage follower is you can get nice control over the output voltage at the emitter. If you dont need that however, then a common emitter configuration is better.
If you need to control the voltage output then a slightly better circuit would be in order...possibly by using two transistors instead of just one.

 

[Diver300]

You should post your circuit.

I would have expected 12 V DC Power Supply > 10K POT > transistor > 12V DC 130 mA FAN but that leaves out a lot of detail.

If you are using a darlington transistor, you would expect that sort of voltage drop. If you are using a conventional transistor, the drop may be due to the gain if the transistor being smaller than you need.

To virtually eliminate voltage drop, you need an op-amp as well as a transistor. You should note that if you have an emitter-follower circuit, the output voltage will always be less than the supply by about 0.7 V

 

[MrAl]

Hello again,

Here is a representative circuit. It's a relatively simple circuit that uses two transistors and obtains a very good linear pot-setting-to-output-voltage range. It should reach very close to +12 volts output when the pot is cranked all the way up because the output stage is a common emitter. The input stage is a voltage follower for good linearity.

Note the 100 ohm resistor represents the fan so leave that out for the real circuit and just connect the fan.

 

[painstream]

So I was able to figure it out. My power supply was not providing 12 volts, just assumed it did. I do get the 0.7V drop. I'm curious to see if I can get better than 0.7 by some of the other methods posted above. Not sure what the use of both diodes like that MrAl? Also not sure exactly what an op amp does, looks like I got some reading to do! thanks for reply and help!

 

[MrAl]

Hi,


I think you will find the two transistor circuit to be good enough but if you want to look into op amps that's cool too.

The two diodes provide a known voltage drop. As you know now, the emitter of a common collector circuit (voltage follower) can only reach down as low as about 0.7v (from Vcc) or so without an extra supply voltage to drive the base, and the first stage is a common collector circuit where the emitter senses the output voltage of the common emitter (PNP) circuit. In order to sense a voltage then that is only 0.2 volts down from the positive rail (12v) we need to loose some voltage with something that has a fairly constant voltage drop. By loosing two voltage drops (about 1.4v or less here) we can control a voltage at the PNP collector that is very close to the actual positive rail voltage while keeping the emitter of the NPN at a comfortable voltage that can be easily achieved. So with 11.8 volts at the PNP collector we'd have about 10.4v at the NPN emitter, which is much less than the maximum 11.3 without any diodes. In actual real life the gains come into play too, so maybe we'll get a max of 11.7 volts output. With only one diode we also get somewhat acceptable operation of maybe around 11.5 volts.

 

[serkan]

painstream: If you are using the transistor for switching of fan, there will always be a voltage drop of 0.7V so it is a better idea to switch from transistor to the op-amp as op-amp provides the large gain and in the saturation state the output of the op-amp is almost same as the Vcc.

**broken link removed**

 

[MrAl]

Hi,

That's not always true. A two transistor circuit works good without an op amp, and many op amps can not reach up to +Vcc anyway nor can they provide enough drive current without going to a specialized version.

 

[crutschow]

painstream: If you are using the transistor for switching of fan, there will always be a voltage drop of 0.7V so it is a better idea to switch from transistor to the op-amp as op-amp provides the large gain and in the saturation state the output of the op-amp is almost same as the Vcc.

The collector-emitter saturation voltage of a single common-emitter transistor configuration can be much less than the 0.7V base-emitter voltage (on the order of a tenth volt or less) so there's no reason to go to an op amp for that. A power MOSFET can have an ON resistance of a few milliohms or less so can have a very low ON voltage, even lower than a BJT.