Using NPN 2N3904 as switch

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jim0000

Joined Oct 28, 2020
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Do NPN and PNP transistors have the same turn on quality as MOSFETs, what I am specifically referring to is the MOSFETs Vth and how it lists a small voltage range for Vth on the datasheet, but for the device to be fully on Vgs would need to be about 10V for Si MOSFETs at least. I notice the 2N3904 has a Vbe(sat) of 0.65V-0.85V but will it be fully on at that point or only partially on? What Vbe would turn this NPN fully on?

View attachment 282180
 

BobTPH

Joined Jun 5, 2013
8,809
They do not work same way at all. The base emitter junction is a diode. The voltage across it can never be more than the Vbe sat. The current through the base needs to be about 1/10 of the desired collector current for saturation.

You need a resistor between the signal and the base to limit the current, same as an LED but with a forward voltage of about 0.7V
 

Audioguru again

Joined Oct 21, 2019
6,672
The 2N3904 is a weak low current transistor. Its datasheet shows it turned fully on conducting (only) 50mA and a collector-emitter voltage loss of 0.3V max. When it is fully turned on and conducting 50mA then its maximum base-emitter voltage is 0.95V at 5mA.

Most Mosfets can conduct hundreds of times more current.
 

AnalogKid

Joined Aug 1, 2013
10,986
A bipolar transistor and an enhancement mode MOSFET are two very different animals. However, in terms as their performance as a voltage-controlled switch, they are quasi-similar. In terms of the voltage applied to the device to cause it to go from completely off to conducting as much as it can, the bipolar takes a much smaller voltage range from 0 V to 0.6 or 0.7 V. A MOSFET is barely conducting at that voltage. It usually needs something greater than 2 V to begin conducting any real current, and around 8 V to get into the edge of its "saturation" (fully enhanced) region. Note that unlike the bipolar, the FET needs almost zero current from the control signal.

Of course, the actual actions are more complex than this, and the numbers vary widely across part numbers.

ak
 

Thread Starter

jim0000

Joined Oct 28, 2020
130
A bipolar transistor and an enhancement mode MOSFET are two very different animals. However, in terms as their performance as a voltage-controlled switch, they are quasi-similar. In terms of the voltage applied to the device to cause it to go from completely off to conducting as much as it can, the bipolar takes a much smaller voltage range from 0 V to 0.6 or 0.7 V. A MOSFET is barely conducting at that voltage. It usually needs something greater than 2 V to begin conducting any real current, and around 8 V to get into the edge of its "saturation" (fully enhanced) region. Note that unlike the bipolar, the FET needs almost zero current from the control signal.

Of course, the actual actions are more complex than this, and the numbers vary widely across part numbers.

ak
Thanks! You brought something up that I am unsure of. How can you tell how much current a BJT needs sent to it's base from the control signal to make it conduct?
 

Audioguru again

Joined Oct 21, 2019
6,672
How can you tell how much current a BJT needs sent to it's base from the control signal to make it conduct?
For a transistor to fully turn on (saturate) we make the base-emitter current 1/10th the collector load current, as shown on the datasheet you posted.
The datasheet shows the range of "current gain" (hFE). For a 2N3904 for a collector load current of 10mA, when it is linear (not saturated) then the base current is 10mA/100 to 10mA/300, because its spec'd current gain is from 100 to 300.
 

sghioto

Joined Dec 31, 2017
5,378
How can you tell how much current a BJT needs sent to it's base from the control signal to make it conduct?
That would depend on how much you want it to conduct and probably not a simple answer from the spec sheet.
Below is an example of a breadboard test I did showing the level of conduction for different value base resistors.
With R1 at 1K the transistor is fully saturated for a 50ma load.
1670198700964.png
 

dl324

Joined Mar 30, 2015
16,845
Thanks! You brought something up that I am unsure of. How can you tell how much current a BJT needs sent to it's base from the control signal to make it conduct?
The terminology you're using is inexact.

Assuming you're referring to using the transistor as a switch, the base current should be 10 times one tenth of the collector current to guarantee that the transistor will be saturated.

EDIT for brain fade.
 
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Papabravo

Joined Feb 24, 2006
21,159
The terminology you're using is inexact.

Assuming you're referring to using the transistor as a switch, the base current should be 10 times the collector current to guarantee that the transistor will be saturated.
The correct statement is:
Assuming you're referring to using the transistor as a switch, the collector current should be 10 times the base current to guarantee that the transistor will be saturated.
\( I_{C}\;=\;10\times I_{B} \)
 

Papabravo

Joined Feb 24, 2006
21,159
In an NPN BJT the collector current is a nearly linear function of the base current when Vbe approaches 0.6-0.7V
In an n-channel MOSFET the drain current is a quadratic function of (Vgs - Vth)
 

crutschow

Joined Mar 14, 2008
34,281
How can you tell how much current a BJT needs sent to it's base from the control signal to make it conduct?
Depends upon what you mean by "conduct".
If you just want it partially on than that's determined by the current gain (Beta or Hfe) which generally has a large variation from unit to unit (such as 100 to 300).

If you want it fully on as a switch, the you use a typical base current 1/10th of the maximum collector current, as stated by others here.
 

WBahn

Joined Mar 31, 2012
29,976
Do NPN and PNP transistors have the same turn on quality as MOSFETs, what I am specifically referring to is the MOSFETs Vth and how it lists a small voltage range for Vth on the datasheet, but for the device to be fully on Vgs would need to be about 10V for Si MOSFETs at least. I notice the 2N3904 has a Vbe(sat) of 0.65V-0.85V but will it be fully on at that point or only partially on? What Vbe would turn this NPN fully on?
I'm not sure what you would call a "small voltage range for Vth". I've seen the spec'ed Vth voltage vary by a couple of volts from min to max even under the same test conditions, let alone over the entire operating envelope.

Also, the Vth of the transistor depends on the transistors design. While some are 10 V or more, many are considerably less, in particular the "logic-level" MOSFETs that are specificially intended to be controlled directly from logic-level circuits.

As for using a BJT as a switch, this is commonly done, but not by applying a voltage directly to the base. Instead, there is some kind of interface circuit (which can be as simple as a resistor), usually to convert a control voltage into an approximate base current.

The usual rule of thumb is that saturation involves getting the base current to be 10% of the collector current. There is nothing magical about this, however. It is merely the nearly-universal operating point at which device manufacturers measure and document their device's behavior.

Where you really need to begin is deciding how much current the transistor needs to conduct with the "switch is closed" and what the maximum voltage you can tolerate there being across it under those conditions.
 

Thread Starter

jim0000

Joined Oct 28, 2020
130
What is the purpose of both the zener diode and the Schottky diode in this circuit?
It could be a wrong idea, I mentioned this in my other reply but I will reply here as well. I was hoping to counteract the 2mV/deg C change in my zener diode. I had read that regular diodes have around -2mV/deg C so I was hoping to negate the temperature effect. This would need to be experimentally determined for both devices, and perhaps more than one Schottky included depending on the exact change per degrees C of the zener diode.
 
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