PNP Switch Circuit Help

Thread Starter

s200bym

Joined Aug 9, 2017
75
Hi,

I want to use a PNP Transistor as a switch instead of a mechanical relay.

All circuits are +12v

The switched +12v will be used to switch the transistor on, to allow a +12v to flow into the +5v regulator.

Is the below circuit correct for what I want to do? What would be the values of R1 and R2, It only needs to be a max of 2A?

Kind Regards,
Mike.PCB.PNG
 

Dodgydave

Joined Jun 22, 2012
8,609
R1 can be from 4K7 to 1K, and R2 i would use 1K to 100R, it depends on what Q2 hfe is,..

I would also put a 10K across Q2 B/E..
 
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crutschow

Joined Mar 14, 2008
23,825
Q2 is FZT751Q
It's saturation voltage is about 0.35V @ 2A, (below) meaning it will dissipate about 0.7W, which is probably marginal in free air.
The recommend in be mounted on a 25mm x 25mm 2oz copper pad.

Note: that saturation voltage is with a base current of 1/10th of the collector current or 200mA through Q1.

Better would be to use a P-MOSFET with an on-resistance of ≤0.2Ω which would require no heat sinking and no gate current.

upload_2019-8-21_14-0-43.png
 

Thread Starter

s200bym

Joined Aug 9, 2017
75
To be honest the FZT751Q is probably overkill. I am basically controlling a custom watered-down Arduino ATMega328p AU serial pin TX controlling 2x 12v valves. 90% of the time it will be sitting idle. It's only when I switch the valves, that will draw the most current which I think only draws around 100-200mA.

Mike
 

AnalogKid

Joined Aug 1, 2013
8,232
What would be the purpose of the 4.7k across B&E?
To assure that Q2 turns off quickly and cleanly.

4.7 K means that over 2 mA of possible base current is diverted away when the transistor is on. I would increase the added resistor to something in the 10 K - 47 K range.

To get 2 A through Q2 it needs at least 100 mA of base current for a standard transistor, or much less for a Darlington. But the Darlington will have a higher saturation voltage and dissipate much more heat. To stick with a 3904 for Q1, size R1 for approx. 5 - 10 mA of base current. With something like a TIP32 for Q2, size R2 for approx 100 mA of base current.

Note - an *old* rule of thumb for saturating a transistor is to make the base current at least 1/10th of the anticipated collector current. But this "rule" comes from the days when a power transistor had a max. gain of 20. Today's parts are ... better. I think a 1:20 base-to-collector current ratio is a good design point, especially for a part like a TIP 32 or TIP42.

ak
 

TeeKay6

Joined Apr 20, 2019
534
It's saturation voltage is about 0.35V @ 2A, (below) meaning it will dissipate about 0.7W, which is probably marginal in free air.
The recommend in be mounted on a 25mm x 25mm 2oz copper pad.

Note: that saturation voltage is with a base current of 1/10th of the collector current or 200mA through Q1.

Better would be to use a P-MOSFET with an on-resistance of ≤0.2Ω which would require no heat sinking and no gate current.

View attachment 184471
@crutschow
So...we have:
Using FZT751Q: I=2A, Vsat=0.35V, P=0.7W
Using unspecified P-ch MOSFET: I=2A, Ron=0.2Ω, P=0.8W
Why is P-ch better? :)
(P.S. This is a tease.)
 
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iimagine

Joined Dec 20, 2010
388
Just so you know that, there are 2 ways of turning on/off a circuit: source and sink. You could 'sink', connect or disconnect the converter GND
by using just one NPN transistor or N-Mos instead. :)
 

MrAl

Joined Jun 17, 2014
6,835
What would be the purpose of the 4.7k across B&E?

Cheers,
Mike.
I think the suggestion was 10k, but with either value it helps the transistor turn off faster as mentioned elsewhere but also it keeps the collector base leakage current from turning on the transistor a little or a lot even when it is supposed to be off, especially if the temperature goes higher during use.

That was originally a Zetex transistor. Usually high gain and low Vsat.
 
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crutschow

Joined Mar 14, 2008
23,825
Why is P-ch better?
Actually, I originally miscalculated a higher power dissipation in the PNP, and then forgot to remove the dissipation comment about the MOSFET. :oops:
So the small advantage of the MOSFET would be just to save the base current required to drive the PNP.
 

TeeKay6

Joined Apr 20, 2019
534
Actually, I originally miscalculated a higher power dissipation in the PNP, and then forgot to remove the dissipation comment about the MOSFET. :oops:
So the small advantage of the MOSFET would be just to save the base current required to drive the PNP.
@crutschow
Actually, I was merely observing that you are not 100% perfect, merely 99.999%. Have a great day!
 

TeeKay6

Joined Apr 20, 2019
534
@TeeKay6
I was thinking of cutting off power to the whole thing :)
View attachment 184496
PS: Ignore components value, I was just using the sim to draw it
@iimagine
I do not believe your circuit will work as claimed. I believe that with the regulator's "ground" terminal disconnected/floating and the "in" terminal connected to +12V, the "out" terminal will be near +12V. The regulator's circuitry will attempt to produce an output that is 5V more positive than the ground terminal, not the expected 0V.
 
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TeeKay6

Joined Apr 20, 2019
534
This is a little interesting. A comparison between the Data Sheet Vsat curve and the Spice Model Vsat.
Notice the Vsat does not look as good with the Spice Model.
All of these are when holding the collector current 10 times the base current.
@MrAl
Yes, an interesting catch. It is likely that many other models do not closely adhere to actual devices. As Robert Pease warned, don't implicitly trust a simulation, test with real components.
 

crutschow

Joined Mar 14, 2008
23,825
This is a little interesting. A comparison between the Data Sheet Vsat curve and the Spice Model Vsat.
Yes, there's quite a difference.
Where did you get your Spice Model?

The LTspice simulation below, with this Diodes model, seems to fairly closely follow the data sheet graph.

upload_2019-8-22_13-32-20.png

upload_2019-8-22_21-59-38.png
 
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