Transistors is not working/opening. How do i fix this?

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
Hey guys,
I have a PBSS4140T transistor in a circuit to control a oven heater.
Only with the same programming as with my previous BC337 transistors, the transistor is not opening.
Here is my schematic:
1706978712302.png

Do you see something i do not see.
The system is connected to the analog pins of an arduino GIGA R1
 

crutschow

Joined Mar 14, 2008
34,464
You show only a partial schematic, so how can we tell what's wrong?
What are the transistor emitters connected to?
What is the output voltage from the Arduino?
What is the transistor load current?
 
Last edited:

crutschow

Joined Mar 14, 2008
34,464
The collector connected to ground is wrong unless the emitter is at a negative voltage.
I think that's just the common connection for the fans and heaters, and are not connected to circuit ground.
I suspect the transistor emitters are connected to ground to turn on the heaters and/or fans, but that's not clear from the schematic shown.
 

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
You show only a partial schematic, so how can we tell what's wrong?
What are the transistor emitters connected to?
What is the output voltage from the Arduino?
What is the transistor load current?
Hey guys, sorry for the partial information I gave. The Arduino board voltage is 3,3V. The load Voltage to the relays is 12v. So there's 12v going through the transistors.

The transistors are placed after the relays and switch the ground on and off back to the power supply.
The HFe is 300 in this transistor.
 

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
More information about the circuit:
- The Arduino can give 8mA maximum from the analog pin to the base.
- The Vsat of the transistor is 1,2V
- The external relay is for 40A, the oven is pretty big and needs big heaters that's why i use a external relay.
- The external relay needs 12V and 25mA to activate.

I calculated the resistance for the transistors on this way:

R-transistor: Vsat/Asat.

Asat: Relayamps/HFe-transistor =
Asat: 0,025A/300=8,333E-5 Amps

R-transistor: 1,2/ 8,333E-5 Amps = is about 14K. So I used a bit more resistance.

Hope this is enough information for you guys.
 

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
hello,

for switching applications the rule is ib = ic / 10
your 20k base resistor is way to high

bertud
Oo thanks man, never knew that, i always used the rule from HFe. Do you know where i can find more of those simple rules that are meant for exceptional cases?

The resistance than goes like this i think:

Ic = 25ma
Ib= 25ma/10 = 2,5ma

R-transistor: 1,2V/ 0,0025A = 480ma

Correct?
 

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
Oo thanks man, never knew that, i always used the rule from HFe. Do you know where i can find more of those simple rules that are meant for exceptional cases?

The resistance than goes like this i think:

Ic = 25ma
Ib= 25ma/10 = 2,5ma

R-transistor: 1,2V/ 0,0025A = 480

Correct?
Sorry, the V is 3,3-1,2V = 2,1V
R-transistor: 2,1V/ 0,0025A = 840R
 

LesJones

Joined Jan 8, 2017
4,190
The discussion so far seems to be related to the transistors not switching on. You post refers to the transistors not opening which is swtching off. If this is the case then the transistors have a collectoor emitter short. This could be caused by the lack of back emf protection diodes across the relay coils. Are you sure that the 40 amp relay only requires 25 mA through the coil ?

Les.
 

BobTPH

Joined Jun 5, 2013
8,998
Oo thanks man, never knew that, i always used the rule from HFe. Do you know where i can find more of those simple rules that are meant for exceptional cases?

The resistance than goes like this i think:

Ic = 25ma
Ib= 25ma/10 = 2,5ma

R-transistor: 1,2V/ 0,0025A = 480ma

Correct?
You are trying to calculate a resistance and coming up with an answer in mA?

And, no, the base to emitter voltage will be about 0.7V, not 1.2V. Where did you get that?

Here is the cirrect calculation for the base resistor:

R = (3.3 - 0.7) / 0.0025. = 1040 Ohms.
 

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
The discussion so far seems to be related to the transistors not switching on. You post refers to the transistors not opening which is swtching off. If this is the case then the transistors have a collectoor emitter short. This could be caused by the lack of back emf protection diodes across the relay coils. Are you sure that the 40 amp relay only requires 25 mA through the coil ?

Les.
Yes, the steering ampere for the relay is 12-25ma. The steering voltage is between 2-32v. I use a 3 fase SSR relais for 40Amps.
 

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
You are trying to calculate a resistance and coming up with an answer in mA?

And, no, the base to emitter voltage will be about 0.7V, not 1.2V. Where did you get that?

Here is the cirrect calculation for the base resistor:

R = (3.3 - 0.7) / 0.0025. = 1040 Ohms.
Thanks for the information.

Why don't you actually don't calculate with the HFe (DC current gain) in this situation? Is it because it is a switch and the resistance than need to be lower?

Also:
I was looking at the datasheet of the transistor and it said 1,2VBEsat, but that's with 1 amp going through Ic and 100ma going to Ib. Now that i look at the graph i see that's indeed 0,7v at 8ma to Ib.

Here's also the datasheet:
 

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crutschow

Joined Mar 14, 2008
34,464
Why don't you actually don't calculate with the HFe (DC current gain) in this situation?
Because the stated value of hFE is for linear operation of the transistors, not saturated.
See below from the transistor's data sheet:
The hFE value is for a Vce value of 5V (linear region), not saturated.
The VCEsat values are shown for a base current 1/10th of the collector current (I think the 1mA value is a typo and should be 10mA).

1707145215304.png
1707145270403.png
 

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
Sorry for the confusion. I was trying to calculate the resistance in ohm.
It
Because the stated value of hFE is for linear operation of the transistors, not saturated.
See below from the transistor's data sheet:
The hFE value is for a Vce value of 5V (linear region), not saturated.
The VCEsat values are shown for a base current 1/10th of the collector current (I think the 1mA value is a typo and should be 10mA).

View attachment 314451
View attachment 314452
Thanks a lot guys, you're pretty smart.
So HFe for linear operations, and 1/10th of Ic for switch operation always right?
Is this for every transistor, or changes this for PNP transistors or other kinds?
 

dl324

Joined Mar 30, 2015
16,943
the transistor is not opening
The collectors are connected to ground.

Oo thanks man, never knew that, i always used the rule from HFe. Do you know where i can find more of those simple rules that are meant for exceptional cases?
The rule of thumb for many general purpose transistors is beta=10. This doesn't apply to power transistors or high beta transistors.

The data is in the datasheet. For BC337, it's 10. For BC547, it's 20. For BCX59-9, it's 40. For Darlingtons, it'll be more like 200-250. For power transistors, it's less than 10.

On Semi BC337:
1707150713491.png
1707150779702.png
1707150674817.png
 

Thread Starter

Rensieboy223

Joined Feb 3, 2024
37
The collectors are connected to ground.


The rule of thumb for many general purpose transistors is beta=10. This doesn't apply to power transistors or high beta transistors.

The data is in the datasheet. For BC337, it's 10. For BC547, it's 20. For BCX59-9, it's 40. For Darlingtons, it'll be more like 200-250. For power transistors, it's less than 10.

On Semi BC337:
View attachment 314490
View attachment 314491
View attachment 314489
Thanks for the information,

I probably used the wrong kind of terms in my schematic but the transistor comes after the relays, so the relays has 12v in, than has gnd out. The output from the relays than is switched by the transistor, so the transistor is between the gnd of the relay and the gnd of the power supply. This should work right?

I could also put the transistor before the relay but i read that this puts more current through the transistor to supply the relay what is unnecessary.
 

crutschow

Joined Mar 14, 2008
34,464
The output from the relays than is switched by the transistor, so the transistor is between the gnd of the relay and the gnd of the power supply. This should work right?
Yes, that should work.
To avoid confusion, a better term for the "gnd" of the relay, would be "com".
Gnd implies that it is already connected to ground.
 
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