Help, transistor wont switch the relay

Thread Starter

Dwall

Joined Dec 12, 2018
4
Hi, i have built a circuit which needs to swith in different resistances, i needed to increase the time of the pulses so have dropped the circuit voltage to 6 volts and changed the relays but now the transistor wont switch in relay 2. Im using a bc32740ta transistor, it works if i lower the 10k on the base or remove resistor/cap delay from the relay earth.
Ford idle up cct 4.JPG

Any help please?
Thanks
 

Dodgydave

Joined Jun 22, 2012
8,389
We don't know the voltage of your circuit, but the transistor will take 4mA base current for 800mA Collector current, (hfe 170) i would use a 3.3K base resistor.
 

Thread Starter

Dwall

Joined Dec 12, 2018
4
Sorry, runs on vehicle voltage (12-14.5 volts). If i change the base to 1.5k then the time delay is all wrong, and resistance pulses dont work.
 

Tonyr1084

Joined Sep 24, 2015
3,578
How is relay #4 energized? To me it appears to be connected to a voltage divider that is connected to ground and the other side of the relay is connected to ground through a 10KΩ resistor / 470 µF cap. I don't see any way it can ever be powered.

Relay #1 jumps out a 2.2KΩ resistor in the bottom resistor network. Not sure why that's important, but I don't fully understand what you're trying to achieve. Relay #4 (assuming it works) jumps out all the resistors on the bottom network except for the 120Ω & 180Ω resistors.

Your transistor (PNP type) is connected to a voltage divider consisting of 100Ω and 2.2KΩ. Assuming the transistor turns on - relay #2 is powered from that voltage divider, through a diode junction (of unknown value/function) through the junction of the BPJ transistor, through a 10KΩ resistor to ground. Now, I'm not real good with following diagrams but I'm having trouble seeing how this circuit is supposed to work. Also confusing me is the way you've drawn relay #3. It appears to be in the energized position while the battery switch is open.

Where did you get this diagram from?
 

danadak

Joined Mar 10, 2018
3,577
"Normally" you want to force 1/10 the collector current (relay current) into the base.

So not knowing the circuit parameters and part numbers we cannot tell you
what the base R should be for the drive transistor.

Reason for 1/10 is to force transistor into saturation to minimize its power
dissipation.

Regards, Dana.
 

dl324

Joined Mar 30, 2015
8,904
Apparently some BC transistors specify Ib=0.05*Ic for saturation mode, but the datasheets I have from OnSemi and Fairchild for BC327 state Ib=0.1*Ic.
upload_2018-12-13_8-42-15.png
upload_2018-12-13_8-41-43.png
 

Thread Starter

Dwall

Joined Dec 12, 2018
4
How is relay #4 energized? To me it appears to be connected to a voltage divider that is connected to ground and the other side of the relay is connected to ground through a 10KΩ resistor / 470 µF cap. I don't see any way it can ever be powered.

Relay #1 jumps out a 2.2KΩ resistor in the bottom resistor network. Not sure why that's important, but I don't fully understand what you're trying to achieve. Relay #4 (assuming it works) jumps out all the resistors on the bottom network except for the 120Ω & 180Ω resistors.

Your transistor (PNP type) is connected to a voltage divider consisting of 100Ω and 2.2KΩ. Assuming the transistor turns on - relay #2 is powered from that voltage divider, through a diode junction (of unknown value/function) through the junction of the BPJ transistor, through a 10KΩ resistor to ground. Now, I'm not real good with following diagrams but I'm having trouble seeing how this circuit is supposed to work. Also confusing me is the way you've drawn relay #3. It appears to be in the energized position while the battery switch is open.

Where did you get this diagram from?
Relay 4 is energised from relay 3's nc o/p, this provides a pulse when the switch is opened, this part works perfectly.
This is a circuit i drew myself it is to switch different engine rpms in on a transit, different pulses of resistance either arm the speed controller or select one of 3 speeds, hence relay 3&4, to disengage the rpm on removal of power. The old circuit worked fine but ford have changed the arm to a pulse rather than a switch so trying to modify my circuit, they also appear to need longer pulses.
 

Thread Starter

Dwall

Joined Dec 12, 2018
4
The trouble im having is i can get the transistor working with a smaller resistor, but this completely messes up the time delays and relay pulses?!
 

Tonyr1084

Joined Sep 24, 2015
3,578
I'm having trouble following the logic. I've redrawn part of the circuit in an attempt to understand how it's working. Please let me know what I'm missing - IF I missed something.

[edit] two posts since I began drawing my schematic.

Relays.jpg
 
Last edited:
The trouble im having is i can get the transistor working with a smaller resistor, but this completely messes up the time delays and relay pulses?!
Anytime you change the resistance in an RC circuit, you must also change the capacitors as well, if you need to maintain the same time constant.
 

Danko

Joined Nov 22, 2017
793
Add 3 components (red) to your circuit and it will work with resistor R = 10k, 22k, 33k etc:

upload_2018-12-15_20-16-37.png
 

Audioguru

Joined Dec 20, 2007
11,251
Why do many people wrongly talk about hFE when a transistor is used as a switch? hFE is used in a transistor amplifier circuit that is not switching because it always has lots of collector to emitter voltage. A transistor switch saturates with a very low collector to emitter voltage then it needs a lot of base current.
 

danadak

Joined Mar 10, 2018
3,577
Common terminology is forced beta. Most common is to force beta to 10,
eg. Ib = Ic / 10. That's well into sat for most transistors.

Regards, Dana.
 
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