Hi all

I've built the attached circuit... It's a timer - the 4020 counts the pulses from a 555 and shuts off a relay at the end of the time period.
I'm using a 12v relay with a 120ohm coil in place of the smaller one supplied with the kit.
It all works fine - however I measured the voltage across the coil just now, and it reads 9V, I'm running from a lab PSU at 12.1V. The relay seems to operate fine, although I haven't put a load on it yet.
I figure the reverse polarity diode and PNP transistor should drop the voltage by 1.4 between them, but I have a 3V drop.... The BC328 should handle the coil current no prob.
The only mod I have made to what you see is an indicator LED in parallel with the relay coil.

Should I be concerned? I think I'll get rid of the reverse pol diode, but that won't buy much....

This relay will be used to switch a hot water system, so I am after reliable service....

cheers
Lee

 

Your 12V relay coil has a lower resistance than the kit relay.
The 39K R4 is not driving the transistor Q1 into saturation, so you measure 3V across the transistor and approx 9V across the relay coil,, is the transistor getting hot.??

BTW: the diode D2 is a back emf clamping diode and will not be conducting while the relay is energised.

 

First of all, NEVER remove the freewheeling diode! It does not introduce a volt drop because it is reverse biased as far as the power supply is concerned. It is a crucial component which circulates the back emf from the relay coil when the transistor turns off. It protects your circuit and is not there for the looks.

What you are experiencing, is a transistor that is not fully turned on. (Possibly because you are using a bigger relay which demands more collector current). Within limits, you may decrease the value of the base resistor to say, 18k or 22k. Look at the data sheet of the 4020B to see how much current it can sink before you reduce the base resistor to much and destroy the 4020B.

Furthermore, give the LED its own supply via a contact of the relay to further reduce the loading on the transistor.

 

Hello,

I did have a look at the 4020 datasheet and the 4020 can sink 1.3 mA at 10 Volts and 3.4 mA at 15 Volts.
To take the safe side, we can assume a base current of 1 mA.
The circuit works at 12 Volts and has a CE junction on top.
The lowest resistor value will be (12-0.7) Volt / 1 mA = 11.7 kOhm, the next higher value will be 12 kOhms.

Bertus

 

Your 12V relay coil has a lower resistance than the kit relay.
The 39K R4 is not driving the transistor Q1 into saturation, so you measure 3V across the transistor and approx 9V across the relay coil,, is the transistor getting hot.??

BTW: the diode D2 is a back emf clamping diode and will not be conducting while the relay is energised.

The transistor is pretty warm, hot but not enough to hurt....

First of all, NEVER remove the freewheeling diode! It does not introduce a volt drop because it is reverse biased as far as the power supply is concerned. It is a crucial component which circulates the back emf from the relay coil when the transistor turns off. It protects your circuit and is not there for the looks.

What you are experiencing, is a transistor that is not fully turned on. (Possibly because you are using a bigger relay which demands more collector current). Within limits, you may decrease the value of the base resistor to say, 18k or 22k. Look at the data sheet of the 4020B to see how much current it can sink before you reduce the base resistor to much and destroy the 4020B.

Furthermore, give the LED its own supply via a contact of the relay to further reduce the loading on the transistor.

I wasn't going to touch the snubber diode, I may remove the one in series with power supply... the transistor can take 800mA from datasheets, the coil should take up to 100mA, and the LED 8mA, I'll try decreasing the bias resistor on the transistor first.... thanks....

I did have a look at the 4020 datasheet and the 4020 can sink 1.3 mA at 10 Volts and 3.4 mA at 15 Volts.
To take the safe side, we can assume a base current of 1 mA.
The circuit works at 12 Volts and has a CE junction on top.
The lowest resistor value will be (12-0.7) Volt / 1 mA = 11.7 kOhm, the next higher value will be 12 kOhms.

Thanks - will try that in the morning....

 

You are moving over 100 mA through the transistor. Usually on the datasheet there is plot of typical gain vs. collector current. With your relay coil DC resistance, that tells you the bare minimum base current needed for the edge of saturation. From there I would look up the maximum continuous base current, find the half-way point between the two currents, and size the base resistor for that. It probably is within the range of the 4020.

ak

 

This graph from the BC328 datasheet seems to show various voltage drops/power dissipation for various loads/base currents (I think!)....
I'm giving the base approx 0.3mA at present it seems, the graph reveals the increased voltage drop and power dissipation, so if I ramp it up to 1mA the curve seems more favourable.... correct me if I'm wrong!

 

Well again thanks for the help gentlemen....
I replaced the 39K bias resistor with a 15K, now the relay gets 11.3V and the transistor runs cool.... happy days!