Driving the Songle SRD-05VDC-SL-C
Driving the Songle
This is an area that many people may encounter; I have modified my circuit and it doesn't work !.
This example is using a PZT2222A NPN transistor to drive a simple LED ( approx 25mA- Its 20 years old I have no idea without measuring).
https://www.aliexpress.com
But in a flash of inspiration I want to do more and control bigger loads.
Queue the Songle relay.
https://www.aliexpress.com
The great thing about a relay is that it can be controlled by a low voltage and switches a high voltage. It looks like 10A at 250Vac and 110Vac and 10A at 28-30Vdc. Be wary these characteristics can be very different between relays from different manufacturers. I suspect the 10A is not an inductive load.
The Circuit is like this
T1 works fine LED 3 comes on and off. But when I add in RelayPTH, It doesn't work ... Why?
We have to first understand the basics of the PZT transistor. The transistor is a current controlled device. The ratio of base current (IB) to collector current (IC) governs the control.
When I first calculated for the LED it was 25mA. LED3 has a series 180R resistor.
So IC here is 25mA. Next I look at the HFE of the transistor for a 500mA IC we are looking at a measly 40 for 25mA the min HFE is 75 but drops to 35 at -55C. So lets take 40 as the worst case.
Hfe = IC /IB IB = IC/HFE = 62.5uA.
We can assume the VBE is around 650mV ( 0.65V).
The input voltage is 5V - 0.65 = 4.35V
R = 4.35V / 62.5e-6
69.6K or 68K.
I had 3K which meant that the transistor was driven hard into saturation. So no problem with a 25mA load.
The songle relay has a coil resistance of 68R to 70R
That means IC will be (5V-0.65V) /70 = 62 mA.... A lot different to my LED!
Lets do the reverse HFE calculation again.
IB = (62e-3 /40) = 1.5mA.
Same again = 4.35V input R = 4.35/1.5mA = 2.9 K
Well I have a 3K in place, its pretty close to the best value. Well to be honest perhaps a 2.7K will work better. But also remember I have 2.5mA LED as well.
IB is now 1.6mA that really puts us down to 2.7K for sure.
If I assume a lower coil resistance of 68R, my IC is 64mA + 2.5mA
IB therefore = 1.67mA
4.35/1.67mA = 2.6K
But it still doesn't work????
Well I had forgotten that I have two transistors in series driving the base of T1 and acting as an AND gate.
That means 2 x0.65V needs to be taken from 4.35V. = 3.05V at the base resistor R4 of T1.
That's the equivalent of driving from a 3V system.
R4= 3.05V/1.67mA = 1.826K or 1.8K.
But I have a 1.6K in place and it still doesn't work.
Lets do a quick test. Unsolder the R4 from its connection to the emitter of T3 and connect to 5V.
Does it give a satisfying click of the relay?
Oh yes very much a real positive click.
So what could be the problem?
This is an area that many people may encounter; I have modified my circuit and it doesn't work !.
This example is using a PZT2222A NPN transistor to drive a simple LED ( approx 25mA- Its 20 years old I have no idea without measuring).
https://www.aliexpress.com
But in a flash of inspiration I want to do more and control bigger loads.
Queue the Songle relay.
https://www.aliexpress.com
The great thing about a relay is that it can be controlled by a low voltage and switches a high voltage. It looks like 10A at 250Vac and 110Vac and 10A at 28-30Vdc. Be wary these characteristics can be very different between relays from different manufacturers. I suspect the 10A is not an inductive load.
The Circuit is like this
T1 works fine LED 3 comes on and off. But when I add in RelayPTH, It doesn't work ... Why?
We have to first understand the basics of the PZT transistor. The transistor is a current controlled device. The ratio of base current (IB) to collector current (IC) governs the control.
When I first calculated for the LED it was 25mA. LED3 has a series 180R resistor.
So IC here is 25mA. Next I look at the HFE of the transistor for a 500mA IC we are looking at a measly 40 for 25mA the min HFE is 75 but drops to 35 at -55C. So lets take 40 as the worst case.
Hfe = IC /IB IB = IC/HFE = 62.5uA.
We can assume the VBE is around 650mV ( 0.65V).
The input voltage is 5V - 0.65 = 4.35V
R = 4.35V / 62.5e-6
69.6K or 68K.
I had 3K which meant that the transistor was driven hard into saturation. So no problem with a 25mA load.
The songle relay has a coil resistance of 68R to 70R
That means IC will be (5V-0.65V) /70 = 62 mA.... A lot different to my LED!
Lets do the reverse HFE calculation again.
IB = (62e-3 /40) = 1.5mA.
Same again = 4.35V input R = 4.35/1.5mA = 2.9 K
Well I have a 3K in place, its pretty close to the best value. Well to be honest perhaps a 2.7K will work better. But also remember I have 2.5mA LED as well.
IB is now 1.6mA that really puts us down to 2.7K for sure.
If I assume a lower coil resistance of 68R, my IC is 64mA + 2.5mA
IB therefore = 1.67mA
4.35/1.67mA = 2.6K
But it still doesn't work????
Well I had forgotten that I have two transistors in series driving the base of T1 and acting as an AND gate.
That means 2 x0.65V needs to be taken from 4.35V. = 3.05V at the base resistor R4 of T1.
That's the equivalent of driving from a 3V system.
R4= 3.05V/1.67mA = 1.826K or 1.8K.
But I have a 1.6K in place and it still doesn't work.
Lets do a quick test. Unsolder the R4 from its connection to the emitter of T3 and connect to 5V.
Does it give a satisfying click of the relay?
Oh yes very much a real positive click.
So what could be the problem?
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