# How do I chain transistors so that the first transistor switches on the second

I am trying to get more hands-on experience with transistors. I have a simple switch working:

I wanted to find out how to change the circuit so that a second transistor would be switched on by the first one, driving another LED.

I tried connecting the emitter of Q1 to the base of Q2 with and without a current limiting resistor but with this schema, nothing works at all:

I have tried a variety of different transistors too, including S8050.

• You must have done something wrong. At least one of the LEDs should have lit up in that configuration, and probably both. (That's not to say that it's a good circuit, but it should have done something.) – Dave Tweed Nov 8 '19 at 23:46
• What about BJTs (as switches) interests you, right now? The basic rule is that if you supply about 5% to 10% of the current needed by the collector to the base, then the BJT will act pretty well as a switch with perhaps a few tenths of a volt drop between emitter and collector. Other than that much, what else are you trying to grasp at? – jonk Nov 9 '19 at 5:59
• @jonk I want to build up my knowledge to the point that I can design and build a RF remote that switches on a relay in my car – Aethalides Nov 9 '19 at 9:11
• @Aethalides Sounds good. I won't bother to write further, here, towards your goal as you already have an answer. But, I'll try and remember for next time. – jonk Nov 9 '19 at 19:03

If you want to set up an arbitrarily long chain of identical sections, the best way is probably to use two transistors per stage:

simulate this circuit – Schematic created using CircuitLab

Q2 drives the next stage in exactly the same way the the switch drives the first stage.

Dave's nice answer suggests that the PNP switch (Q2) could drive a LED as well as the NPN switch (Q1, Q3, etc).
Thus, there are two different cascaded sections, alternating. Each transistor acts as a current source for the single LED that it's collector drives. The LED acts as a voltage reference for the following stage's current source. Stage 0 requires a larger resistor, since it is not driven by a transistor current source.

If you run this on a simulator it can work, because each transistor and each LED matches its mates, and everything runs at the same temperature. If you choose a resistor value very large, each stage LED current is less - somewhere down the cascade LED current is too small to turn on the next stage.
A resistor value too small is the better choice - stage gain is limited by the dynamic resistance of the LED...LED current reaches a limiting value...in this case about 11.7mA.

simulate this circuit – Schematic created using CircuitLab
The LED used here is RED. With about 11mA flowing through it, roughly 2V appears from anode to cathode. The transistor base-emitter junction drops that by about 0.67V to 1.33V. So resistor R2 has 1.33/100 amps flowing...most of this reaches Q1's collector, and most of that flows through LED D2.

A BLUE or WHITE LED would likely require a higher supply voltage to keep the transistors from saturating...you need a DC supply at least twice the LED's turn-on voltage.
If the switch SW1 is grounded, Q1's base is pulled low, and Q1 supplies no current to LED D2. So there is no base current available to Q2...this current source is off...and so on down the chain.

It is interesting to speculate about a very long chain. Suppose LED D1 is left open-circuit, rather than grounded with the switch. This LED might now act as a photo-detector. Its photo-current flows into Q1's base, and is amplified by Q1 into D2. Much further down the chain, accumulated current gain may be enough to visibly light up a LED. Thermally-generated "leakage" currents could act similarly.

• I'm not sure I understand all of that. I thought the idea was that a transistor needs to be saturated for it to provide the maximum amount of current amplification it can manage and a current limiting resistor is needed to prevent burning out whatever circuit is on the collector . You're saying that in the case of the blue or white leds you want to prevent the transistors from saturating. Why is that please – Aethalides Nov 11 '19 at 10:57
• Saturated switches are most common. Not always: ECL logic use unsaturated switches. This scheme is also not saturated, but unsuitable for use as logic since turn-off time is much longer than turn-on time....this switch chain favours being "on" rather than turning "off". You seem to grasp switch function OK. This scheme is a bit complex because of the mix of NPN & PNP switches. CMOS logic also uses a mix of PMOS & NMOS switches too (a different arrangement with NO resistors). Try to get a grasp of how pull-up & pull-down transistors work in concert. – glen_geek Nov 11 '19 at 14:12