Transistor Question in RS Flip Flop Circuit - Page 1
Forum section: electronics / beginners

author: eev_carl
author_rank: Regular Contributor
timestamp: January 12, 2018, 02:01:19 pm
content: Hi,I'm implementing this circuit, but not getting the expected results from LED 2. I have the following truth table. LED 1 is behaving correctly.RSLED1LED2GND+VONOFFGND-VOFFON+VGNDOFFON-VGNDONOFFThere is a voltage drop on LED 2 when it's supposed to be off. However, it doesn't appear to be enough, only dropping from 2.57V to 2.47 at the base of Q2.Is there a way for me to adjust the Q2 drop so that will fall below the voltage needed to light LED 2?I'm not using the optional Zener diode limiting path.Thanks in advance,Carl

author: Paul Rose
author_rank: Regular Contributor
timestamp: January 12, 2018, 02:35:54 pm
content: I normally like Forrest Mims stuff, but this is not the typical way to demonstrate a discrete R-S flip-flop.I'd go with something like the upper right schematic on the Wikipedia pagehttps://en.wikipedia.org/wiki/Flip-flop_(electronics)You can put LED is series with R1 and R2. Maybe 470 ohm for R1 and R2 and 5k to 10k ohm for R3 and R4Any NPN for Q1 and Q2 ( 3904 or 2222 )

author: eev_carl
author_rank: Regular Contributor
timestamp: January 12, 2018, 03:24:06 pm
content: Thanks,I have a NAND IC here too if I really want a flip flop. I think my question is more on the transistor configuration than the flip flop. How is Q2 supposed to work? Is Q1's base voltage supposed to drop so much that it shuts off Q2?I tried the 470 for R1 and R2 and increased R4 to 10k. It made LED 1 dimmer, but LED 2 didn't seem to be affected by the new values.-Carl

author: rstofer
author_rank: Super Contributor
timestamp: January 12, 2018, 04:59:39 pm
content: Think about the voltages when LED1 is on. That puts the emitter of Q1 at about 2V and the collector at about 2.2V. As a result, the base of Q2 is about 2.2V and the emitter is 1.5V (more or less). This may be enough to light up LED 2.You need to remove the uncertainty around Q1 (and Q2) so put the LEDs above the transistors (in series with the 1k resistors) and move the transistors down such that the emitters are grounded. Even so, the base of Q2 will never get below about 0.2V. That should keep Q2 shut off.The zener diodes are there to protect Q1 from a large reverse base voltage (assuming +- 15V supplies) as well as excessive forward base voltage. I would not omit them.

author: Paul Rose
author_rank: Regular Contributor
timestamp: January 12, 2018, 06:24:04 pm
content: QuoteThis is good advice. Having the load in the collector circuit is normal practice, and it simplifies biasing.Different color and types of LED's have different voltage drops. If you put the LED in the collector circuit, you won't have to worry about the effect of the voltage drop of the LED might have on biasing. You'd just adjust the resistors to control brightness.If you are running lower supply voltage, youmightnot need the zener. Check transistor data sheet. The max reverse-biased emitter-base junction is something like 6V. Not good to exceed this. Your posted schematic says something like "see note below", but you can't see the note itself in the picture. I expect it says something like this.

author: eev_carl
author_rank: Regular Contributor
timestamp: January 12, 2018, 07:07:38 pm
content: I have some LM431s on hand. Would I just wire REF and Anode together for a Zener diode?

author: rstofer
author_rank: Super Contributor
timestamp: January 12, 2018, 07:28:46 pm
content: Quote from: eev_carl on January 12, 2018, 07:07:38 pmThe examples I see in the datasheet show the Ref connected to the cathode (Figure 3):http://www.ti.com/lit/ds/symlink/lm431.pdfIn zener mode, the current seems to be about 10 mA. I suspect R2 at 1k is just at that point (worst case, assuming 15V supplies). I MIGHT change to a 1.2k or 1.5k just to guarantee I didn't get above 10 mA.It is important to measure the voltages around the circuit after it is working (or before). The learning part is being able to guesstimate voltages and rationalize what is happening. You start by guessing that, in saturation, the voltage between the base and emitter is at least 0.6V and that the voltage drop between collector and emitter is about 0.2V. Now prove it!Vbe can be higher than 0.6V but that's about the minimum required to get the transistor into saturation.BTW, these values come from the datasheet:http://www.st.com/content/ccc/resource/technical/document/datasheet/a0/75/d9/bd/fc/ca/42/15/CD00003223.pdf/files/CD00003223.pdf/jcr:content/translations/en.CD00003223.pdfCheck out Vce(sat) and Vbe(sat) on page 2.

author: Paul Rose
author_rank: Regular Contributor
timestamp: January 12, 2018, 09:48:56 pm
content: It may seem a bit silly, but consider that a blue LED has a forward voltage drop of about 3.3v, and a red is about 1.8v.A blue and red in series would be near 5v.You could use a blue/red series string as a clamp. Use two of these strings in "reverse parallel" in place of the series pair zeners and you would have an approx. 5v bipolarity clamp.Just to be clear, you would be using this for the forward voltage drop, and not the reverse "zener action".In the circuit you showed, there should not be enough current to overdrive the LEDs, due to the 1K from the 741 output.If you don't have blue, three reds would be about 5.4vI probably wouldn't want this in production, but in a pinch for a prototype, it is OK.

author: eev_carl
author_rank: Regular Contributor
timestamp: January 13, 2018, 04:37:48 pm
content: I think you're on to something with the LEDs. I don't have any bicolor LEDs here. Is there a way for me to convert a single color LED to behave like a bicolor LED with respect to Vf?It seems like the op amp is working. When I disconnect the transistor / LED subcircuit, my multimeter follows the truth table. Also, floating S retains the setting.

author: Paul Rose
author_rank: Regular Contributor
timestamp: January 13, 2018, 05:52:54 pm
content: QuoteJust put 2 regular LED's in parallel, but in opposite directions.Hopefully my ASCII art schematic below is readableCode:This instead of this| || |+-+-+ || | |- v Red v D1 zenerRed ^ - -| | |- v -Blue ^ - Blue ^ D2 zener| | |+-+-+ || || |Ground

author: eev_carl
author_rank: Regular Contributor
timestamp: January 13, 2018, 08:53:25 pm
content: Great suggestion. I abandoned the transistors for now. I think Q2 was supposed to be an inverter, but the collector voltage from Q1 never seemed to get low enough to turn off Q2's base.I'll order some bicolor LEDs sometime and try again.For now, here's the revised schematic. The op amp produces a +3.7V when the RS logic is ON and a -3.7V when the RS logic is OFF. Pulling out the lead to S will also retain the saved value.EDIT: I replaced a version of the schematic that was missing a 10k from power to - on the op amp. Also, the S and R labels were reversed./

author: Zero999
author_rank: Super Contributor
timestamp: January 13, 2018, 11:47:53 pm
content: Now the LEDs are clamping the output of the op-amp and the current limiting is the only thing which is stopping them from being damaged. Move the 47k resistor to the left hand side of the 1k.

author: eev_carl
author_rank: Regular Contributor
timestamp: January 14, 2018, 12:24:37 am
content: I posted an updated schematic. There was a missing 10k resistor and some labels were mixed up.Can you describe the two nodes you'd like for the 47k resistor? It already looks left of the 1k to me, meaning connected on the op amp side to its output.

author: Zero999
author_rank: Super Contributor
timestamp: January 14, 2018, 08:33:29 am
content: Quote from: eev_carl on January 14, 2018, 12:24:37 amThe 47k & 1k resisor is connected, as I described, in the schematic I posted previously.