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I am working with a circuit that grounds when a LED is to turn on. I have an 8V supply voltage for the LED, which cannot exceed 10mA. The signal to turn on the light is a very weak 6.5V, so Vbe is fine.

The real issue seems to be that the extra PNP voltage drop seems to kill the voltage for the LED.

What techniques could I use to turn on the LED without burning PNP?

Other components are acceptable.

Ideas? Is this a good place for a FET?

High PNP Vbe

More info

The reason I don't want to connect direct to the switch is because its connected to the push-to-talk circuit on the radio and need to be careful with that line. I need to "read" it, but shouldn't draw much current from it.

ICOM 9700 MIC connector

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    \$\begingroup\$ I have the feeling the circuit you're working with might be more than a switch. What are the actual devices? \$\endgroup\$ Commented Jul 18, 2024 at 3:19
  • \$\begingroup\$ Why do you need the transistor at all? Just connect R1 and D2 in series with the switch. \$\endgroup\$ Commented Jul 18, 2024 at 3:28
  • \$\begingroup\$ @vir is right. Reload :) \$\endgroup\$ Commented Jul 18, 2024 at 3:40
  • \$\begingroup\$ @DaveTweed, updated post \$\endgroup\$ Commented Jul 18, 2024 at 3:40
  • \$\begingroup\$ Since you updated the question, my answer might not be what you need. What are voltages of PTT (pin 5) with respect to battery ground (not PTT ground) with and without it "pushed"? What is the voltage of PTT ground (6) with respect to battery ground? \$\endgroup\$ Commented Jul 18, 2024 at 4:03

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As pointed out in the comments (Thanks, Tim), the original circuit will not work; the transistor isn't properly biased. So to save face, here are three additional circuits that will work:

schematic

I'm not sure if it was a requirement or not that the LED was ground-referenced.

Note: the last circuit is clearly the simplest, but may be tricky to make work reliably in practice - it will be dependent on the gate threshold voltage of the MOSFET and the accuracy and variation of the "8V Line" and the "6.5V ATT" signal. In other words, It might not be fully "off" when the signal is high and the LED may be sensitive to detect this.

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  • \$\begingroup\$ Is the original request turn LED on when PTT line is pulled high (not active) or when PTT is grounded? This is good circuit for LED turns ON when PTT not active, and my answer is good when LED turns on when PTT is active (at ground). And by the way, the N-FET needs to be Enhancement Mode (not Depletion mode), otherwise PTT has to be -3V to turn off LED, correct? \$\endgroup\$ Commented Jul 18, 2024 at 5:45
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    \$\begingroup\$ I'm afraid this won't work: the D-NMOS never turns off. Pin5 would have to go negative for that to happen. It mostly works with PMOS (as a source follower), but the LED may still turn on while line is high. \$\endgroup\$ Commented Jul 18, 2024 at 5:59
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    \$\begingroup\$ @MOSFET, do you agree with assessment that suggested depletion mode N-FET as low-side switch would be ON when PTT line is at ground (Vgs=0), and still be ON when PTT line is at +12V (Vgs=+12). \$\endgroup\$ Commented Jul 18, 2024 at 6:54
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    \$\begingroup\$ @KJ7LNW, In my opinion the first circuit of this revised answer is the best, which has the N-FET inverter, followed by N-FET low-side switch. That one works well even with the revised PTT high voltage of only 6.5V. The third circuit, which is simple with just high side P-FET switch controlled directly by PTT signal was the MOSFET equivalent of my original PNP answer, but with the revised question stating that PTT might only pull high to 6.5V instead of 13.8V, then when PTT is not active (high), the M2 P-FET Vgs is 6.5V-8V =-1.5V and could be very close to turning on accidentally. \$\endgroup\$ Commented Jul 18, 2024 at 7:51
  • \$\begingroup\$ @BK303 Yes, I agree. My original reasoning was based on a 13.8V spec that was changed in the original question. 6.5V would have been the ground reference for the circuit, 13.8V would have been the LED power supply. Therefore, you can get Vgs = -6.5V with respect to system ground. But that's all irrelevant now. \$\endgroup\$ Commented Jul 18, 2024 at 14:49
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My understanding is can't modify the control signal PTT circuit, and LED should be ON when PTT control active (low), OFF when PTT not active (weak pulled to was 12V, now 6.5V). A good option is diode isolate the PTT control (with new D3) so that only the ground connection state is relevant (can then ignore the details of V2 PTT pull-up source). Add a weak pull-up (new R4) from 8V power supply to PNP Q2 base to make sure transistor is off when PTT not active. Move the R1 load current limit resistor to the low side of Q2 (still in series with D2 LED), so Q2 emitter remains at fixed voltage (easier to control Q2_be voltage). Add new R5 between Q2 base and control signal to limit the base current, probably about 4.7K Ohms based on (8V - 0.7 - 0.7)/1.5 mA. R4 weak pull up could be 470K. Original PNP high-side switch with diode between base and PTT

EDIT -- ADDITION Upon revisiting this, the PTT switch is already providing a good path to ground that can easily handle the 5 to 10mA for powering the LED, so that can be used directly as an open-drain control to turn on LED from low-side. The fact that +8V pin 2 supply is higher than the PTT open circuit voltage can easily be accomodated by the diode drop through the LED plus a second silicon diode in series. In fact, likely the PTT is internally pulled up with diode in series to same +8V supply, rather than a separate 6.5V supply. See circuit below. new circuit LED with low side through diode directly to PTT

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  • \$\begingroup\$ Regarding OP question "is this good place for a FET?". I'd say yes, because FET is voltage controlled, not current controlled so will have less current disturbance to your PTT control signal. You can use either the PNP above or a P-FET as a high-side switch to turn on the LED. If replaced PNP (Q2) by enhancement mode P-type MOSFET (aka "P-FET"), then you build very similar high-side switch as my modified circuit above. The P-FET source goes to V1 (can omit R4), the gate can go directly to PTT (can omit R5 andD3), and the drain controls the current into D2 + R1 on low side of FET. \$\endgroup\$ Commented Jul 18, 2024 at 5:32
  • \$\begingroup\$ If using P-FET instead, just make sure the FET can tolerate the reverse bias of Vgs when PTT not active of +(13.8V - 8V) = +5.8V. If that Vgs not within data sheet spec, you can keep R4, R5, and D3 in the P-FET circuit too, and R5 value can be small value like 100 Ohms (or maybe can be omitted completely). \$\endgroup\$ Commented Jul 18, 2024 at 5:34
  • \$\begingroup\$ OP keeps changing the question, now the PTT not active bias (V3) has been reduced to 6.5V, so the answer above needs to be re-evaluated because V2 is no longer greater than V1. I may delete all of this and start over. \$\endgroup\$ Commented Jul 18, 2024 at 7:01
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Firstly, I doubt your circuit would produce \$V_{BE}=-5.8V\$. I suspect your transistor is damaged, isn't what you think it is, or is wrongly connected.

With such a small resistance at the base, most of the collector current will flow out of the base, instead of through the LED. You'll also need to be more explicit about how you set base potential. Both problems are fixed here:

schematic

simulate this circuit – Schematic created using CircuitLab

R1 and R2 set base potential at the mid-point between 0V and the battery voltage, for about +7V. Those resistances have been chosen for a base current which is very small compared to collector current. This places the emitter slightly above that at about 8V. Emitter and LED current are then:

$$ I_{LED} \approx \frac{V_E}{R_3} = \frac{13.8V-8V}{800\Omega} = 7mA $$

LED current will vary slightly with battery voltage, since emitter potential will vary with the base. To mitigate this, the base can be held stable with a zener diode:

schematic

simulate this circuit

Unlike the resistor divider, whose voltage varies with the battery, the 5.1V zener diode here keeps the voltage across R3 stable at 4.4V or so, over a wide range of battery voltages. This keeps LED current constant:

$$ I_{LED} \approx \frac{5.1V - V_{BE}}{R_3} \approx \frac{4.4V}{560\Omega} $$

That seems a lot of work when the following would do exactly the same thing:

schematic

simulate this circuit

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  • \$\begingroup\$ Current for the LED needs to come from the 8V line, and can't exceed 10mA. The 13.8V battery isn't really a battery...its a signal line from the radio that weakly floats high to 13.8V because of the magic of "the black box", and pulling it down will trigger transmission. I think the way it is drawn would lock the radio in a transmitting state. \$\endgroup\$ Commented Jul 18, 2024 at 4:03
  • \$\begingroup\$ @KJ7LNW Well, the original question didn't tell me any of that. Please include this information in the question. \$\endgroup\$ Commented Jul 18, 2024 at 4:06
  • \$\begingroup\$ I agree, sorry about that. \$\endgroup\$ Commented Jul 18, 2024 at 4:27
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Another variation to add to the pile: we can use a MOSFET as a crude differential amplifier, i.e. it remains off until Vg is below Vs, and then turns on incrementally from there.

By setting the source Thevenin resistance, we also limit LED current at the same time:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

EMC Considerations

Mind that a real implementation will need input ESD protection (can be a 9-15V zener from GND to M1-G), and a passive pull-up wouldn't hurt (1M is fine, from M1-G to +8V). Another zener from GND to +8V might not hurt either. Filtering shouldn't be necessary, but in the presence of strong RF fields, consider building it on ground plane, with a 10nF bypass on +8V and V1, and a 100k series resistor from V1 to M1-G.

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