For the circuit shown, I'm trying to figure out if I need a base resistor or not.

enter image description here

What I'm looking to accomplish is a simple logic inverter to hold a power amp's active low shutdown at logic low until the tube warms up to a stable temperature.

I'm planning to leverage an existing inrush current limiter to drive the shutdown imput. (The idea for the inrush limiter is to extend the life of the tube filament by letting it come to temperature slowly, based on what I read in this article: https://audioxpress.com/article/constant-current-for-heater-tubes-extend-the-life-of-your-amplifier-s-vacuum-tubes)

The tube filament and inrush limiter form a voltage divider. When the circuit is at room temperature, the ratio should be about 20:200, making the voltage at point A logic high. When the circuit comes up to operating temperature, the datasheet for the inrush limiter says its effect will be negiligible (I'm assuming near zero reistance) so the vast majority of the voltage wil be dropped across the tube heater. This makes point A logic low.

This arrangement is exactly opposite of the active low requirement of the power amp's active low shutdown input.

Researching transistor inverters, I came across several that look like what I have in the schematic, with Rb, Q1, and Rc.

My question is this... Do I need the base resistor?

I'm thinking no, because I'm technically using voltage divider bias, with the tube filament and the inrush limiter forming the resistor network. But, the current flowing through these elements is about 150 mA. Normally, the voltage divider resistors in this type of biasing are calculated to pass a lot less current. Is there any possible harm to the transistor with these small resistance values and this large current?

If I think in terms of current amplification, and the transistor's beta of about 200, the ratio of Ib to Ic should be 1:200. Ic is limited to about 1mA by Rc. Does this arrangement dictate the base current can never be more than 1/200th of 1mA? It would seem so to satisfy the 1:200 current amplification. Or, is it a one way function where Ib controls Ic, but limiting Ic does nothing to limit Rb? In other words, Rb is still needed to limit the base current to a reasonable value? This does not seem to fit with the idea of voltage divider biasing, but I have never tried it with such a low resistance voltage divider.

I would prefer to not blow up a transistor to find the answer. Can anyone tell me if the Ib:Ic ratio holds true both ways, or do I need Rb to limit base current?

  • 1
    \$\begingroup\$ Yep, you definitely need Rb. No time to write a proper answer though, sorry. \$\endgroup\$
    – TonyM
    May 19, 2021 at 19:15
  • 3
    \$\begingroup\$ You're missing the fact that the BE junction diode is in parallel with, and therefore shorts out the bottom resistor of your divider making it no longer a divider. \$\endgroup\$
    – DKNguyen
    May 19, 2021 at 19:26
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    \$\begingroup\$ The base-emitter junction in a transistor is a forward-biased PN junction which has certain characteristics in common with other PN junctions such as those in diodes, namely the ability to pass a lot of current with only a small change in applied voltage. The current gain only "limits" collector current with a specified base current, not the other way around. \$\endgroup\$
    – vir
    May 19, 2021 at 19:32
  • \$\begingroup\$ Okay, I think I'm seeing it. So when the filament is cold, there's a 20:200 ratio in the voltage divider and about 11.5V at point A. I drop about 0.7 over the BE junction and the rest goes across Rb. If I size Rb at 2.4M I get about 5uA. With the beta of 200, I should get about 1mA through Rc. \$\endgroup\$
    – Dave H.
    May 19, 2021 at 19:34
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    \$\begingroup\$ @jonk My understanding of the inrush current limiter is that it heats up on its own due to the current passing through it and it does not need to be in contact with the device its limiting current for. I think the typical application is to limit inrush current for an empty capacitor charging up. Here's a sample data sheet: cantherm.com/wp-content/uploads/2018/08/MF72_AUG_2018.pdf Concerning the constant current filament, I've read conflicting reports saying it's not a good idea. And, the inrush limiter seems like it will solve two problems for me with one part. \$\endgroup\$
    – Dave H.
    May 19, 2021 at 20:42

3 Answers 3


I suggest you use any Nch FET to act as a constant current source as long as RdsOn <=1 ohm or so and not use an NTC thermistor. Inrush current limiters (NTC’s) are not intend to be run continuously as they transition at high temperatures which reduces lifespan.

I can use a simple 5mm RED AlGaAs LED as a Zener of 1.9V @ ~5mA to bias the NPN current limit to 150 mA with only a 0.5 Ohm filament current sense of 75mV (standard current sense voltage) to drive the base with about 630mV.

The pullup from LED 2.9V to base Vb=630 mV was simulated with a dim light bulb as load that is 80 Ohms hot.

With slight variations in every component, the LED to base R should be tuned between 2.1k and 2.8k if you prefer full power or reduced a tad, to extend the tube life even more x2 for every 10’C drop at some tradeoff with thermionic gain.

I invite all pro designers to peer review this design including @jonk.

enter image description here

One can easily soft start this with an e-cap on the gate of the FET. The constant current power risetime is 2s/div.

This can reduce 12.6Vdc ac ripple slightly by choosing to dissipate some power in the FET.

  • \$\begingroup\$ The data sheet for the inrush current limiter specifies "Filament Protection of various types of lamps" as one of its applications. [cantherm.com/wp-content/uploads/2018/08/MF72_AUG_2018.pdf] I should also mention I'm using DC 12.6V for the filament. The power amp is running off a 24V switch mode power supply. I'm planning on an LM317 to supply the heater voltage. I toyed with the idea of using the LM317 in constant current mode, but I've read conflicting information on constant current being a good approach for filaments, regardless of the original article. \$\endgroup\$
    – Dave H.
    May 19, 2021 at 21:30
  • \$\begingroup\$ This design would use a cap and pot for bias fromLED with a 1s risetime tinyurl.com/yhhraufo. With an OA I can cut out the delay to Vt. Lamps have an 11:1 R range at 3500’K but this filament is only 2:1 so the stress force is only 2:1. With PTC effect clamps on the current ratio. The force is a constant step on a rugged coil. Look at the filament power risetime. \$\endgroup\$ May 19, 2021 at 23:17
  • \$\begingroup\$ Looks like a current mirror. Is that what it's doing? \$\endgroup\$
    – Dave H.
    May 19, 2021 at 23:46
  • \$\begingroup\$ No. Mirrors share common gates or bases, this is a negative feedback loop sensing current yet with a very low dropout unlike typical designs. I just thought of it. Also because of low LED and base current any device works. But FET needs power for 1s \$\endgroup\$ May 19, 2021 at 23:55

Thanks to all those who have commented, I see the Rb is definitely required. When the circuit is cold, the filament to inrush limiter voltage divider ratio is 20:200. With a 12.6V supply, the voltage at point A is about 11.5V. 0.7V is dropped across the base-emitter junction. This leaves 10.8V across Rb.

If I'm trying to get 1mA of collector current with a transistor with beta = 200, I need 5uA of base current. Ohms law says I need a 2.2M Rb. But 1mA is a pretty conservative collector current, and not every 2N3904 in my kit has a beta of 200. Assuming I pick one with a low beta of 100, I now need twice as much base current. A 1M resistor is now a better choice. If beta happens to be 200, I get 2mA of current. A low beta of 100 gets me 1mA of current. Both get me about a 22V drop over Rc. But, I should go a little higher still, because the voltage left across the collector-emitter junction in this arrangement is 2V and that barely qualifies as a logic HIGH.

Ultimately, I could use a 500k Rb for about 0.02mA of base current. Even with a low beta of 100, this results in 2mA of collector current and a saturated transistor. For a beta on the high side (say 300) the current is 6mA, still well within the maximum current and power for the device.

EDIT: In my original attempt at a solution, I made the mistake of treating the circuit as an amplifier. Based on comments, I now know I should look at it as a switch and use more base current than I had originally calculated.

An article from Nuts & Volts does a great job of explaining transistors as switches. https://www.nutsvolts.com/?/magazine/article/may2015_Secura

I need to ensure my transistor in the ON state is operating in the saturation region. The 2N3904 data sheet I have lists two values for Vce in saturation. (Listed in the section entitled "On Characteristics".) One of these is Vce = 0.2V at Ic = 10mA and Ib = 1mA. As @Jonk pointed out, this is what I should be designing for.

With my circuit that drops 10.8V across Rb, a 10k resistor will supply the 1mA base current I need.

According to the Nuts & Volts article, Rc is chosen as (Vcc - Vce) / Ic(max). My 24V supply gives Vcc. The data sheet gives the 0.2V Vce and 10mA Ic(max). 10mA is more than I need just to drive the power amp's shutdown input, but this is the lowest value on the data sheet and I hesitate to deviate too much.

I hate to waste 10mA for nothing, so I recalculated with an LED thrown in to indicate a "preheat" state. Now my circuit looks like this:

Updated Schematic

Designing for the 10mA Ic, I can calculate the Rc value by first subtracting the Vce of 0.2V (from the data sheet), and another 2V for the LED, from the 24V supply. This leave 21.8V dropped over Rc. For a 10mA current, I need about a 2.2k Rc.

With this circuit, when the filament and the inrush limiter warm up, the filament resistance increases while the inrush limiter resistance drops. The voltage at point A eventually drops below Vbe and the transistor shuts off.

As Ic drops, the voltage at SD rises, eventually reaching a logic HIGH at the power amp's shutdown input, and the power amp switches on.

If all goes well, the vacuum tube is warmed up gently, saving wear on the filament. The power amp stage will remain shut off until the tube is ready, saving the listener from any strange noises coming from the speaker while it warms up.

  • 1
    \$\begingroup\$ Well, no. You should consider the BJT as operating as a switch. And you do NOT use the active mode beta, or anything close to it, as a yardstick. You will want to flood the base with no less than 1/20th of the collector current and, by habit, many engineers use instead the value of 1/10th. There's no specific hard number here. For example, if this were a 2N3055 then you might prefer to use 1/2 or 1/3rd, instead. Datasheets will clue you in about this when you read them (though they do NOT paint huge arrows pointing where to read it.) For small BJTs like yours, you could get away with 1/20th. \$\endgroup\$
    – jonk
    May 19, 2021 at 20:25
  • \$\begingroup\$ A Nuts & Volts article from 2015 has enlightened me greatly on designing a transistor circuit to serve a switch. nutsvolts.com/magazine/article/may2015_Secura \$\endgroup\$
    – Dave H.
    May 19, 2021 at 20:51
  • \$\begingroup\$ Well, you may be on your way then. If you update your answer with what you learned and make it fulfill your own question well (no one posts a serious problem), then you can select the answer as your own and that's okay at this site. \$\endgroup\$
    – jonk
    May 19, 2021 at 21:01
  • \$\begingroup\$ Feel free to critique my design :) \$\endgroup\$ May 19, 2021 at 21:14

Okay, I think I'm seeing it. So when the filament is cold, there's a 20:200 ratio in the voltage divider and about 11.5V at point A. I drop about 0.7 over the BE junction and the rest goes across Rb. If I size Rb at 2.4M I get about 5uA. With the beta of 200, I should get about 1mA through Rc.


That is called dangle-biasing, and is extremely unreliable. The beta of the transistor changes with temperature, atmospheric pressure, age, base current, and collector current.

You want the transistor to act as a saturated switch, not a linear amplifier. For this circuit, the collector current is just over 1 mA if Q1 is truly saturated. The rule of thumb is that for firm saturation, the base current should be no less than 10% of this, or approx. 100 uA. My vote, use a 22K base resistor. This creates a collector-to-base current ratio of approx. 2:1, which will firmly saturate the transistor without over-currenting the base-emitter junction.

Separate from that, you might want to put 1 or 2 small signal diodes (1N914, 1N4148, etc.) in series with the base. This is to assure that the transition voltage of the transistor inverter stage is approximately in the middle of the NTC thermistor voltage range from cold to hot. If you do this, add a 100 K resistor from the base of Q1 to GND to assure complete turn-off.


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