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I'm designing a circuit with an NPN (2N4401) transistor connected to a PNP (2N4403) transistor to amplify the current inputed to the NPN base terminal. Also, the general design I am working with is depicted here Trigger

What I am wondering is what resistance values I would need in order to get roughly 200 mA at the PNP transistor's collector terminal. I am using 5 V at the "Trigger" pin, so 1 kOhm at the base would provide ~4.4 mA. Amplifying this would bring me to ~264 mA (with Hfe = 60). Now going through the PNP would amplify this current again to a ridiculously high value.

What is the correct way to find these resistance values?

Thanks!

Also original question posted with this ckt: enter image description here

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  • \$\begingroup\$ What is VIN? And why do you think 200 mA is adequate? \$\endgroup\$ – WhatRoughBeast Aug 9 '16 at 19:09
  • \$\begingroup\$ Vin is roughly 11-12 V, and 200 mA is what my company has found to be adequate to fire the SCR \$\endgroup\$ – Seth Shill Aug 9 '16 at 21:02
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I am using 5 V at the "Trigger" pin, so 1 kOhm at the base would provide ~4.4 mA. Amplifying this would bring me to ~264 mA (with Hfe = 60). Now going through the PNP would amplify this current again to a ridiculously high value.

Your current gain formula only works below transistor saturation. Once it's saturated the current that flows depends on the limit set by the external components. Given that R36 is 1k the max current you can get through Q2 is \$ \frac {V_{IN}}{1k} \$. e.g., For a 12 V supply you can't get more than 12 mA through Q2.

Similarly for Q3, the most collector current you can have is \$ \frac {V_{IN}}{R_{23}} \$.

What is the correct way to find these resistance values?

Start at the end and work back.

  • Let's assume that C25 and C11 are charged to Vin and that the SCR trigger voltage is low relative to Vin.
  • With Q3 hard on we can calculate R23 using \$ R_{23} = \frac {V_{IN}}{I_{TRIG}} = \frac {V_{IN}}{0.2} \$.
  • Minimum base current to turn Q3 hard on is \$ \frac {I_C}{h_{fe}} = \frac {0.2}{60} = 3.3~mA\$. You should put a good safety factor in here to cover variation in transistors, tolerance of resistors, etc. As discussed above the maximum current is given by by R36 and, since that's a reasonable value (12 mA in our example calculation) let's go with that.
  • We now need to drive Q2 into saturation so R27 could be calculated the same way. Most of us don't bother and run 1 to 5 mA through the base to be sure to turn it hard on. (If it were a battery circuit you might think differently.)

Caution

You are charging C25 and C11 through R28, 10k. The charge time constant \$ \tau = RC = 10k \times 20u = 200~ms \$. If you give frequent or long pulses on the SCR then voltage on the capacitors will fall and your circuit may fail to operate until you give it a rest.


See also my answer to How to calculate currents and voltages with transistors circuit where the OP was having the similar confusion.

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  • \$\begingroup\$ Thank you for the Caution. The pulses on the SCR are actually spaced > 1 second, so there is plenty of time to recharge. I used you suggestions in LTspice simulation and got what I expected! \$\endgroup\$ – Seth Shill Aug 10 '16 at 12:35
  • \$\begingroup\$ Good. Wait another day or two to encourage other answers and don't forget to accept one if it answers your question. You can upvote others that you think are good too. \$\endgroup\$ – Transistor Aug 10 '16 at 12:45
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This is not a very good circuit (though it may look like one).

Presuming your Vin is something like 12V, your current will be limited by the gain of the transistor, so highly variable with unit-to-unit and temperature variations, and likely to exceed the maximum current or SOA (Safe Operating Area) of Q3. Ideally you should either operate the transistors in saturation or at some Vce/Ic that is controlled (to a first order) by the circuit design not by transistor parameters. That can be accomplished either by increasing R23 or by adding a transistor and resistor to rob base current from Q2 (to make it into a constant current source of sorts).

Then you can operate Q2 well into saturation (with Ic/Ib between 10 and 20) and either saturate Q3 (with a higher value R23) or operate it with a reasonably controlled collector current of (say) a few hundred mA. The maximum trigger current of the thyristor is 280mA at low temperature. The type of current source I suggest increases at low temperature so that is well.

Something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Rs would be around Vbe/0.3A so around 2\$\Omega\$, which sets the gate current. R2 should allow about 15-30mA base current at the end of the gate pulse (when the capacitor voltage has dropped). And R3 should result in a couple mA of base current for Q2 when the input is high.

For relatively high current thyristors, it's better if they get a clearly adequate trigger pulse, both in magnitude and in duration. 'Tickling the dragon's tail' with inadequate pulses can lead to current crowding on the thyristor die and an unfortunate early demise.

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  • \$\begingroup\$ Wouldn't this cause Q3 to "overcurrent" the trigger current, that is Ic3 will be too large (roughly 440 mA from what I got from simulation)? Unless that's what you're suggesting. \$\endgroup\$ – Seth Shill Aug 10 '16 at 19:34
  • \$\begingroup\$ 440 sounds rather high. I just did a quick simulation 12V/1K/2 ohms with 2N4403s and got 340mA which is pretty close to the 300mA I wanted (for a back of the envelope calculation). \$\endgroup\$ – Spehro Pefhany Aug 10 '16 at 20:37
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The trick to figuring out the circuit is to realize that you don't want to use the transistors as "current amplifiers" in the usual sense. What you want to do is use them as switches.

Start with the PNP. With a 12 volt supply and a fully charged pair of caps, let's say we can turn on Q3 so that the voltage across is very low - a few tenths of a volt. This is called being in saturation, and the rule of thumb is that saturation will reliably occur with current gains of 10 to 20, as Spehro has answered. Let's be conservative and assume a gain of 10.

Figuring about 2 volts drop across the SCR gate and Q3, that leaves 10 volts across R23, which suggests that 50 ohms would be about right to produce 200 mA through the gate. This in turn establishes 20 mA as a base current for Q3. (Gain of 10, remember?)

The same desire to operate in saturation applies to Q2, and hence the voltage drop across R36 will be on the order of 11 volts. That is, 12 volts minus 0.7 volts Vbe(Q3) and minus 0.2 volts Vce(Q2). 11 volts is close enough. So R36 should be about 11 volts/.02 A, or about 560 ohms.

Finally, since Q2 is operating with a gain of 10 and a collector current of 20 mA, base current should be 2 mA. R27 should be about 2k.

If all of this seems rather imprecise, it is. There is no need for high precision given the relatively large range of current gains associated with saturating a transistor. Using a conservative estimate of a gain of 10 allows for very rough calculations to do the job. And since there are no high voltages or currents involved, the possibility that rough numbers will produce non-optimal component values simply doesn't carry much of a penalty.

You'll also notice that this approach is better than trying to tailor current levels via transistor gain rather than resistor values. There is a very good reason for this. If you attempt to operate the transistors linearly, in principle you could do away with R36 and R23, which might seem like a good deal. It's not. Individual transistors will have quite a broad range of gains, so you would have to select your transistors to get proper operation. And then you'd have to deal with the fact that transistor gains will typically vary with temperature, so all your calculation and selection would not guarantee proper operation in varying environments.

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  • \$\begingroup\$ Great comments! This complements the other two responses greatly, especially where my understanding of the deeper theory of application lacks. Thanks so much! \$\endgroup\$ – Seth Shill Aug 10 '16 at 12:57

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